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This chapter focuses on the fundamental concepts of linear equations in two variables, their graphical representation, and methods to solve them algebraically.
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Unit 3

Pair of Linear Equations in Two Variables 

 Exercise 3.1 

Q1 :

Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Answer :

Let the present age of Aftab be x. And, present age of his daughter = y Seven years ago,

Age of Aftab

Age of his daughter

According to the question,

Three years hence, Age of Aftab = x + 3

Age of his daughter = y + 3

According to the question,6BYXW8RMbZO4DaxSXerGJ5Ba aMHyuGDQuqMObEjB UpywmazQJCn2QYLXx 3o3v62k1ABd7m

Therefore, the algebraic representation is

fHFI3UBVJljYvJoocIHWKPqpW mFAk9iALUYV2Lj2Q2plGVHNntWg21KWGP0eQEACbo6PNCjOsvcYY

For     (3)SGIrsbSeZxj4bcaOlMc0cHqQotAmHUK13HJjKnp4dCharpKuaVZeM7AzeN9IU70DXq vk oYpovjcZ3Gr9WKZ7wGv sEIW73UWTj YswjCO49mwaSWH3tE

The solution table is

x– 707
y567

For (4)

CUjR2BcijrLAbLLWJ N3k8wIKlbOMUVFoXaysce DxV30JXAUlYNpYYKDkHTilW4FxjeTjbj5vr8Dceu9ZTSTZHvKL46pj24uI0U0gL8HUJht8ycMIHKT wcYQ

The solution table is

x630
y0– 1– 2

The graphical representation is as follows.HQpccv70SOtPQCzMZpHRlGSnOdf 6WjlBFjV q0Sbv 9c92sdx28PPSJPiILBDhHuM7YLhmBnE mAFPhd FGTg5vMywEhgISpD2d U3Mm0K6EzaqbKIZPis7NiGe6yLJEZhdBk

……(i)

……(ii)

⇒ Subtract i – ii 

c2w6PvJboZykfNQAqSJ 3InybzqVkegHtJMpTP3BqZWi9TprHGugipOpWyvi 0mYKr jMmIUOrr t1IQHs7qYBSHL4VFwaOKoLtA7lvMWfYE WBH6AZO1Kijkzl096lynt5bbMc
cpoDdI8TLpq1VHe1XFlWA4oH8By51F3K60RK7vZ80RdPVnB4B5OTXNyMManqcMO7Z23ZNY9VnJLnPKq4JAQY57qAuWYy
d5i29VXW7SHOgnJ9yLar ao5N3nBL2bVYaJ3Z RmR6Xy oT8SFDPY7R LUb imtjqyDANFHPwtUUMQNkWRn6D5isenGjS9QJcueZVHuhOl9MycWpju6qHsRI0V0Lkv9FH4vD58

Put in

r7aqiuQRok3lLwFAMUbSW VR oqpmScb24Etl4Fy qnBsmQO7EWIYewcGpMp0HY2W9qip4aFzgg6A fmFEVX2OlaMtMtOM 3OiNLKtF48kn6CXXygCtEq1CiWJR Vr156RDnDbA

So

Q2 :

The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.

Answer :

Let the cost of a bat be Rs x. And, cost of a ball = Rs y

According to the question, the algebraic representation is [Math Processing Error]

For kdWTuO42 9tOSAWoAUw46U1fbjQovHl8W2qGzrEGYR3R9gg7Lp33tnNK AFCR4qhOB

The solution table is

x300100-100
y500600700

For x+ 3y = 1300, x = 1300 – 2y

The solution table is

x4007001000
y300200100

The graphical representation for first line is as follows.

gv1N0TrYA19l909 6z17So TUQwFTTR v6MDzMjExvhz3I1 cJf gXt0pYnonYB4dKMSh5Qaxo8frkDstb5H0Mxdwmdx1QKbtWQM48bfEt yGtYA bR 7vtWN9p5u8oVc qfcFg

And graph for second line will be,

JIwdqFNl6 4ojCpjtc58Nn9869GNlOXk1E1CEFhT faVUHLzJ3itS fIAfuKQerF3NTl tEMSvBPTqkrlxlKUc5 RoC npA2vIBcEfXL2kTb9GpvBIGu8LTctqOh6iAVppXSs9s

Q3 :

The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

Answer :

Let the cost of 1 kg of apples be Rs x. And, cost of 1 kg of grapes = Rs y

According to the question, the algebraic representation isP45bID0 cMdC 2IHajN3Xg5H7Y9lZwFehOKWdFqV6nrSmGQ9Ld1 GdOjXOg wFLssOTcvf3RfeOH7 qY2Va3Mpj3DzUZMzZFtsMFfjjZa pvOsPJ093If2 UwR8n J8RBoyu9j4

For  V6Wyrf7jez0j32KHz 2slcns0Denm38iy4W6qmSeeBmHzW8B2 MmN9YsmARsk04vAHl7egT66aJgQuqLhZGnmkL3BxOThp4LxwzAEt,

bWAKBIQp6Qm 11BZadK9Gyox0I

The solution table is

x506070
y604020

For 4x + 2y = 300,

The solution table isvdEFO3xINHHaMvamqihYwootz0jbiGtCfHAHrG bSPD8or0eDqSy GwFYMjK dvodeSR5AfyFMH9DZJD NjY2ZdJtNW6NNfRRCyu4x8vaH3VUHBHYw G3sL6V6CKF8HbdqArt9U

x708075
y10– 100

The graphical representation is as follows.
ZHMqWayv0IuzcmPVYtDE3DGXBnSsIBxPoAnSWPd1G4KpwQSzawxd NWDxzjPgbQtoCOVO9X tA qiJYQG4SLxuhKnsEUee Kp1c

Solving the equation algebraically

Eq-1:

Eq-2:

….(i)

….(ii)

Subtract (i)-(ii)

0 = 10 (Not possible)

Hence, no solution

Exercise 3.2

Q1 :

Form the pair of linear equations in the following problems, and find their solutions graphically.

  1. 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
  2. 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

Answer :

(i) Let the number of girls be x and the number of boys be y.

According to the question, the algebraic representation is

x + y = 10

x – y = 4

For x + y = 10,

x = 10 – y

x546
y564

For x – y = 4,

x = 4 + y

x543
y10– 1

Hence, the graphic representation is as follows.NHWFZK7WsiLlBr TR9cgrwZz0QeWnpHWz8s3o aAesmcNuOap OFawkaJcRIdFHedqxgTy2yQeA4 cfOr XFVf5Tn1HcQHRX4PfItlJnIbf5lLQUtodws3uN2AVpBPM694gZqqo

From the figure, it can be observed that these lines intersect each other at point (7, 3). Therefore, the number of girls and boys in the class are 7 and 3 respectively.

(ii) Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y. According to the question, the algebraic representation is

5x + 7y = 50

7x + 5y = 46

For 5x + 7y = 50,wzNQDS5lSW086fg7oI5sl4h7BNoWMxuL75FuRgcXyN7GA6M2IypV5WJQt11s7vwcFDHJsySrh89EaUgX gi7rAG3z PC 36mhZNT8JCMJpdEaP2pOwLYC4GBRaISRS COJt Vcg

x310– 4
y5010

7x + 5y = 46mAxER85caU059OrhExHzcsF8W8BD64JNEad9Ma3g9gARTLqZ2d6SxZDksXSXOg95lc ThZ

x83– 2
y– 2512

Hence, the graphic representation is as follows.BmwIMAj3M6nk9uOmxqY0EdG4mSESQVnZou3rK2mEQwpEL2G9keBatOwr

From the figure, it can be observed that these lines intersect each other at point (3, 5). Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.

Q2 :e LfEj6YaRqESgJp3t4GbyxOJPF0gxIpxAC3Dsm7M3LQR9spoL2KujaFg iUx f 13H cvpe0khuDlq0rkqZeweCPjSNvRETaIFH8S9GrG9U3pp1hqwcp3j3 nA WnWwMcUVHfIIVlIjoXdsdqNVyivX58yzL 7Km RKADuv3uBzJwhzeX3dDZ65iHY21S7W HhSXIfWjt7bJOc67ieg2tp7MwvJ3uXErRsAe1itKzKTuMfry53AnHUFshcqmDqpdCeTURlRMztmJed7oqI7 E7j9g pq9NZmRnxjoEQVzOOy6OiUijntxuzahZq gXPl3Gw5 5X0S8E3nLXk9XEXAkEMd4zje2wT mLtSY

On comparing the ratios , find out whether the lines representing the following pairs of linear equations at a point, are parallel or coincident:

Answer :

(i) 5x – 4y + 8 = 0 7x + 6y – 9 = 0

Comparing these equations with

and we obtain

wAwKabI TpdKQC5DAahWZ4HaQGluwULDc1BBGfccwrEaQb5X5df6 loKLCiCXmTAzeS1rzJsGpdXwD75Vlvy9 oXMfB1IIhMlwB9 I4alNChGRFGiQA8UYQn8oOyeoJ8Lb4pIqM

Since 

Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point.

(ii) 9x + 3y + 12 = 0 18x + 6y + 24 = 0

Comparing these equations with

and  we obtain

Since  

Hence, the lines representing the given pair of equations are coincident and there are infinite possible solutions for the given pair of equations.

  1. zJBNE1XnI7 8rARKwbMO2QkXte iFuQQ0J1l SRIM80A OWF1bwgHQdM3uLO6W5fVqx5y6eUyCHyUn8jPdZu 4ZjB HsWmdfxaODbF5UxCB4VrH wz17lp1prZWUd8rD4odIwU
sq41LI9C7KSl27TN7wyHXl 9VH36b6CGStl Cb CV1RKQBvurFMKWhlx igRsc5yZAcNj5J3vD oxxsnfEyY iLyWxh0Myy3DF6DxNtErHyw0M46nY xf6J auIaa68QJ6gn89E

Comparing these equations with

and  we obtainUDwVP4xcXVixPAu4ck91Bqjptohdjvcd6y BawLCNyZwsY1NXQ2NlGTZwdNij4iCg1Utlene19R021mII72J5I7wggxdMdJuy54yh3NCL4ovSfTI7u7lpNNfK27G IwyCp7kpFY

GCenSvZEnGf0LYb JcNJC3eIcyBlyE qO80J47 8Y04VlkuO 9CnHxWWXm0yE6Qa7wZXX UDvkZW8dEoQgKnUolFGCTe5HKUC1f5 mt5VNSVNollYjB4AjXqc9pPtUnlrIsQ8FM

Since 

Hence, the lines representing the given pair of equations are parallel to each other and hence, these lines will never intersect each other at any point or there is no possible solution for the given pair of equations.

*Q3 :

On comparing the ratios find out whether the following pair of linear equations are consistent, or inconsistent.yYOjWTb5CO SJdnX5 dLZ2N1kJgcX USr4DP806EbZ6gqdH qZQqZnSWXkT0X46S5CZOZ7fla7vhkV42reCZPbWgTjlLd6bDGia q8YOqmNmco1m 8GoJ13DDe85TW3mWVll 4I

Answer :

(i)

xbaXpPWvk4Im8Dld3Aix1OPXfZdEAzv4VfcAmz6mrORWpwKtDjkX5wCxDZtLk0h0 YW3C7yOVqjjX428a1v8XIil4miDeYnr5lPM3mnN88957

DQr50NKou2CJ1ZmAguYhKUDsT7FxJIQ3xOefzAQKyZy9Av16t6my7lM5sur9HTHLXVdZeo7ZlX8aSskk0FDuySB2tQDHEcSYUVveJY1LYa0yKGsJ0q14cQxTVz UfqE9isFp0bQ

These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(ii)

DUpY8tEESrg G5R vdwQzQxM9aN04wC2cLAU65 nWqEwHauaEoEQXruh9DOk5f1a1u2H8NiXb4dxySoSaLLbRP ygdoGcZ3PV8sbjgZ DmNXttYfdLBvA1pRwzD0TbiFlcqOQuA
0vRC4XCCk3qa entFGrolfDo4yZmgaOmIpkwLEWuKctv1DEn7VMOWoLprJ REWymVuLbA77hCsJJhFI0WnmzVucWlQTQnc4UuvmR7o

Since  41NDCxJMHan5ccD4tlCFz1EyWMbePPRduZx9tboJ CM z9mnmgsotMqY73RR8AoMdRBuOow Hs6jw6o0mF bO6 1ITo6tvTAh0cZZTE9nbPOfTxtZ2wmI9 arwE70liOuRD2lt4,

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii) wHRcEnrLzK2gOB1UqVhtyBVmUcfiz7hBiCctodk8BnaPvfhLL3wQqRGBzF6fD3es8n81vxxmBVHMmvWw478KAQizbYu8Rgv58cfn vsO6v1XeaIAtSf Asvy6kEmVDw9 RT qhM2I9Z2ntM

Since  0gxvo4yvdv6vyYKsfqyKlnqDLhwTKrdFLp5OTOp3eqZp7nWG4Xu41I5pqdxqWJYFVPnYoKYaB10 g3o88dOB0XhneizrvuiFKhUgag5cB4 jKUtAuqAiz3W0H cWfa9f5r1NKYQ,

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(iv)5x – 3 y = 11

– 10x + 6y = – 22JoPtzTPYIjjt30YxVtiGfwOdO1 cUcBIrFdAj wFMJ RRddUT09611HGLL kw0aYNbZWVAAGcdL9y8v47bV3VTxR9YO wXW9Bz lEz4izpEqu51j0ood rRqDBzH5YsPp4rO44s

Since  Pg7OczMCtcT6EE2U1QQSkJS5sRjufwhGX7Y9tqgVMbH0ouMskESBBgLX47HFymNwipdujX6MMwj6tQlwuirCD8hnl5COEXnstLhTR i2roNN6i2 rOyTa9Q9ooC4f1UFz24di1o,

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

(v) gL81pbcUL 1FdrVb4gT3kiA4ALJzYprKgBsJ1BGl9zIqHwpXGGrL6YndOvwIZVkOyTCqurgA absHBeNykXhXkbK768gbCxjDJJ SBWMKxE86NTVOYmBV5YlTIOmxf RI1PwGAF9aWuoOCLPFxwUJU7wBF2jQ73lX9pBPxny bsodXUAdsB75iUuKp HGaXy3ahjqC37fiBbD

Since c8ZGy43XZE4rjK0kJEQ5l0EgaeXTXbgC ihQv K7dzWCUu9g9VtqFwNQtWeBc7ahnCNFnT iQRAl8AdvY3FUXzB8 WCweJACRTFxoUucZYB

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

Q4 :

Which of the following pairs of linear equations are consistent/ inconsistent? If consistent, obtain the solution graphically:3fJxjTsYnvMCAOuLXvuAUbfJuxJDPnZ81FNw3xfuabfmQYesgVT0rxK0SZVKM49OVzQQ5uglD7txdFbZ9rVgFlX3LbFY7wuD Uz6uWG k86xN7r3 U nnoByvSGeXhHzTrgwR I

Answer :

(i)x + y = 5 2x + 2y = 10JBLqcZZ5kAaqZ1s7qUS1TH9TvgZzdlveyZhUJbV8TKYxUvhH4 7ZCLBQEOGgN2508YK3TFP ikN6zDIUdxlpvXeUMVrZzWEzAE3Yn15 Gkr

Since  Pg7OczMCtcT6EE2U1QQSkJS5sRjufwhGX7Y9tqgVMbH0ouMskESBBgLX47HFymNwipdujX6MMwj6tQlwuirCD8hnl5COEXnstLhTR i2roNN6i2 rOyTa9Q9ooC4f1UFz24di1o,

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

x + y = 5

x = 5 – y

x432
y123

And, 2x + 2y = 10wIW

x432
y123

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite solutions are possible for the given pair of equations.

(ii)

rzuZ6Cr41po1ZhMFh0Nubuy0Uh CMkx1t9G1w4TfrILOzwA6kq8aJeKCzuRNfK2eTTlyny4E8OWakWdpfrGitAqEZWZOwOzq2u r2ykKM6RDzXKN4hK7ukSrzvNgE3ZvjOGJXg
HNF OIZB1n5LwD7CfHcILmXt6omVY0a22OmPcpIxiETnuV6w MHxSktNcYuAE1nJwAy0bnbDwh c DJq6r17JEQb2qH0ceIe M KmeJn5TYbA2d7eFWsHoQLROe6Ssp5m 41Ph8

Since  41NDCxJMHan5ccD4tlCFz1EyWMbePPRduZx9tboJ CM z9mnmgsotMqY73RR8AoMdRBuOow Hs6jw6o0mF bO6 1ITo6tvTAh0cZZTE9nbPOfTxtZ2wmI9 arwE70liOuRD2lt4,

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii) 2x + y – 6 = 0 4x – 2y – 4 = 0cfTMJGqLYdjAeB3B9AhG amHL7bOYJP9LyiYIV2 oTupHdJOa6qwnOWcB7oDySwvOsI 64fMvOg25 w6Zc6LX X6pecjWgR7IzQD7HcPpFwlJQjNsh3vvRPcWni2tVQaKhbM58

Since

 0gxvo4yvdv6vyYKsfqyKlnqDLhwTKrdFLp5OTOp3eqZp7nWG4Xu41I5pqdxqWJYFVPnYoKYaB10 g3o88dOB0XhneizrvuiFKhUgag5cB4 jKUtAuqAiz3W0H cWfa9f5r1NKYQ,

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

2x + y – 6 = 0

y = 6 – 2x

x012
y642

And 4x – 2y – 4 = 0

x012
y-202
C:\Users\varun\Desktop\JPG.JPG

Q5 :

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Answer :

Let the width of the garden be x and length be y. According to the question,

y – x = 4 (1)

y + x = 36 (2)

y – x = 4

y = x + 4

x0812
y41216

y + x = 36

x03616
y36020

Hence, the graphic representation is as follows.

4UEosFByC5EsafyOqVj2vGCGqDQiVF972YHEcDxZh67KkNh1a0ZLbdPoWmFpS4z8DV7qHr MEjzWWQlvXpkFk6

From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.

Q6 :

Given the linear equation 2x + 3y – 8 = 0, write another linear equations in two variables such that the geometrical representation of the pair so formed is:

(i) intersecting lines (ii) parallel lines

(iii) coincident lines

Answer :

(i) Intersecting lines:

For this condition,0gxvo4yvdv6vyYKsfqyKlnqDLhwTKrdFLp5OTOp3eqZp7nWG4Xu41I5pqdxqWJYFVPnYoKYaB10 g3o88dOB0XhneizrvuiFKhUgag5cB4 jKUtAuqAiz3W0H cWfa9f5r1NKYQ

The second line such that it is intersecting the given line is

FYuuZEsEMdy88fD0QeH27hHW0VN2OjVwZsynYXW AIYrP5eQiIL68eM ivjwSKdl7Sd fb1R6fudWDTTeZwm y2Rs5boicL4lMd7k6ebrT2V46yvApWKNWKBZ6eT68ulGI1fK5c.

(ii) Parallel lines: For this condition,
41NDCxJMHan5ccD4tlCFz1EyWMbePPRduZx9tboJ CM z9mnmgsotMqY73RR8AoMdRBuOow Hs6jw6o0mF bO6 1ITo6tvTAh0cZZTE9nbPOfTxtZ2wmI9 arwE70liOuRD2lt4

Hence, the second line can be 4x + 6y – 8 = 0

jCoKbVT ZRV6nyZguw9PAupgEOTQz XVXO9LkpFPEcODc5JfAB0TGPW8emDOw3zWE6ouRb HztY9m4wkWNdl3zQgRWXTkMoklKorLhSkSulZtqcx6FNYEJpyhV1l R51pZwZ54

(iii) Coincident lines:

For coincident lines,tMdZ4jrjljJ7qk11qbd5scJgs8SQ7zFHaFhiAsFg celoeL16oE Y2l97MQ8EgKvug23heIfR3doIDMtUhCqZIbFTP Y1K5jsPg7OczMCtcT6EE2U1QQSkJS5sRjufwhGX7Y9tqgVMbH0ouMskESBBgLX47HFymNwipdujX6MMwj6tQlwuirCD8hnl5COEXnstLhTR i2roNN6i2 rOyTa9Q9ooC4f1UFz24di1o

Hence, the second line can be 6x + 9y – 24 = 0

Q7 :

Draw the graphs of the equations Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Answer :

x – y + 1 = 0

x = y – 1

x012
y123

3x + 2y – 12 = 0Aa4gH7RQUhXBz3JLJ5NQL6qiK6KtKcWthyeoZJP92xWT71RDuR7itlY1kHu0e3ENvE13LMYaSGm8

x420
y036

Hence, the graphic representation is as follows.

fJ2Rsa5COr6vn MUrJvjs T3OUA1Pl9caio9eL9gHMYvmT2PtSAOB66iv4m0XAXGclEvL qKqoqf1jyvDIM5mUyRc1G9M2yt0pSnvhe8ViZy39m9qZe7TU3 yjKLZurSl1KIis

From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and x-axis at ( – 1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), ( – 1, 0), and (4, 0).

Exercise 3.3 

Q1 :

Solve the following pair of linear equations by the substitution method.c4dB 3jcZM T8M7MPL3jd8Z4wL16TiO99qMPO DXWJnyuaEtCrHAVbB6UO 1HIVNsoQDOtfh mtiKZ842M6MyHwA8vZLOXCdIulpEF6GX3KpXTpRDrycERU9tBHkRlTT w1mpho

Answer :

(i) x + y = 14 (1)

x – y = 4 (2)

From (1), we obtain

x = 14 – y (3)

Substituting this value in equation (2), we obtain

WV6wyXvW2nFoH1IJQhdeA73akqqVhgwk0qWJRFRdnLDQ BFSvKHbaqxB0e7HL zsdRZRLzOhB20 Fz2A5e0gMUO39X6nnWUJZyH q FzDvKoIii4sNu1 GIIDmZdpVEuITbRQHo

Substituting this in equation (3), we obtain8u5jDQ0ihffnOcBPq87IVS6mT5az

(ii) B7SmR0mVx38XXBeYMzleZxBu 626OBxs bAS0RTg6Eu rpfbLDl7mcNsQJacAZNzGfyqJYWGDJIIxj P gfpGIkUygj5npQXTqTOJH48WL tg0PS8A1tKGDvzgERrYnoV18SmJs

oy517 KleanAXlaO alYCrWvv5IfcOhH88wL3fvz4RAcZ O2z7uVMpY1JU8I8yJeBPC7Ba7pVO0ilChYsD325WEjbdS4CfPg5Qc5yb3qv uLg9yJcoiQdzcoCG6KtHvBmJq3 U

From (1), we obtainm4RUVg

Substituting this value in equation (2), we obtain

1z5WSuR75TQKtTuulunk6QJDEiW0ZxkcCeusggKKiHa 8qLELg33uT41gPlkMLmDyD1sG9Yz1qIWWK9ns fS Fsnr3ABxt4TaYnSsKCvCP9ra6fn6j gJpUCG7 dmtL hJQK4Dc

Substituting in equation (3), we obtain

s = 9

s = 9, t = 6

(iii) 3x – y = 3 (1)

9x – 3y = 9 (2)

From (1), we obtain

y = 3x – 3 (3)

Substituting this value in equation (2), we obtainaj7Chf8C7fvo8i6UP CF7AFfUGFfu7QGyo01rJfLg7NnoWMuQUbhfYFRrO QFo9dTETBNMqtjX4bo6qJs0FfSfynCFtZ7Z3VuACmTY5lXo8IO2METS3ccXqnxKUgUei RVYRq6g

9 = 9

This is always true.

Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by

y = 3x – 3

Therefore, one of its possible solutions is x = 1, y = 0. 8cJ71ouZEujqiK2XGtHvn6CrxCAroWblPzwMAQ6ZRcwm4ujDS8eqpjHbKxXS6F6q0mB7GUNNjfFXnDX6BzRme4iJj OA79ZiB4CEuEa3ZmkRmWw9nCEODdb

(iv) (1)

From equation (1), we obtain

Substituting this value in equation (2), we obtain

1fdnAswCcqRXckBgJgGDm7w8G60NCoXpUkZ GTE4R3Ggyauu1XL U0sLdee9iE96G74ep4QHwqeiyq7zClTWdtgUIcfg7AqbNCFcLfG9NaWLyue1ggatYUKn6ucpQQ5frkju9o8

Substituting this value in equation (3), we obtainN7aWMyHs

(v)

BE1ORcrTeTv

From equation (1), we obtain

OxZRcOaYe W6TrxfflWH4UfzhoBYy8fSdn JBdOteZpTTO4uo7JIA5mM5fy7DUtvUWUU3rrcACxKZefsE9Z3ys8eEz7UD5W6ZYgpiXp Yyhvz EFnlBLCiwV3oj0cx 51 oyiU4

Substituting this value in equation (2), we obtain

EpeG79NG6 6jIStxl8Z3EGIPovP9LZV5y oQSZ5UBxOH2u64nkchLjla0gdJQUK XbjbZZIqk43q5avVuUiXN2zoj2DZX5ShJOUcWq7Ca2Mw4iO 4ziehBl5XTpKaFfNfyyPUmw

Substituting this value in equation (3), we obtain x

(vi) …….(i) …….(ii)

GWkJXS7ATDFYfEmdODq7DrAHW20d8eZlSQtpZLVqBMcsTsIkAaltvTZysfRgYAnRNKiUTrGdmrX2Q6GkPuFhUyS WDJbA0gsnh4cKSLjWi0JM3IQIskvjXdRdDsUoNVuhvrw0gw

……..(iii)

Putting (iii) in (i)

Zah4jFBG0RzduAXqvYodSS5aRX3YfEHYoVJ28TW3upI3vqZNtrXjvxvCLUIDrglY UE7tSvdtgfyRCjghHALhIRwys0bjacot Ts98PdTHia pzyLPaFms klI6ckIPKCo jP0
DvYfxK1 XlMX0OVxJ POxv94NVVHA h xobYgGMrXlYUa3TcKX0W2lET24ZHBm8SVGKPpaCWNT6 JlY0RXu2NqWw7p5MW7oZEaRG0d dWiMT38Ypopl5hATxfj

Putting in (i)

and

Q2 :

Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Answer :DVcgB71Dn1X1YllPVOXCLqwzNCUXihtJc8NYd8cT0jN1oM SaNSMFUDkouHX9dmJO88DqV8VBn1MIUdKHgZQgUs9ArxlxTgPMWLUxLZ2rPmZUcltq5r9pVRddRwp0Z6RNqNPy s

From equation (1), we obtain

Q1VVq8AGCpAR8ZHhYwPqREclGKQSUXv xzDJU9banIUWFo7S5JqHVBfL4zTxsbLYongZVGQUVRHOP8vZqa5vY3t7RLOnLBJn3ZMgzDmc9qgvpNmF3wvAAWo bLItNQpPdrNR1Mc

Substituting this value in equation (2), we obtain

XVTpBudHsmQIZ HAEJw6TK B N2mwALMCBETxE 8ZPPdkRGyGNjvQJUkD3EWaHYiSPT7fHVLlEb2sG5ZWsGsAYkEa06wHxbgCBFAu0 g8Owv1U8lQNy3C07zk5Q 7zyfnu9NPrk

Putting this value in equation (3), we obtain

KQAkQXfBEn5HplHf0yEeSXPS5l6PnqAhjeRJGe UEhM8N1qS2tnIJ 01ojGG3ulLXkB1J07SPhrsuatluB UAE ir2dO5Hzo85V7DCbumpRXl Y840piFqdB1vjNscVM0lOF04

Hence, x = – 2, y = 5 Also,
uq5CemBaG0hxuNYp8jSNZNBYt95ao46veZCe4MDz0rKd2 HDxPiRM0ACPlST4vXL6mUzfNAr DfrOvH 3TMuKNHCsF8Ym1my6r0jvg5QmAoVWgUv41qLk5TCp20 tH VSO9WayQ

*Q3: From the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.

*(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

*(v) A fraction becomes if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes Find the fraction.

*(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Answer :

  1. Let the first number be x and the other number be y such that y > x. According to the given information,

Enyi5LdvcCU9S02jMyyXLEwu1lNAhM9KIGKI7n7sVCTMUwn8XMn4MZRSx5HrXOLgvJm

On substituting the value of y from equation (1) into equation (2), we obtainbvaXYNQtuF1dkLNrjFVwZ MwJ 2 DSNt fJELawlIJxRGbjCypnD odFOWSpZJvoV0qNX MIEbAR toAaykTlz71IneYH6xAL vhJQ8CcZK0fkmZVo3mn42VxWi mf ZQaHl1k

Substituting this in equation (1), we obtain

y = 39

Hence, the numbers are 13 and 39.

  1. Let the larger angle be x and smaller angle be y.

    We know that the sum of the measures of angles of a supplementary pair is always 180 º. 

According to the given information,
Ri2OVeGMk0M4bTaMsqYP9pweNkyjEsdfwly8v2MtJJZBX2fK2cn47KL1r9znCcuno2rGFBmMx1HyQsXJJR WFeoSuehdYw2IYQ6cLRA1FYbRIwtfME0rsy0G8w88NmXan3Arhng

x = 180 º – y (3)

Substituting this in equation (2), we obtainmtpeJvVbUEYLV274fdy PcVT617LbhqajN86wvqSR3 barIwoB1kZy901pI2U3yn17g r0Pqgj3LQB37BTzp3DxCCf6kXXSiNFb

Putting this in equation (3), we obtain

x = 180 º – 81 º

= 99 º

Hence, the angles are 99 º and 81 º.

  1. Let the cost of a bat and a ball be x and y respectively. According to the given information,

B EaTZqN8 NcMXgXZM8GrP1I LdVPt5i1oLqnYWRb5pLW x1A6ChpiDjexpe9pYoyChaOc1 A yQNA JDkNRvvp27wGBej6t1Xuwf aqcezVTSkrgDR0G7Dhhy6sKovGI05ZnOo

XF7wqKNf2FvXuVeKMGNeazyZxirAIWomiCG4ifu6owNKsEJu4EuBivmopGRENYciPgijNRQkmF092HkIWU JbLXir5LWAdRNFcbg0y

Substituting this value in equation (2), we obtain

wCAjkB69LKxyHM4zgWk vTcywmX1MPUzlVa0qvJOkE7BlT41X8dPHu o35AjkftPgsN9SIIvCU9U EQi69shIms3WpkHH9Z8JxAa5wbUTByHOx680RSauSY 0HCjrBN05saXyHQ
R2Xa LeLMRZ5Y89J Rv3gIIZEWP9sRzx6qZPTlnbBvYhu42Kz6Sq7qAvhr2YiLf

Substituting this in equation (3), we obtain
r9DVmMBeIOFSNIZdMb4sh4FbtQCutXhr1Cyt vSuJLKQv4cyf FYobGRsWZdj0Fq4smVDZwTX9dOJdDosAbIVEzESVOlK8wnojDvRIibPcW9E7T3fmDuCcAaItGPYphf95LvyE

Hence, the cost of a bat is Rs 500 and that of a ball is Rs 50.

  1. Let the fixed charge be Rs x and per km charge be Rs y. According to the given information,Nyi21Lsc7Zhe9Ew0fKDSGE2DSc20isGdByi FelnlFenmlFSY
S0pi45 lW JlUyjl4iPXTuZl5pf8AhgNo7hHkXx M kksoPQ1NgKVHJY5ntEeUZE1MQXRwTbEBJuWx7JobA 48pq777nxfQRvVb0LKFnySM sVF0z8wrwoy NnB YZPRTc3KSIE

Substituting this in equation (2), we obtain

0vNrKjc9YG16JOGQvuZuMUld8UrSHsU6pe1rzr1Vbot02yyPNy7VVLV

Putting this in equation (3), we obtain4hQV29uRXb ntd0 JgLosn0C6qOMtWPjXf1ngUmnQdXY4INdoM13EtsIyiJO vQO P934R3eVxrjVqbx5C5j1gUSj8APe0JdsKxqr310okg 5cKy wpNPyc7NgDOir5eILnizCw

Hence, fixed charge = Rs 5 And per km charge = Rs 10 Charge for 25 km = x + 25y

= 5 + 250 = Rs 255

  1. Let the fraction be  According to the given information,

From equation (1), we obtain

…..(3)

SBUxTRBwAuoGsVOzH EGf9vEzfK9ZRO1lCKpBSPpdcLlI4S4pz2fjTulZtr3CbnPQ2tv81cD j7qYfJ7Jy2PYPwunp iSCRjQKtZ32gc2Eww0Wc8WeO9RFZl Q7M2 IGmCXtw3Y

……(4)

Putting value of y in (3)

aHJRz76xZTVM6CI4DSP 0 a92gqwAmkwY9YaM5GvN2kAOdackILqvO64kthUPliYqnkGp6HD7t5fJDkyWgZL3cnvOtjMqSVpW4pipcWl0 fxxJzUxuU21BDVq6 xs cUAMQQGQw

Hence fraction is

(vi) Let the sage of Jacob and his son be x and y respectinvely 

vda4gVGbnV0jsvdleVuQTWYDuyQa7NftXcC upFKzqlkoHmvPxjqA9GnpJCr fLy9A3CEF0zd1JIYf O870lZ3fn1O1NJ2 NXf0899TGBZk8Drm7cGQi8X2urbJ0se9beFORZ6Y

…..(1)

saIwT8m Qw1YBEm6nxJDwaxfUl8nVHoUbaS2y SKbDvBJlZRANTgSmqODkMB4VQjA8w3M2xosj AvmMm 17cn ErGXKEFYtT0LbOuZzbv bFY5TBLJ149G2DbWI60U1r5ACNB6w

…..(2)

From (1)-(2) we get

gq1xA 3ld9Tgtn4mtxE8o e KdmAnblk1BOBLdxIqlXvheFX8K4bTtXrS1cAi8aqOABygAQCyqEqd bY5WlxwyZiCx83V i0sl76Ae3nRknAZ0w7cvSHLQRGdSfM2oQtTg9b9GM

Putting

Current age of Jacob = 40 yrs his son = 10 years.

Exercise 3.4

Q1 :

Solve the following pair of linear equations by the elimination method and the substitution 

method:V0VFXJNPoD1A2qzvfb22qwrRtdMcBcum4HL2QJmxRn4TyJUzyllj2ycim8ris0ZaAyD36p95Z0gD0bY5EpM3mMtPM1TsqLKx9pP5ecQT0GvZcWAHHHw7qqnDMSXTtQ ZF8 ylPYAq Sky1dWeJDF8W9rHhzVq3cI
EMoJ1BobPfs6qyzgH70i969ZDkIS4KByFv5CmGNAWDLjZlIcOSLRGF0WZtT6QO0IpmPO6M0H0cVvZCJix37WyKRAFG0esz 8iecBRHVCbnxVY4 tlNan 9Wa7hSyPpbDDdUKTjUxvH4oKD jjLO67Hc4SqyCuO1O6YNTPOzGM SQUpP0C7hGIq3jWoMkvI9zgZjTRhb9az8zY76TvEb0x8rSD0sMKVDz 6XCIpu2RXIBw218NZQ2vdYFZLEBTf539 KBPkOku82kMo

Answer :

(i) By elimination methodtT844yZWLRK3BbwWEDsXKu7P5buih1SbDaqh1 Bj5sf5MVtThHZXnAATDaQiqsrJT 346SV9cXtH7Ic SfZ

jnEYkOvWTVaSI2TtA RT PQvmgbVb vJoTRoqvLOcEE6nIX73 N2DgkG asH1BDu4zgo7UjSka5rwRxJgKW2AOR5M0 u89JbyLCO t0OcuI99hibhsUXHC7wHq09G0EfACaMG1c

Multiplying equation (1) by 2, we obtainEWkF0zcRNEioZrKifFroe DKTvoyvrkwOQzBHP80Ku4rFvjiStHEs4 J1knTK7Re btEDya34bjiQ qjpY6buuu6O OcNRIurvDISChQ6sfuFHr1kW7Ern4tVqo

Subtracting equation (2) from equation (3), we obtain

RIMIUFkWaOqYR5y AWpLwfBQj rbzCSvSxa4nm8UNGFHQO2NTAnXg1T06DWVxWxsmN5ChSiHCbZ2U9gOhad2e82 n3Hm RmjAdntJmcBcCdoNdUI63lal8bOII3MxYvUn k0aw8

Substituting the value in equation (1), we obtainC6fiF6ICIRYOhdWiZeqJrp6XeWtxqsgy5pA4eKzeT256nwN012T3g 4rJwfmBzIL2RMvY7zE8gdoFu do0HZyTtsj1gSDm1Ilx86ytFJkY8qRBvxq 2j4cLgPsvI IovL6wPLUI

By substitution method

From equation (1), we obtain (5)

Putting this value in equation (2), we obtaine 0TD6xY1fq61MwSJVksrV73a6DrFZ5TLXtR6

– 5y = – 6

8ze

Substituting the value in equation (5), we obtainC6fiF6ICIRYOhdWiZeqJrp6XeWtxqsgy5pA4eKzeT256nwN012T3g 4rJwfmBzIL2RMvY7zE8gdoFu do0HZyTtsj1gSDm1Ilx86ytFJkY8qRBvxq 2j4cLgPsvI IovL6wPLUI

(ii) By elimination method

GV j4qiNbudoA0nvi9xtVfeR8 X LkezmHA7q3typEk30tkJlYedS5QvpdukIqFFj mnEigB Itx4VNEz1miZrRQEAhStoftFSggAnhW
L7rDEyidGHif6TS8EobDESvgGrxKosnGwY6O

Multiplying equation (2) by 2, we obtainnm5Gce7r03DOcUmONpph3ZMMFqMvoWm18P7asctvnsRpRx9Hzq9uWoHnAx2w7SR4z6ex8esDXUK2WIz

Adding equation (1) and (3), we obtain

siqwCyvQXM3j0oKa pEwHM te6wJmdzcH0SVlW7 9OQrEGP45xq4P 7wHIsi6WH9BaBhNxkuku2gHzAtN8 69 utevAWxtgK2OpjcGHvK8WleQV7ntiQYRk0v75jBPTD4q1 c

Substituting in equation (1), we obtainR nSV6CFwByOwelbb1XkWsClwehplCIw7XIqQyw352Myr1HKoS5dKto8NYru5pSr1vp4EYZekKtg1DzAqNFRUoxAoiwZ4GfmECIs1VLBWWc7XlAQ VNraRFn9WaI3BgBReJoSrM

Hence, x = 2, y = 1

By substitution method

From equation (2), we obtain

fzSk7RmVcvLxDi69m4XQPdK7Acpt EaldGDLk3SlIVSJj06Uh5MAah4zEmiLPfYGTkyuAyyvb8b2JaaU6LRg8rUeX2d3klpbIpwkmvZwqFV HOT1 IAWB2FgooEDgnscn6iKLE (5)

Putting this value in equation (1), we obtainIh1UMBNZ1MBd lZi hcmDNvtYzVz3 R5M5O1RYFOPP2HgDt4PYx1imZMuhEMShJYJiNWDn n T s3 fVvfSCK0uHQHxnqHDAzy82DRsHDhWZtRoKDJqP6RlI0vJ0v

7y = 7

Substituting the value in equation (5), we obtaincmLFGUiv8QX6S4PRy2qHJBlpB4b1 eIP5ANHe4icl IEDUJwdGrnAZlV7uZDcTp8ji0yw089oV4DcX2Zud7joek7RLoV2buqrFXUmC7BXqBq8DhnGSTSzWmPj5An8bMVTO Pmc

Q2 :

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

  1. If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction?
  2. Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
  3. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
  1. Meena went to bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
  2. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer :

(i) Let the fraction be According to the given information,

Subtracting equation (1) from equation (2), we obtain

x = 3 (3)

RyimbxYDTLYLmpTxWDcK5UzOLYve4ejrZ2arB0TUxObB4QbxnB4bCw

Substituting this value in equation (1), we obtain

u2w6fV17wWRrK7mUW62K41TlQG2NXfNSANAtlVWxa19wY6pwUtouAJvdl1z9QvegeIkmPkAdRRJbY ihhiUBCERCdbhJ3gw6qngX9X7n1tYzM3do9XEijvEBO5JZ3WyOAiD7URM

Hence, the fraction is  

(ii) Let present age of Nuri = x and present age of Sonu = y

According to the given information,

jPEttUjUh REnLZo2rVIiM6QzpPX3JC8MkfMorrImVDalXpLLtmNFwJoeR7VFodAszVbhsuiGCGY2bHf3swf2gZgRst

Subtracting equation (1) from equation (2), we obtain

y = 20 (3)

Substituting it in equation (1), we obtain

RFdNiUH4JJZyt uplB8J 2CLJ0v1kDs4NkjLke2gDBj80 m5 F86aANfNIm08yV Rnd6QFZ0vaYXkmhYqcGISpqk9pL

Hence, age of Nuri = 50 years And, age of Sonu = 20 years

(iii) Let the unit digit and tens digits of the number be x and y respectively. Then, number

Number after reversing the digits

According to the given information,

x + y = 9 (1)

o52WXT8Ea7e8fvLPB0VTkgdOL9 YaK 9V3LXZbv A3f3zfsg5wRTigu0Sc8tIgT0H8mMEUptfFAlez8dQ3StZvDP7KG
Q2GqMgSvZkIlQuuwLMEsF8epDlILNpPlJ4053h0zlH7 gQ0BvOo4jripXbqUjAqlyMh 5oTNVNhCzfZ61 Ye7aMOodvK ejcxXZnTDJcXAdY DTZb3mKyk8OmXzzTZ5hXWMNceM
D2VO aDnFfEGGF49mCE32TjgqM6U9F7P8mjiE 5tmVUksOq g60N1csmcfs9FCAkEmfK26F9

Adding equation (1) and (2), we obtain 9y = 9

y = 1 (3)

Substituting the value in equation (1), we obtain

x = 8

Hence, the number is

(iv) Let the number of Rs 50 notes and Rs 100 notes be x and y respectively.

According to the given information,XKk x9tDbxOa8MGOdl7dOmpK9ISaFd ac7dyHAG22IGBVbzSRsjS RCeoDCtxB4xCrzxHEM14 wVpmrWORNvsUU8C65TNiHm oAdXQ717 jIApqp6AQ9cfMtIPjBVR0c zfIXU4

Multiplying equation (1) by 50, we obtain

POn2Q94eZLt68oA2 D1PmLlRq pJ h4Qra8ab5TKh9BhOL6GEdSIuaaaFclaWfOwp qzR OZpmUrIa aufY 9NX89bq40Nkr1uNoSX0Q9tS cdhnKfVfa3A3bjwnsvr20zZrWEg

Subtracting equation (3) from equation (2), we obtainQNo8Uzu201d ZnxUufTTQnSYWt2g2aatMZRIgrN20iWOpWjmVB5bNld99iyJ63ZSHIc5 pU6t8Nj9Stzns

Substituting in equation (1), we have x = 10

Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.

(v) Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively. According to the given information,

Subtracting equation (2) from equation (1), we obtainxknyHDM8G5qo6LqSiRSnGpjhvZWuhmhFwaKSJbKUq62fgqSQrwQI9PBzRo7uJsIY1iSzsaTDJBYs5I70ocdpSzafr2A5uxj2N2 VPUP50EhihdNcpKn1sYutoElmdMlw it3Gzk

Substituting in equation (1), we obtain

cvc1Oxkn4XExRDFPw6VylkoBr4hhCFoqWvPN65FZLFvjgmYf3serrSMKM9feacS5ULugtjD7ynUHZeX4Dbjrb1Bz h5SJIN9R1cD1Tmi9H4747U0 gpyUzS6YU7 PXv8jji6ngI

Hence, fixed charge = Rs 15 And Charge per day = Rs 3

Exercise 3.5

Q1 :

Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication 

method.
NU2veS0XIePKezDY1tXJFfzrj5Fj1qLRE8sB4AMYPTws3wnROdFEbHMKYpC1KGN1

Answer :

Voiox446URKICVdfakisX4N7zmIG0iDX5oraDTFiOcARxW9 Kli9EopyoGF39XwNVa n2o8GR35HWK Q1

Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.pKicRLWZ7SojR3akNH gg2fn688YO iPYo5XeOSKLSXCgWNCRzYPd8FguBU0Jl79EL6r4Wvq8 UqTqn1wa1iLVXDZQAaClC3ekzPFGEDrqTNt v uAG7irQg8R488Xkt6pzPlL8

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication method,

∴ x = 2, y = 1

zpbqnjX3b8mwFPJgDa5x0ghzLV71KFlzViNfpLQHJOVT08O RixYVXm5c rGfzE9NJOUo1eZizaGB78ReCcxQTsTWYES1sI8zefiSgdZf V4LUziGZgqeqk3uyhPtfyIwtMPJQI

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

Ad6oEqOErWkak80gyx9mz h04PJEvVQDhR1iKZVEgPKUggNUxkfDsctQ48 w3BQhd hHooQqrISVc9oAdHEZLabAttHbQLihvzPzJhb m7C1eGQqaWA6EnX v RvDRbNqHXAGo

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication,t5sOd7ZeB3Gv1sraii eYxTU1MP8U

JI5zyRGT7Nb0psjHdl8mRnyyQ0jIy8mIR7 i3 G5JDaY323vEXo0XdFKFYhHriMBo4BFadUl 3dhw05jMKjXwaFqV44IAY6zYy5BZQwuXkyfFg6F5sF1qJy7o5uJyGD93zCt9LQ

Q2 :

(i) For which values of a and b will the following pair of linear equations have an infinite number of solutions?

4WaPHVBzSAv0pTzLVykDqRHSc0e6iJrM 6X7ff80I05VxDB3DE5wKJ2 SY2FJT9tKPiKlFaX1Hq7KXAUonNXphev usNNKJItCWMWJ7yZuMzbOdalL1FpaAUATlMqzkDJRPDVlE

(ii) For which value of k will the following pair of linear equations have no solution?l6 t85jedFkTZ7K4MM

Answer :

For infinitely many solutions,L4nQH3 Vsz ID9EKHANyOvPg5P3EH2BMLjwlA6ZzPA nJ7JSa0GS84d1bew wgfjPeBHVKfW0EtugU3ytuyXmtjBdGBYkC3Ok9aYaIFKD2RdbDYBGjSnwNpfTTP9QDrG9f aaM

9s4DjVepyOEtuAHL9EJYrH7QIP5X9BZ5ipbdnAKEc6bVCZM63mbpqqLoFKyV4yXAK79hDowUFJlP03M0dK1sSr1fCUBkM3l557CqPvjoPNlH7p3SPV60LLglwdp3CQqn0YvhpaQ

Subtracting (1) from (2), we obtain

Substituting this in equation (2), we obtain

7uq50qOvoP6fj5WmpWHzsOXBlTXNt5ktAFeS0BZkCybvODrFq39AdLUbi wz2ZUlb9CBCr8BVTYbRDRY467fSDtLaTfR1sXYSNsEc wn2JJzpN6HUFFBEx j jcOjyXQ4FZ4K2E

Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions.eO0r1NTTr4HX5v 8JEGbCl634LHsO21OGhNR6xqYYPko3LuL7JuoA1IgruGymw

For no solution,

t2THEDT33qqzSibs7MgYbTwP9nRYLlbk pMdAxS7vvq3lsrxHzCdgWPZ79W6q6ufKbQktBqmhPM86WHBtnIgE BglBkvMKNEEn3SH1LlbfpdcKunMC71LryHyGzFmzGI4V2vS w
kU uc2u8dR X4vxuY5kYz2zJlB3P4pO0KSY7TdgkRsTtgQuApvarVvzXMQ531nDpXlZZvIQ3rEstgCDDbIJ h9cVoQgkROkbw83 GbwokN9rkddGtGNiv0R9weWZIIWVuizB1g4

Hence, for k = 2, the given equation has no solution.

Q3 :

Solve the following pair of linear equations by the substitution and cross-multiplication methods:

FpgWJvHDeulPgGMuWa1eiTA6xn80ie1Vr4KLAkatps9F9JKXw qcI28AWnj wCv9Rpm9ah2 uwOexmITnD4O0EV1gG

Answer:Mq5Htvyz3aLH1X4jjtuEwujG2wJ9 hsXj3XsqOMQpB x9E1up1qUMvZF0Oi51blBjQ6IcAgvw9z84XN6r4oNtPLOgsDU8QzZs9Eh0EXQjsTMzRC2sc7sZSzcZKzq9BWJ tpmGNI

From equation (ii), we obtain

Y5DJFpJaKJUMWWF5TJ9jkJuhmCQSybY2SxQNbi3EbJAXb1QhTp2bLiikX3Dg3wVT1u76Mr7brz9EEw c AIu4ZH9uH2cJ2hjSsZd2BnSFv9mmOlwBzhcAE2D 2lI3uo5AQXw2 U

Substituting this value in equation (i), we obtainPm6rKIQ OJTKO0Ft3beX4tBWrZevDTW8ZJqyelqU4ZIa8QCBcrxNBLWVkZJXdeXmKu V000Dd39AmZDTV8AHy9wrWhXjjZa5egpumfUTR812g3i zCGFCQttM6nLv6R1uVD2b2Y

Substituting this value in equation (ii), we obtain

x6CJFrIf ur4UYtEg1Er2YhcH2BUS7ePIED9Y5bf0QVHI4Ht7IKNHm7gNZJeK81LKSOSPcr3pmd3fDcKl ESqShE94RaGeO UiSAQvp41FejzfcsTrXvQYsXdod1TFtNCAdn8c

Hence, IrYZd2nIZrFs

Again, by cross-multiplication method, we obtainofrXpBQe7MFg7coHNz6R7CGJvoHWny1m6LSvptI mrPH67JHgKFeuH3OXTgJcuE8tBKNUgaJG0Zt2R AHc8mmykirdbtCoMk a9yKwLEagLvc8Z771elF3P1lQuasQrrsOggf9o

Q4 :

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

  1. A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.
  2. A fraction becomes when 1 is subtracted from the numerator and it becomes when 8 is added to its denominator. Find the fraction.
  3. Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
  4. Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
  5. The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Answer:

(i) Let x be the fixed charge of the food and y be the charge for food per day. According to the given information,

Subtracting equation (1) from equation (2), we obtain

H4ivigxmvroYKHfK7t yMCoPqKqg6Y xDCPuUakARsUNjhPGeonfY5 EQid2YbyO4uV1TBZTO6Cw5KSFpaQAo2A8Z V44zi

Substituting this value in equation (1), we obtain

HGfJRw24IQ82V27DLxDsEVEySZ6aG9f0AH1zvLZuRf2Aa4yHudPFQoFdccJcQwU9MgiyiOVQ 8SW9MObZ5vQUg CVZl6Q3El1zysA hQgskpKKgfr D2Eh6AGN5VhOr3LTQrhLw

Hence, fixed charge = Rs 400 And charge per day = Rs 30

(ii) Let the fraction be

According to the given information,

udvxyPD G6TTB8CPwelY Q 3 s3mg50VzaDIH J6Hrw5Z ZziiG gGjOkykU95 yPorCbLBF0KIbsMx3zONBx1qXyFAssrX2FS oA4BND8mdU4qlC7cnUPu8RwwMHywmHtNdYiU

Subtracting equation (1) from equation (2), we obtain

x5y4STXALPu0R0PWkbEKoa2v2HKAOq8C80KIGGjvTYy4BoXdnPAdDVnTtsyluneEqKbvMSg2eXUsE28z0cK4L3 FNfIFQG3fGAzUlkF9dK sWY1ZTPe30CZVZY6M26vKq8x0 0

Putting this value in equation (1), we obtain

r7AbIuNzOk3bAqDcE6CyCQmNSnjqHeRpWdCquwHRy RdsfUJqvvkA5KJUmtKXco4bXIBoOZwtDTVB9hgTOEDhznr6SDubsAbZU82kM5L F6TICZ2IWT30UlPOTd2Gh7dml8riSc

Hence, the fraction is

(iii) Let the number of right answers and wrong answers be x and y respectively.

According to the given information,kVQjr8EWHj0Pvh6jzCuevE4qeYhlp1 zgDseR9R 5U1v oPP9QokSoOzqKIxki4GWCAHMSD6qlmCvMyxglCQXkC I2VRVKLQ

Subtracting equation (2) from equation (1), we obtain

x = 15 (3)

Substituting this in equation (2), we obtainE46M7L WT9ITfRfG2QLOWCuusWApMYL8f02O2Dy07pa23ZBaacfolVA3 Z NwU6FQOk82r3S8RXkNm9kc7mbC4UsSB4oJlLlgoA46 N1KF0VdSQsSQpILwQPKhGBYnM ejLXAw

Therefore, number of right answers = 15 And number of wrong answers = 5

Total number of questions = 20

(iv) Let the speed of 1st car and 2nd car be u km/h and v km/h.

Respective speed of both cars while they are travelling in same direction = km/h

Respective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = km/h

According to the given information,

PobLZ1Ro4 xB2zhpw3 uPdBmv53wqYOR5otVlH8Y5opR V eJ69K9LvoThyFQbHO3S wTfVHFAMPOewpNuGpnEGg

Adding both the equations, we obtain

Substituting this value in equation (2), we obtain

v = 40 km/h

Hence, speed of one car = 60 km/h and speed of other car = 40 km/h

(v) Let length and breadth of rectangle be x unit and y unit respectively. Area = xy

According to the question,

JM6QIYa8dbjjlZ2kFr7sBzVVR rKg0siQ cdYvXDNGPwNOUo 6FWRUG9b6YXLR1PFhjdMAgwfirojZAPqdaVnhFnxzdzaAmA8AApGpwoKASio9CygZe2Ajw jWKnLMM qZ49Ozo

By cross-multiplication method, we obtain

hhm10hpex6RU MLUWH14GWuU5Vr07dSZysfjiWqcWjcDGf

 Exercise 3.6 

Q1 :

Solve the following pairs of equations by reducing them to a pair of linear equations:

uq3P3V AEpQs3COKnbblo9xvwu 1MThBkmQgYV2zURP0BchVDQ1BXIeZ6KU6ssDXxiA 2VVHVTAt7BDcJG5FkYp6nHSlm1yMBPHQ1NP Hsj8KXMz ud43T7QCQO2J8dD6 q7mNY
ll65wpxDMmzPHXaKB4tJIMhz5SuamVHpr3e 0AgD5 TfF1jnfpXb3NwPixgw EbavnuKRHM8PgYjd4z6 ftmSHF 7HgSAWBcZfZFw Fk3k yOEWgkZinUmTzAyLS5ekiKueTwAA

3WIe GBOWmvvvuahTMW0cICaTuL0HNyqCfaj jMTsI8suQar6j6pNUd92D4Y6gWJJswcYOrHZuFdTExPF 43 TQhT0qapKjjDYdnCZSMBajCqAonuZgqjtNKKBV 0 97uA zgnI
vFPjkE7u8zkoKbuxGGTCpSosr7yFxUWVQxr4ZUQgPOTkQB8DN4yhOxykBihBncBggXny 4JKPZtQAbRl5VOcuu1Di5yoC49 53DaDPGeaLOASC0pIS uImJCDTs4vSh9PzGTMEU

Answer:togHRNgVKEl8Jx2eeLO KSVfnf ux0g5YohixllkPyUq6ooDsNSKArKg9MPOmu K1ETBeF0p1Es8dOmraMB0r1gOD7

Let    and then the equations change as follows.

OaXiAusIunFmvGWpQXtVxyI5e6fMjFxnlljVVgUbDYX dFEbtOlOzYdUq5fHPuVOvxcJJeX4b6oSYFhvVRS Cy IkPbMHmQpQc7z3F3vH FYjFZJbL41uWZnALTFwg fNkQK oE

Using cross-multiplication method, we obtainz

MZm5OsxHU1ELHig3rO mNZvP4ENpO2NyChYsAAjCivt8354FxxvpujZNeQ4kIC3vFH6h051 WxR0Wawm0RW j7mizIscPL80OuajDNnzHAjPJueBYo4qjaxXiFAPJGX WT Ll7M

Putting and in the given equations, we obtain
4ESzjVcJTtGEU2lwWktQj37Ba6jqf2xPXv0Ei8EDpas1UJfJmfzKGaumuslNcThaOrZ6Ex7jwyrATGi xIHF

Multiplying equation (1) by 3, we obtain     …..(3)

Adding equation (2) and (3), we obtain7XPDTHtOgiIuPXLKKIQFts4g3oG3fm3BiKOQU kZuM2HYaMeEAB5FU7pAMUn0Dhih

Putting in equation (1), we obtain

rm4 rvyEen9NU kreXNmdQEmCnO9mhDSx9Z9nygpCUjx7bvRu9n hJ5I4Ma6ny XOfzBTr9Vyt1zH8kWsqHIFSVnQ8MOpvk34gbSd1jZ9SRpBozDz eeHbVro3fjFKbNzS ASPQ
OvtfX3sTyAcn2AQLvQxiQwf8vOUwLGzCTbbBsWO3F53w 72OTrKJt coB1nXEXMvSi5aGA02EVs2TYXj0kERULIutmb2BOeb1uPhNvGt6birCKcui5P5zqpzkka2K4soc25fjcE

Hence, GjulQ9 HjKiHZdiN0KXa20LcanwCg385

uq3P3V AEpQs3COKnbblo9xvwu 1MThBkmQgYV2zURP0BchVDQ1BXIeZ6KU6ssDXxiA 2VVHVTAt7BDcJG5FkYp6nHSlm1yMBPHQ1NP Hsj8KXMz ud43T7QCQO2J8dD6 q7mNY

R89f62LrPgSrzlMc7cnqmcRL3PFxjX44ghquY38Q0X2Dgo2kkLRVMG6Ytw2S3DUdGcfnjOGQISCHzOsy SH0g8xrqm B6MCjquk76OPP0kbSYH5foFKJD5uEbgt l6NsNU1izHI

Substituting in the given equations, we obtain

By cross-multiplication, we obtain

diXkYhfrfhd0Tp8imcXJZ2amZIGlXEZGtmioJKAAe1W4sTncm7UeqrjnUQ4I779a3exblDkC4vot8AJLfDMDiVsurmfvaLLclPqdlvmFVaYYK3CAzG00k DKAXP uAGf0UdemM
ll65wpxDMmzPHXaKB4tJIMhz5SuamVHpr3e 0AgD5 TfF1jnfpXb3NwPixgw EbavnuKRHM8PgYjd4z6 ftmSHF 7HgSAWBcZfZFw Fk3k yOEWgkZinUmTzAyLS5ekiKueTwAA

Putting  and  in the given equation, we obtain

Multiplying equation (1) by 3, we obtain

*Q2 :

Formulate the following problems as a pair of equations, and hence find their solutions:

  1. Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
  2. 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
  3. Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Answer :

(i) Let the speed of Ritu in still water and the speed of stream be x km/h and y km/h respectively.

Speed of Ritu while rowing

Upstream km/h Downstream km/h According to question,

Adding equation (1) and (2), we obtain

H9bVlFTY3UavfvInI5uAH RCDxjs kdEKHce6KTHUfoxxMb6IsqkW OemB3utqOm7 enmYFvqJSYjzz7zWcSBy2TBbH3ya2oaQpIF 5Pw5lE 5MLV6ogqhVVjLIrFRIIbByuT g

Putting this in equation (1), we obtain

y = 4

Hence, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h. 

(ii) Let the number of days taken by a woman and a man be x and y respectively.

Therefore, work done by a woman in 1 day

Work done by a man in 1 day According to the question,

SAM Vlzg4zuL4u715nmkbj 5Moq2wnDzOV071VvXnBfkLMmVUVI3mEkK2Jbefdu5PjkCSB HEjEHPTMVyRsK8OvmXJFMaa6ANToaZ5os9PMf06gJNrqVsqEKnaSKHU scU0ZHK0

Putting in these equations, we obtain
WhqZPZBbSMxhYpaol9qt89Tj2EcTfvHKO083zvpT8r3EkwN2DL SjqmeKVEBMaeeIyCf9eLG5DS artZ6sMfjsFUVJGh8mjTM0VHDqe4yuJ6LoOzkMk4NqfTyfTtowKOAkd1vo4

By cross-multiplication, we obtain

elRZUg99BMz0ugV7BlKu4AySXpftw49qo3luzv
lbPZuV0DwlmaZC6o5Pa7Ba6B359ZMmVVMO1rrytRsvmSJQF9yDS6wgo6h 8NpxAGagJnIZUdZMvJECNJumVR2EmD2 E8E0wPscibnsidqnbY XsfBrV4earfWL9M7qM2OwZjrQ0

Hence, number of days taken by a woman = 18 Number of days taken by a man = 36

(iii) Let the speed of train and bus be u km/h and v km/h respectively. According to the given information,

q8pZkUUJMSVEdDgCB37pLKHZE3YpG9UAz3AUBfs5QHvIe4QuxPTyGMBWMdV

Putting and in these equations, we obtain

LVcdaf3gfJJigNuLpEAGjV1 QXTC1LuBf1fegbZuT3yMWhBNgXM7neUBBl

Multiplying equation (3) by 10, we obtain

Subtracting equation (4) from (5), we obtain

L68qG2kPm1tVT9961YzOo0KB ino2RdwReuYojnuoX89rHa6HJlgrx4qK0xqs1 jpILeWivKvlAyIM3RyBwtaCxvbUZug0IZReZsPXjgmzZmec0 Rb fTKdkU0gG1FC8VUFKmjc

Substituting in equation (3), we obtain

AN0l 88rOkzc y0FWK54nCG7ghULLA0qO7hSUAT2J7XnScnGioIeR4vQIoeKS INhm95QCv6ZcAzOG9ho3f1bXDYnAf0NcK8RrkYlvLcam7WSImlPH5UuS0LFKuQIaMnT Rtgow

Hence, speed of train = 60 km/h Speed of bus = 80 km/h

Exercise 3.7 

Q1 :

The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differs by 30 years. Find the ages of Ani and Biju.

Answer :

The difference between the ages of Biju and Ani is 3 years. Either Biju is 3 years older than Ani or Ani is 3 years older than Biju. However, it is obvious that in both cases, Ani’s father’s age will be 30 years more than that of Cathy’s age.

Let the age of Ani and Biju be x and y years respectively. Therefore, age of Ani’s father, Dharam years

And age of Biju’s sister Cathy years

By using the information given in the question, Case (I) When Ani is older than Biju by 3 years, x – y = 3 (i)
S 7jm2p1x0XCN b8CfrtEtu0HSUxSa72rkd UJQxotp21Urf5tGMXQcSBlXc5eayA55VB5b9

4x – y = 60 (ii)

Subtracting (i) from (ii), we obtain 3x = 60 – 3 = 57
c47rnuRhALh1Elacf HYiCG560VBDvunYAmknAVoq3XMkWXHd5zxE

Therefore, age of Ani = 19 years And age of Biju = 19 – 3 = 16 years

Case (II) When Biju is older than Ani,

y – x = 3 (i)S 7jm2p1x0XCN b8CfrtEtu0HSUxSa72rkd UJQxotp21Urf5tGMXQcSBlXc5eayA55VB5b9

4x – y = 60 (ii)

Adding (i) and (ii), we obtain 3x = 63

x = 21

Therefore, age of Ani = 21 years And age of Biju = 21 + 3 = 24 years

Q2 :

One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II)

[Hint: x + 100 = 2 (y – 100), y + 10 = 6(x – 10)]

Answer :

Let those friends were having Rs x and y with them. Using the information given in the question, we obtain x + 100 = 2(y – 100)

x + 100 = 2y – 200

x – 2y = -300 (i)

And, 6(x – 10) = (y + 10) 6x – 60 = y + 10

6x – y = 70 (ii)

Multiplying equation (ii) by 2, we obtain 12x – 2y = 140 (iii)

Subtracting equation (i) from equation (iii), we obtain 11x = 140 + 300

11x = 440

x = 40

Using this in equation (i), we obtain 40 – 2y = -300

40 + 300 = 2y

2y = 340

y = 170

Therefore, those friends had Rs 40 and Rs 170 with them respectively.

Q3 :

A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Answer :

Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel was d km. We know that,FTF4w0CWJz7dmFVsHLlhBMGFcL7nOei6FCkCdjCpumd38JaSsps0wvdHGHxsaGs

TRf 9m1l s3RRG Ooubdy5Dq8b73wYWXm Q7mlg

Or, (i)

Using the information given in the question, we obtain

bXfwHnqj onfaw7PAOYl4ark97TIBGPyf N9OpeivVV21ljmxnxGrYdzZraagQY

By using equation (i), we obtain

(ii)PEtHF7gDLJdL91yQiPQLdXo6GHq A3x6n9mg85TUyVHw66e6Vk5ZmuNUueB9CpLjLZY0OranRbo9SWh62QCXU6UjCB0

By using equation (i), we obtain 3x – 10t = 30 (iii)

Adding equations (ii) and (iii), we obtain

x = 50

Using equation (ii), we obtain

zTrw5W 13tpRzFMaSgeES88zNn 0lRFA6JF7s YOZRzmHJQQ xmnxJwW7DKHmcLm8ruxfZYnO29vVp2ro lqVJQyC7cWe9 smzwE62BerzWdwJbdxRRvmCyFIjgBM6LA oIdtrI
XIPCOaaV4gU0nbnq4Txbx204M6wGL4gyQoeGrKLsdQlkfq4PL7jNB5vsXhSeobFxb9sHFVCh Oh1JNJcyVpADS8iP3BS4yGPRL9P0ok PdUUIv4qsjr6HfUxw97CY asGJfFS0Q

t = 12 hours

From equation (i), we obtain Distance to travel

ICkmfyOruVpYuW9gcn29ucqFChCp BqHcz L6frnTOXgjdbRV7AMMjCOZwI 0Macjm45S6nTWsBwV3 BlyMkFBnMuohGS9Hckv7TZtHVWY7UowoRIGs1uEQXecKHwkOZDeigBdY
Ey6JPcDGRGmc7IS3JEFFR0u0zqjIOO uaY6OYFewMJ3 zmFk3mlwhldesiwhyJgTqegYIU1dvY8TtMvyXe79aOHhUnw KHuED4czT7E 1vZPp kvCJ0YlvUmArnAEIndejYJZPY

Hence, the distance covered by the train is 600 km.

Q4 :

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Answer :

Let the number of rows be x and number of students in a row be y. Total students of the class

= Number of rows x Number of students in a row

= xy

Using the information given in the question,

Condition 1

Total number of students

LmPXivl5PZYuyzlp6CvR58zxzTFfDpmy7gWitvq pOscUZGOhHcBEqeVkrV14

(i)

Condition 2

Total number of students

(ii)

Subtracting equation (ii) from (i),

9HFh3Kr9IFgYj1RF0TaJEHZjMKku9etvthFztp6ne9yIlE4cDWtgPqpcrwxS uIIj4UsNDZhRuePUbLflRNUuHrmjWUvtOrkbJK6P7eCcgrty01yN1
fwnWTdL4rdJiiadgzKxAweqleCQcEiPksKqlkAp U Q kkK1Q3Ht2m265LZ45qcz2eN4e5Z1lQjnnUW 0XV8YiplmHePaNyOBpIzJM8zidPRdhYl4KDI1iINMhBU164 unbymHQ

By using equation (i), we obtain

1gGnFyZUssm YvLLQuUtRVp acpthEo0TtlfMvot5oaY4LjrskT2ofVqC9orSU25azjwsPSBzmypzeFlCmsoSe0Mpnf429E3PSuKCpPRZoJ7j z2zQkvpac8l4q A1d9yXHnv8g
MYFaL TgsD6hTI5tOkR DhXIb7cVVwgAzFRrBPHwBVuz6McKOloQN5Bbh2Mrg1fITyrYErMznUVc52bvGHqWwBOOP4qD0y5hrB3UbpA7Nr5onEJB1hiSkesKfD7LOL fIj 7wc

Number of rows

Number of students in a row

Number of total students in a class

Q5 :

In a ΔABC, Find the three angles.

Answer :

Given that,

Lxj74WFGmAMcxqlq1AIW1lU6dGHr9otE5kCMcGmxqQ09Z0c0AIwbyQnh
DiHedsmHGut0oelmxIJyccNbNIjOsTEvw4veWLy6lLQ8SMWEPqf6XjdMpPbS8apv56I66q1 joCt8uN3DBwT pGciOlJaroPhuL9 1temaCAwfKk2b6g9Umg61OCW 2ECTlOV2I

v9eVG5CSWJtBpuxPPgNCPc5Hlx KBymTg90W hNhsB GM5x2hohaxU6PLNI 3ulmHW07lzwbSeRfRtMVmOPo … (i)

We know that the sum of the measures of all angles of a triangle is 180°. Therefore,

vj3hTQSIyCxxoTyJQfduzy57Xt0iZy1MP9LC vxaWs2LXQxt5aW0QkWANZu1jPNJjyIAIvCz4iBh7EIgiMObu3wvkuWiDL867Y5

∠ A + ∠ B + 3 ∠ B = 180°

∠ A + 4 ∠ B = 180° … (ii)

Multiplying equation (i) by 4, we obtain 8 ∠ A – 4 ∠ B = 0 … (iii)

Adding equations (ii) and (iii), we obtain 9 ∠ A = 180°

∠ A = 20°

From equation (ii), we obtain 20° + 4 ∠ B = 180°

4 ∠ B = 160°

∠ B = 40°

∠ C = 3 ∠ B

= 3 x 40° = 120°

Therefore, ∠ A, ∠ B, ∠ C are 20°, 40°, and 120° respectively.

Q6 :

Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.

Answer:

Or,

The solution table will be as follows.

x012
y– 505
eJbKSV82tYdO0RrYsJIZlsAN9iZ

Or,

The solution table will be as follows.

x012
y– 303

The graphical representation of these lines will be as follows.

J6NRa9avMdjB6zyvqpbvRdnDLYeLa z 1m1eLzT9fRBt1CHAOzHcpOZaHxVnH2lbZvXr2m6hqAufFIxPyyxWG1KVMu h171p2UAFeV aRqPRu1o00dLFDiygIR 9C1sg0NY2JrE

It can be observed that the required triangle is ΔABC formed by these lines and y-axis. The coordinates of vertices are A (1, 0), B (0, – 3), C (0, – 5).

Q7:

Solve the following pair of linear equations.

(i) px + qy = p – q, qx – py = p + q

(ii) ax + by = c, bx + ay = 1 + c

(iii)

(iv)

cF9oHVXYv4bEVCB0NDFZhcjbnpexcqS1P
c68sG8isn41SjELNl18xINxIZaUAEh56v t IHGO NQQj1NgOWSTfD9JeOIsdnNY0W18cJ6mOIJq06JzV0JJhFepA9NxLziBCQf5k2ZFKj3vVBOxnY BRE OJpbE8qnYUsDPy7o

(v)

O9laI1UVeb9hCKqhg3Bp MPBMh6qvS0hyUAPZvhpBXyRPvX6B99OuMv0tJznCExCsZ4K2rDSTjuAWuKbL5p CAee2NRTHGAy22r AHCCBxfIDQ0yZvOSX fRB

Answer :

(i) … (1)

… (2)

Multiplying equation (1) by p and equation (2) by q, we obtain

… (3)

… (4)

Adding equations (3) and (4), we obtain

l519v3wHJEPIzGaTQmJZlw9N2ICXQtTwEmuQbpiDjhSokiK PeWHRc9nWrLZnMFSwDURYByluceZI7eG41w ygp9wVBLxtKOecY u5EPgXhQP G0nxG3bIC PY
Qb78h1HnqBtLWVsyhizaAzmEapQn JvuYEGdW

From equation (1), we obtain

2bp2Ozx1IZ lHblFX9 dpMYwCemBDc7hPtIUgE6YAzSvSByvRzSMGpuPZ90b7oGSwhkDUJ0NvwJQ7eZpLjTRx2A5kio9 E R5S1kQku9VnLYh YNOliAg tMGnADlAK0EwjzI74

(ii) … (1)

… (2)

Multiplying equation (1) by a and equation (2) by b, we obtain

… (3)

… (4)

Subtracting equation (4) from equation (3), (a2 – b2) x = ac – bc – b
ESHl8pil X OfPMnpYDuk sz xtBpmy34rMGv77MpcllOiNfuc7yLhhJnz vsqvv0YJjBaEszaS2lWSjDmW8WMH4Ia4UYm0KqtKpOtMHESXcoOeLuFlOrAqQVfLqswI4jPqjVNk

From equation (1), we obtain

qHcvw1aG4EyXltF1ufFlDbBkzo3zxdkjyE 3XopjyuEiwJu58q9g HL6puzyWtWraV 2dLtGh1AiSKYoKp01v2f6akTgloOrYKjUOcE247lmN CaN8ocIuw1w

1ycSniYiqEjmGDePNTL0Q2FjxlS1iWz32RhWRjAGGlZOYC2LZT2ULzhe ci6ZMRDZiae9CcecOMt2NNsQcZizyO7MqUqfm2QFZ0GSOClUp4mjkSHeap ZJ

(iii)

Or, … (1)

… (2)

Multiplying equation (1) and (2) by b and a respectively, we obtain

… (3)

… (4)

Adding equations (3) and (4), we obtain

51X 4vQstAoYdLFCUBfZE2E0wg7ju0TB fYKI3YNzRkxAlFzXCSTa YiKesSiNRgBSRhhB1xYQKht9LvKw6lnwluGj0DvaC143NG lp5a2cYNp2Te 8OGYMg2gZttRLpBSwJbOQ
bgmwoTtwYppSXJHBPu3YsYlcSETz3hjgK crwAoHzasCgR2U877jKKJN4qu6CzY 6gfMXpMPTXzRFtGBQB 1CWgfVOKcQ 8JBJW4 fg tnUj EGYMzQnB5fZaMeM4l64oaGc7I

By using (1), we obtain

lBDT3mA h2zYpGIjr5YkOGN HWT1VMfr646G0jFV81FJaDZTF iJV6alNsXLZQLT7mkhrlGCMAn8HZSrs6k87CBrX2gAilPpkqD1gSC9yuoI4swdPytROvjhwNLr gHBDD6ye E
4Aqs5E75b6m3wx09vhi16paw1M2y8LGkoanw5r2Q7Szhw4xwtbxgbPA6MP0oEFigs2SmG00qjBI8XMaS6Hd33VTu0av 7EXhyfMQnVachT6dtV xG0jbgz7QgtlN w15OUNtOP0

(iv)

… (2) Subtracting equation (2) from (1), we obtain

0kuy8rzthvFGn34snZ0bsq01sVIjKbn9hYDicHVQaGZ0M4Mf KD8i4bn2wFVvMg6KVuK3LCOXYTDhppAbH5QZi2qDWuuf9OFD9mDIXDQJygSFp4OfzQhllsquhPYeq5dmDNelpE

Using equation (1), we obtain

arC41Pjq0xaZw5526wdxmdO MJVaB7k3dTK3HUtUlAi WZIrhkr2MH3PYr1dU3bk16Wbh8PEMPd 7t OEJNMWqvxJDKmZB6V4oA4qQEYAeJLSC mADfitLAL5EUWsvOvbVpGgs
7lMw gEIegOqIRxUZ0vazps3Km7yQnhX

(v)

Q8 :

ABCD is a cyclic quadrilateral finds the angles of the cyclic quadrilateral.

Answer :

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°. Therefore, ∠ A + ∠ C = 180

4y + 20 – 4x = 180

-4x + 4y = 160

x – y = – 40 (i)

Also, ∠ B + ∠ D = 180 3y – 5 – 7x + 5 = 180

– 7x + 3y = 180 (ii)

Multiplying equation (i) by 3, we obtain 3x – 3y = – 120 (iii)

Adding equations (ii) and (iii), we obtain

– 7x + 3x = 180 – 120

– 4x = 60

x = -15

By using equation (i), we obtain

x – y = – 40

-15 – y = – 40

y = -15 + 40 = 25

∠ A = 4y + 20 = 4(25) + 20 = 120°

∠ B = 3y – 5 = 3(25) – 5 = 70°

∠ C = – 4x = – 4(- 15) = 60°

∠ D = –

Unit 3

Pair of Linear Equations in Two Variables 

 Exercise 3.1 

Q1 :

Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Answer :

Let the present age of Aftab be x. And, present age of his daughter = y Seven years ago,

Age of Aftab

Age of his daughter

According to the question,

Three years hence, Age of Aftab = x + 3

Age of his daughter = y + 3

According to the question,6BYXW8RMbZO4DaxSXerGJ5Ba aMHyuGDQuqMObEjB UpywmazQJCn2QYLXx 3o3v62k1ABd7m

Therefore, the algebraic representation is

fHFI3UBVJljYvJoocIHWKPqpW mFAk9iALUYV2Lj2Q2plGVHNntWg21KWGP0eQEACbo6PNCjOsvcYY

For     (3)SGIrsbSeZxj4bcaOlMc0cHqQotAmHUK13HJjKnp4dCharpKuaVZeM7AzeN9IU70DXq vk oYpovjcZ3Gr9WKZ7wGv sEIW73UWTj YswjCO49mwaSWH3tE

The solution table is

x– 707
y567

For (4)

CUjR2BcijrLAbLLWJ N3k8wIKlbOMUVFoXaysce DxV30JXAUlYNpYYKDkHTilW4FxjeTjbj5vr8Dceu9ZTSTZHvKL46pj24uI0U0gL8HUJht8ycMIHKT wcYQ

The solution table is

x630
y0– 1– 2

The graphical representation is as follows.HQpccv70SOtPQCzMZpHRlGSnOdf 6WjlBFjV q0Sbv 9c92sdx28PPSJPiILBDhHuM7YLhmBnE mAFPhd FGTg5vMywEhgISpD2d U3Mm0K6EzaqbKIZPis7NiGe6yLJEZhdBk

……(i)

……(ii)

⇒ Subtract i – ii 

c2w6PvJboZykfNQAqSJ 3InybzqVkegHtJMpTP3BqZWi9TprHGugipOpWyvi 0mYKr jMmIUOrr t1IQHs7qYBSHL4VFwaOKoLtA7lvMWfYE WBH6AZO1Kijkzl096lynt5bbMc
cpoDdI8TLpq1VHe1XFlWA4oH8By51F3K60RK7vZ80RdPVnB4B5OTXNyMManqcMO7Z23ZNY9VnJLnPKq4JAQY57qAuWYy
d5i29VXW7SHOgnJ9yLar ao5N3nBL2bVYaJ3Z RmR6Xy oT8SFDPY7R LUb imtjqyDANFHPwtUUMQNkWRn6D5isenGjS9QJcueZVHuhOl9MycWpju6qHsRI0V0Lkv9FH4vD58

Put in

r7aqiuQRok3lLwFAMUbSW VR oqpmScb24Etl4Fy qnBsmQO7EWIYewcGpMp0HY2W9qip4aFzgg6A fmFEVX2OlaMtMtOM 3OiNLKtF48kn6CXXygCtEq1CiWJR Vr156RDnDbA

So

Q2 :

The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.

Answer :

Let the cost of a bat be Rs x. And, cost of a ball = Rs y

According to the question, the algebraic representation is [Math Processing Error]

For kdWTuO42 9tOSAWoAUw46U1fbjQovHl8W2qGzrEGYR3R9gg7Lp33tnNK AFCR4qhOB

The solution table is

x300100-100
y500600700

For x+ 3y = 1300, x = 1300 – 2y

The solution table is

x4007001000
y300200100

The graphical representation for first line is as follows.

gv1N0TrYA19l909 6z17So TUQwFTTR v6MDzMjExvhz3I1 cJf gXt0pYnonYB4dKMSh5Qaxo8frkDstb5H0Mxdwmdx1QKbtWQM48bfEt yGtYA bR 7vtWN9p5u8oVc qfcFg

And graph for second line will be,

JIwdqFNl6 4ojCpjtc58Nn9869GNlOXk1E1CEFhT faVUHLzJ3itS fIAfuKQerF3NTl tEMSvBPTqkrlxlKUc5 RoC npA2vIBcEfXL2kTb9GpvBIGu8LTctqOh6iAVppXSs9s

Q3 :

The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

Answer :

Let the cost of 1 kg of apples be Rs x. And, cost of 1 kg of grapes = Rs y

According to the question, the algebraic representation isP45bID0 cMdC 2IHajN3Xg5H7Y9lZwFehOKWdFqV6nrSmGQ9Ld1 GdOjXOg wFLssOTcvf3RfeOH7 qY2Va3Mpj3DzUZMzZFtsMFfjjZa pvOsPJ093If2 UwR8n J8RBoyu9j4

For  V6Wyrf7jez0j32KHz 2slcns0Denm38iy4W6qmSeeBmHzW8B2 MmN9YsmARsk04vAHl7egT66aJgQuqLhZGnmkL3BxOThp4LxwzAEt,

bWAKBIQp6Qm 11BZadK9Gyox0I

The solution table is

x506070
y604020

For 4x + 2y = 300,

The solution table isvdEFO3xINHHaMvamqihYwootz0jbiGtCfHAHrG bSPD8or0eDqSy GwFYMjK dvodeSR5AfyFMH9DZJD NjY2ZdJtNW6NNfRRCyu4x8vaH3VUHBHYw G3sL6V6CKF8HbdqArt9U

x708075
y10– 100

The graphical representation is as follows.
ZHMqWayv0IuzcmPVYtDE3DGXBnSsIBxPoAnSWPd1G4KpwQSzawxd NWDxzjPgbQtoCOVO9X tA qiJYQG4SLxuhKnsEUee Kp1c

Solving the equation algebraically

Eq-1:

Eq-2:

….(i)

….(ii)

Subtract (i)-(ii)

0 = 10 (Not possible)

Hence, no solution

Exercise 3.2

Q1 :

Form the pair of linear equations in the following problems, and find their solutions graphically.

  1. 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
  2. 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

Answer :

(i) Let the number of girls be x and the number of boys be y.

According to the question, the algebraic representation is

x + y = 10

x – y = 4

For x + y = 10,

x = 10 – y

x546
y564

For x – y = 4,

x = 4 + y

x543
y10– 1

Hence, the graphic representation is as follows.NHWFZK7WsiLlBr TR9cgrwZz0QeWnpHWz8s3o aAesmcNuOap OFawkaJcRIdFHedqxgTy2yQeA4 cfOr XFVf5Tn1HcQHRX4PfItlJnIbf5lLQUtodws3uN2AVpBPM694gZqqo

From the figure, it can be observed that these lines intersect each other at point (7, 3). Therefore, the number of girls and boys in the class are 7 and 3 respectively.

(ii) Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y. According to the question, the algebraic representation is

5x + 7y = 50

7x + 5y = 46

For 5x + 7y = 50,wzNQDS5lSW086fg7oI5sl4h7BNoWMxuL75FuRgcXyN7GA6M2IypV5WJQt11s7vwcFDHJsySrh89EaUgX gi7rAG3z PC 36mhZNT8JCMJpdEaP2pOwLYC4GBRaISRS COJt Vcg

x310– 4
y5010

7x + 5y = 46mAxER85caU059OrhExHzcsF8W8BD64JNEad9Ma3g9gARTLqZ2d6SxZDksXSXOg95lc ThZ

x83– 2
y– 2512

Hence, the graphic representation is as follows.BmwIMAj3M6nk9uOmxqY0EdG4mSESQVnZou3rK2mEQwpEL2G9keBatOwr

From the figure, it can be observed that these lines intersect each other at point (3, 5). Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.

Q2 :e LfEj6YaRqESgJp3t4GbyxOJPF0gxIpxAC3Dsm7M3LQR9spoL2KujaFg iUx f 13H cvpe0khuDlq0rkqZeweCPjSNvRETaIFH8S9GrG9U3pp1hqwcp3j3 nA WnWwMcUVHfIIVlIjoXdsdqNVyivX58yzL 7Km RKADuv3uBzJwhzeX3dDZ65iHY21S7W HhSXIfWjt7bJOc67ieg2tp7MwvJ3uXErRsAe1itKzKTuMfry53AnHUFshcqmDqpdCeTURlRMztmJed7oqI7 E7j9g pq9NZmRnxjoEQVzOOy6OiUijntxuzahZq gXPl3Gw5 5X0S8E3nLXk9XEXAkEMd4zje2wT mLtSY

On comparing the ratios , find out whether the lines representing the following pairs of linear equations at a point, are parallel or coincident:

Answer :

(i) 5x – 4y + 8 = 0 7x + 6y – 9 = 0

Comparing these equations with

and we obtain

wAwKabI TpdKQC5DAahWZ4HaQGluwULDc1BBGfccwrEaQb5X5df6 loKLCiCXmTAzeS1rzJsGpdXwD75Vlvy9 oXMfB1IIhMlwB9 I4alNChGRFGiQA8UYQn8oOyeoJ8Lb4pIqM

Since 

Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point.

(ii) 9x + 3y + 12 = 0 18x + 6y + 24 = 0

Comparing these equations with

and  we obtain

Since  

Hence, the lines representing the given pair of equations are coincident and there are infinite possible solutions for the given pair of equations.

  1. zJBNE1XnI7 8rARKwbMO2QkXte iFuQQ0J1l SRIM80A OWF1bwgHQdM3uLO6W5fVqx5y6eUyCHyUn8jPdZu 4ZjB HsWmdfxaODbF5UxCB4VrH wz17lp1prZWUd8rD4odIwU
sq41LI9C7KSl27TN7wyHXl 9VH36b6CGStl Cb CV1RKQBvurFMKWhlx igRsc5yZAcNj5J3vD oxxsnfEyY iLyWxh0Myy3DF6DxNtErHyw0M46nY xf6J auIaa68QJ6gn89E

Comparing these equations with

and  we obtainUDwVP4xcXVixPAu4ck91Bqjptohdjvcd6y BawLCNyZwsY1NXQ2NlGTZwdNij4iCg1Utlene19R021mII72J5I7wggxdMdJuy54yh3NCL4ovSfTI7u7lpNNfK27G IwyCp7kpFY

GCenSvZEnGf0LYb JcNJC3eIcyBlyE qO80J47 8Y04VlkuO 9CnHxWWXm0yE6Qa7wZXX UDvkZW8dEoQgKnUolFGCTe5HKUC1f5 mt5VNSVNollYjB4AjXqc9pPtUnlrIsQ8FM

Since 

Hence, the lines representing the given pair of equations are parallel to each other and hence, these lines will never intersect each other at any point or there is no possible solution for the given pair of equations.

*Q3 :

On comparing the ratios find out whether the following pair of linear equations are consistent, or inconsistent.yYOjWTb5CO SJdnX5 dLZ2N1kJgcX USr4DP806EbZ6gqdH qZQqZnSWXkT0X46S5CZOZ7fla7vhkV42reCZPbWgTjlLd6bDGia q8YOqmNmco1m 8GoJ13DDe85TW3mWVll 4I

Answer :

(i)

xbaXpPWvk4Im8Dld3Aix1OPXfZdEAzv4VfcAmz6mrORWpwKtDjkX5wCxDZtLk0h0 YW3C7yOVqjjX428a1v8XIil4miDeYnr5lPM3mnN88957

DQr50NKou2CJ1ZmAguYhKUDsT7FxJIQ3xOefzAQKyZy9Av16t6my7lM5sur9HTHLXVdZeo7ZlX8aSskk0FDuySB2tQDHEcSYUVveJY1LYa0yKGsJ0q14cQxTVz UfqE9isFp0bQ

These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(ii)

DUpY8tEESrg G5R vdwQzQxM9aN04wC2cLAU65 nWqEwHauaEoEQXruh9DOk5f1a1u2H8NiXb4dxySoSaLLbRP ygdoGcZ3PV8sbjgZ DmNXttYfdLBvA1pRwzD0TbiFlcqOQuA
0vRC4XCCk3qa entFGrolfDo4yZmgaOmIpkwLEWuKctv1DEn7VMOWoLprJ REWymVuLbA77hCsJJhFI0WnmzVucWlQTQnc4UuvmR7o

Since  41NDCxJMHan5ccD4tlCFz1EyWMbePPRduZx9tboJ CM z9mnmgsotMqY73RR8AoMdRBuOow Hs6jw6o0mF bO6 1ITo6tvTAh0cZZTE9nbPOfTxtZ2wmI9 arwE70liOuRD2lt4,

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii) wHRcEnrLzK2gOB1UqVhtyBVmUcfiz7hBiCctodk8BnaPvfhLL3wQqRGBzF6fD3es8n81vxxmBVHMmvWw478KAQizbYu8Rgv58cfn vsO6v1XeaIAtSf Asvy6kEmVDw9 RT qhM2I9Z2ntM

Since  0gxvo4yvdv6vyYKsfqyKlnqDLhwTKrdFLp5OTOp3eqZp7nWG4Xu41I5pqdxqWJYFVPnYoKYaB10 g3o88dOB0XhneizrvuiFKhUgag5cB4 jKUtAuqAiz3W0H cWfa9f5r1NKYQ,

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(iv)5x – 3 y = 11

– 10x + 6y = – 22JoPtzTPYIjjt30YxVtiGfwOdO1 cUcBIrFdAj wFMJ RRddUT09611HGLL kw0aYNbZWVAAGcdL9y8v47bV3VTxR9YO wXW9Bz lEz4izpEqu51j0ood rRqDBzH5YsPp4rO44s

Since  Pg7OczMCtcT6EE2U1QQSkJS5sRjufwhGX7Y9tqgVMbH0ouMskESBBgLX47HFymNwipdujX6MMwj6tQlwuirCD8hnl5COEXnstLhTR i2roNN6i2 rOyTa9Q9ooC4f1UFz24di1o,

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

(v) gL81pbcUL 1FdrVb4gT3kiA4ALJzYprKgBsJ1BGl9zIqHwpXGGrL6YndOvwIZVkOyTCqurgA absHBeNykXhXkbK768gbCxjDJJ SBWMKxE86NTVOYmBV5YlTIOmxf RI1PwGAF9aWuoOCLPFxwUJU7wBF2jQ73lX9pBPxny bsodXUAdsB75iUuKp HGaXy3ahjqC37fiBbD

Since c8ZGy43XZE4rjK0kJEQ5l0EgaeXTXbgC ihQv K7dzWCUu9g9VtqFwNQtWeBc7ahnCNFnT iQRAl8AdvY3FUXzB8 WCweJACRTFxoUucZYB

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

Q4 :

Which of the following pairs of linear equations are consistent/ inconsistent? If consistent, obtain the solution graphically:3fJxjTsYnvMCAOuLXvuAUbfJuxJDPnZ81FNw3xfuabfmQYesgVT0rxK0SZVKM49OVzQQ5uglD7txdFbZ9rVgFlX3LbFY7wuD Uz6uWG k86xN7r3 U nnoByvSGeXhHzTrgwR I

Answer :

(i)x + y = 5 2x + 2y = 10JBLqcZZ5kAaqZ1s7qUS1TH9TvgZzdlveyZhUJbV8TKYxUvhH4 7ZCLBQEOGgN2508YK3TFP ikN6zDIUdxlpvXeUMVrZzWEzAE3Yn15 Gkr

Since  Pg7OczMCtcT6EE2U1QQSkJS5sRjufwhGX7Y9tqgVMbH0ouMskESBBgLX47HFymNwipdujX6MMwj6tQlwuirCD8hnl5COEXnstLhTR i2roNN6i2 rOyTa9Q9ooC4f1UFz24di1o,

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

x + y = 5

x = 5 – y

x432
y123

And, 2x + 2y = 10wIW

x432
y123

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite solutions are possible for the given pair of equations.

(ii)

rzuZ6Cr41po1ZhMFh0Nubuy0Uh CMkx1t9G1w4TfrILOzwA6kq8aJeKCzuRNfK2eTTlyny4E8OWakWdpfrGitAqEZWZOwOzq2u r2ykKM6RDzXKN4hK7ukSrzvNgE3ZvjOGJXg
HNF OIZB1n5LwD7CfHcILmXt6omVY0a22OmPcpIxiETnuV6w MHxSktNcYuAE1nJwAy0bnbDwh c DJq6r17JEQb2qH0ceIe M KmeJn5TYbA2d7eFWsHoQLROe6Ssp5m 41Ph8

Since  41NDCxJMHan5ccD4tlCFz1EyWMbePPRduZx9tboJ CM z9mnmgsotMqY73RR8AoMdRBuOow Hs6jw6o0mF bO6 1ITo6tvTAh0cZZTE9nbPOfTxtZ2wmI9 arwE70liOuRD2lt4,

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii) 2x + y – 6 = 0 4x – 2y – 4 = 0cfTMJGqLYdjAeB3B9AhG amHL7bOYJP9LyiYIV2 oTupHdJOa6qwnOWcB7oDySwvOsI 64fMvOg25 w6Zc6LX X6pecjWgR7IzQD7HcPpFwlJQjNsh3vvRPcWni2tVQaKhbM58

Since

 0gxvo4yvdv6vyYKsfqyKlnqDLhwTKrdFLp5OTOp3eqZp7nWG4Xu41I5pqdxqWJYFVPnYoKYaB10 g3o88dOB0XhneizrvuiFKhUgag5cB4 jKUtAuqAiz3W0H cWfa9f5r1NKYQ,

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

2x + y – 6 = 0

y = 6 – 2x

x012
y642

And 4x – 2y – 4 = 0

x012
y-202
C:\Users\varun\Desktop\JPG.JPG

Q5 :

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Answer :

Let the width of the garden be x and length be y. According to the question,

y – x = 4 (1)

y + x = 36 (2)

y – x = 4

y = x + 4

x0812
y41216

y + x = 36

x03616
y36020

Hence, the graphic representation is as follows.

4UEosFByC5EsafyOqVj2vGCGqDQiVF972YHEcDxZh67KkNh1a0ZLbdPoWmFpS4z8DV7qHr MEjzWWQlvXpkFk6

From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.

Q6 :

Given the linear equation 2x + 3y – 8 = 0, write another linear equations in two variables such that the geometrical representation of the pair so formed is:

(i) intersecting lines (ii) parallel lines

(iii) coincident lines

Answer :

(i) Intersecting lines:

For this condition,0gxvo4yvdv6vyYKsfqyKlnqDLhwTKrdFLp5OTOp3eqZp7nWG4Xu41I5pqdxqWJYFVPnYoKYaB10 g3o88dOB0XhneizrvuiFKhUgag5cB4 jKUtAuqAiz3W0H cWfa9f5r1NKYQ

The second line such that it is intersecting the given line is

FYuuZEsEMdy88fD0QeH27hHW0VN2OjVwZsynYXW AIYrP5eQiIL68eM ivjwSKdl7Sd fb1R6fudWDTTeZwm y2Rs5boicL4lMd7k6ebrT2V46yvApWKNWKBZ6eT68ulGI1fK5c.

(ii) Parallel lines: For this condition,
41NDCxJMHan5ccD4tlCFz1EyWMbePPRduZx9tboJ CM z9mnmgsotMqY73RR8AoMdRBuOow Hs6jw6o0mF bO6 1ITo6tvTAh0cZZTE9nbPOfTxtZ2wmI9 arwE70liOuRD2lt4

Hence, the second line can be 4x + 6y – 8 = 0

jCoKbVT ZRV6nyZguw9PAupgEOTQz XVXO9LkpFPEcODc5JfAB0TGPW8emDOw3zWE6ouRb HztY9m4wkWNdl3zQgRWXTkMoklKorLhSkSulZtqcx6FNYEJpyhV1l R51pZwZ54

(iii) Coincident lines:

For coincident lines,tMdZ4jrjljJ7qk11qbd5scJgs8SQ7zFHaFhiAsFg celoeL16oE Y2l97MQ8EgKvug23heIfR3doIDMtUhCqZIbFTP Y1K5jsPg7OczMCtcT6EE2U1QQSkJS5sRjufwhGX7Y9tqgVMbH0ouMskESBBgLX47HFymNwipdujX6MMwj6tQlwuirCD8hnl5COEXnstLhTR i2roNN6i2 rOyTa9Q9ooC4f1UFz24di1o

Hence, the second line can be 6x + 9y – 24 = 0

Q7 :

Draw the graphs of the equations Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Answer :

x – y + 1 = 0

x = y – 1

x012
y123

3x + 2y – 12 = 0Aa4gH7RQUhXBz3JLJ5NQL6qiK6KtKcWthyeoZJP92xWT71RDuR7itlY1kHu0e3ENvE13LMYaSGm8

x420
y036

Hence, the graphic representation is as follows.

fJ2Rsa5COr6vn MUrJvjs T3OUA1Pl9caio9eL9gHMYvmT2PtSAOB66iv4m0XAXGclEvL qKqoqf1jyvDIM5mUyRc1G9M2yt0pSnvhe8ViZy39m9qZe7TU3 yjKLZurSl1KIis

From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and x-axis at ( – 1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), ( – 1, 0), and (4, 0).

Exercise 3.3 

Q1 :

Solve the following pair of linear equations by the substitution method.c4dB 3jcZM T8M7MPL3jd8Z4wL16TiO99qMPO DXWJnyuaEtCrHAVbB6UO 1HIVNsoQDOtfh mtiKZ842M6MyHwA8vZLOXCdIulpEF6GX3KpXTpRDrycERU9tBHkRlTT w1mpho

Answer :

(i) x + y = 14 (1)

x – y = 4 (2)

From (1), we obtain

x = 14 – y (3)

Substituting this value in equation (2), we obtain

WV6wyXvW2nFoH1IJQhdeA73akqqVhgwk0qWJRFRdnLDQ BFSvKHbaqxB0e7HL zsdRZRLzOhB20 Fz2A5e0gMUO39X6nnWUJZyH q FzDvKoIii4sNu1 GIIDmZdpVEuITbRQHo

Substituting this in equation (3), we obtain8u5jDQ0ihffnOcBPq87IVS6mT5az

(ii) B7SmR0mVx38XXBeYMzleZxBu 626OBxs bAS0RTg6Eu rpfbLDl7mcNsQJacAZNzGfyqJYWGDJIIxj P gfpGIkUygj5npQXTqTOJH48WL tg0PS8A1tKGDvzgERrYnoV18SmJs

oy517 KleanAXlaO alYCrWvv5IfcOhH88wL3fvz4RAcZ O2z7uVMpY1JU8I8yJeBPC7Ba7pVO0ilChYsD325WEjbdS4CfPg5Qc5yb3qv uLg9yJcoiQdzcoCG6KtHvBmJq3 U

From (1), we obtainm4RUVg

Substituting this value in equation (2), we obtain

1z5WSuR75TQKtTuulunk6QJDEiW0ZxkcCeusggKKiHa 8qLELg33uT41gPlkMLmDyD1sG9Yz1qIWWK9ns fS Fsnr3ABxt4TaYnSsKCvCP9ra6fn6j gJpUCG7 dmtL hJQK4Dc

Substituting in equation (3), we obtain

s = 9

s = 9, t = 6

(iii) 3x – y = 3 (1)

9x – 3y = 9 (2)

From (1), we obtain

y = 3x – 3 (3)

Substituting this value in equation (2), we obtainaj7Chf8C7fvo8i6UP CF7AFfUGFfu7QGyo01rJfLg7NnoWMuQUbhfYFRrO QFo9dTETBNMqtjX4bo6qJs0FfSfynCFtZ7Z3VuACmTY5lXo8IO2METS3ccXqnxKUgUei RVYRq6g

9 = 9

This is always true.

Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by

y = 3x – 3

Therefore, one of its possible solutions is x = 1, y = 0. 8cJ71ouZEujqiK2XGtHvn6CrxCAroWblPzwMAQ6ZRcwm4ujDS8eqpjHbKxXS6F6q0mB7GUNNjfFXnDX6BzRme4iJj OA79ZiB4CEuEa3ZmkRmWw9nCEODdb

(iv) (1)

From equation (1), we obtain

Substituting this value in equation (2), we obtain

1fdnAswCcqRXckBgJgGDm7w8G60NCoXpUkZ GTE4R3Ggyauu1XL U0sLdee9iE96G74ep4QHwqeiyq7zClTWdtgUIcfg7AqbNCFcLfG9NaWLyue1ggatYUKn6ucpQQ5frkju9o8

Substituting this value in equation (3), we obtainN7aWMyHs

(v)

BE1ORcrTeTv

From equation (1), we obtain

OxZRcOaYe W6TrxfflWH4UfzhoBYy8fSdn JBdOteZpTTO4uo7JIA5mM5fy7DUtvUWUU3rrcACxKZefsE9Z3ys8eEz7UD5W6ZYgpiXp Yyhvz EFnlBLCiwV3oj0cx 51 oyiU4

Substituting this value in equation (2), we obtain

EpeG79NG6 6jIStxl8Z3EGIPovP9LZV5y oQSZ5UBxOH2u64nkchLjla0gdJQUK XbjbZZIqk43q5avVuUiXN2zoj2DZX5ShJOUcWq7Ca2Mw4iO 4ziehBl5XTpKaFfNfyyPUmw

Substituting this value in equation (3), we obtain x

(vi) …….(i) …….(ii)

GWkJXS7ATDFYfEmdODq7DrAHW20d8eZlSQtpZLVqBMcsTsIkAaltvTZysfRgYAnRNKiUTrGdmrX2Q6GkPuFhUyS WDJbA0gsnh4cKSLjWi0JM3IQIskvjXdRdDsUoNVuhvrw0gw

……..(iii)

Putting (iii) in (i)

Zah4jFBG0RzduAXqvYodSS5aRX3YfEHYoVJ28TW3upI3vqZNtrXjvxvCLUIDrglY UE7tSvdtgfyRCjghHALhIRwys0bjacot Ts98PdTHia pzyLPaFms klI6ckIPKCo jP0
DvYfxK1 XlMX0OVxJ POxv94NVVHA h xobYgGMrXlYUa3TcKX0W2lET24ZHBm8SVGKPpaCWNT6 JlY0RXu2NqWw7p5MW7oZEaRG0d dWiMT38Ypopl5hATxfj

Putting in (i)

and

Q2 :

Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Answer :DVcgB71Dn1X1YllPVOXCLqwzNCUXihtJc8NYd8cT0jN1oM SaNSMFUDkouHX9dmJO88DqV8VBn1MIUdKHgZQgUs9ArxlxTgPMWLUxLZ2rPmZUcltq5r9pVRddRwp0Z6RNqNPy s

From equation (1), we obtain

Q1VVq8AGCpAR8ZHhYwPqREclGKQSUXv xzDJU9banIUWFo7S5JqHVBfL4zTxsbLYongZVGQUVRHOP8vZqa5vY3t7RLOnLBJn3ZMgzDmc9qgvpNmF3wvAAWo bLItNQpPdrNR1Mc

Substituting this value in equation (2), we obtain

XVTpBudHsmQIZ HAEJw6TK B N2mwALMCBETxE 8ZPPdkRGyGNjvQJUkD3EWaHYiSPT7fHVLlEb2sG5ZWsGsAYkEa06wHxbgCBFAu0 g8Owv1U8lQNy3C07zk5Q 7zyfnu9NPrk

Putting this value in equation (3), we obtain

KQAkQXfBEn5HplHf0yEeSXPS5l6PnqAhjeRJGe UEhM8N1qS2tnIJ 01ojGG3ulLXkB1J07SPhrsuatluB UAE ir2dO5Hzo85V7DCbumpRXl Y840piFqdB1vjNscVM0lOF04

Hence, x = – 2, y = 5 Also,
uq5CemBaG0hxuNYp8jSNZNBYt95ao46veZCe4MDz0rKd2 HDxPiRM0ACPlST4vXL6mUzfNAr DfrOvH 3TMuKNHCsF8Ym1my6r0jvg5QmAoVWgUv41qLk5TCp20 tH VSO9WayQ

*Q3: From the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.

*(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

*(v) A fraction becomes if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes Find the fraction.

*(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Answer :

  1. Let the first number be x and the other number be y such that y > x. According to the given information,

Enyi5LdvcCU9S02jMyyXLEwu1lNAhM9KIGKI7n7sVCTMUwn8XMn4MZRSx5HrXOLgvJm

On substituting the value of y from equation (1) into equation (2), we obtainbvaXYNQtuF1dkLNrjFVwZ MwJ 2 DSNt fJELawlIJxRGbjCypnD odFOWSpZJvoV0qNX MIEbAR toAaykTlz71IneYH6xAL vhJQ8CcZK0fkmZVo3mn42VxWi mf ZQaHl1k

Substituting this in equation (1), we obtain

y = 39

Hence, the numbers are 13 and 39.

  1. Let the larger angle be x and smaller angle be y.

    We know that the sum of the measures of angles of a supplementary pair is always 180 º. 

According to the given information,
Ri2OVeGMk0M4bTaMsqYP9pweNkyjEsdfwly8v2MtJJZBX2fK2cn47KL1r9znCcuno2rGFBmMx1HyQsXJJR WFeoSuehdYw2IYQ6cLRA1FYbRIwtfME0rsy0G8w88NmXan3Arhng

x = 180 º – y (3)

Substituting this in equation (2), we obtainmtpeJvVbUEYLV274fdy PcVT617LbhqajN86wvqSR3 barIwoB1kZy901pI2U3yn17g r0Pqgj3LQB37BTzp3DxCCf6kXXSiNFb

Putting this in equation (3), we obtain

x = 180 º – 81 º

= 99 º

Hence, the angles are 99 º and 81 º.

  1. Let the cost of a bat and a ball be x and y respectively. According to the given information,

B EaTZqN8 NcMXgXZM8GrP1I LdVPt5i1oLqnYWRb5pLW x1A6ChpiDjexpe9pYoyChaOc1 A yQNA JDkNRvvp27wGBej6t1Xuwf aqcezVTSkrgDR0G7Dhhy6sKovGI05ZnOo

XF7wqKNf2FvXuVeKMGNeazyZxirAIWomiCG4ifu6owNKsEJu4EuBivmopGRENYciPgijNRQkmF092HkIWU JbLXir5LWAdRNFcbg0y

Substituting this value in equation (2), we obtain

wCAjkB69LKxyHM4zgWk vTcywmX1MPUzlVa0qvJOkE7BlT41X8dPHu o35AjkftPgsN9SIIvCU9U EQi69shIms3WpkHH9Z8JxAa5wbUTByHOx680RSauSY 0HCjrBN05saXyHQ
R2Xa LeLMRZ5Y89J Rv3gIIZEWP9sRzx6qZPTlnbBvYhu42Kz6Sq7qAvhr2YiLf

Substituting this in equation (3), we obtain
r9DVmMBeIOFSNIZdMb4sh4FbtQCutXhr1Cyt vSuJLKQv4cyf FYobGRsWZdj0Fq4smVDZwTX9dOJdDosAbIVEzESVOlK8wnojDvRIibPcW9E7T3fmDuCcAaItGPYphf95LvyE

Hence, the cost of a bat is Rs 500 and that of a ball is Rs 50.

  1. Let the fixed charge be Rs x and per km charge be Rs y. According to the given information,Nyi21Lsc7Zhe9Ew0fKDSGE2DSc20isGdByi FelnlFenmlFSY
S0pi45 lW JlUyjl4iPXTuZl5pf8AhgNo7hHkXx M kksoPQ1NgKVHJY5ntEeUZE1MQXRwTbEBJuWx7JobA 48pq777nxfQRvVb0LKFnySM sVF0z8wrwoy NnB YZPRTc3KSIE

Substituting this in equation (2), we obtain

0vNrKjc9YG16JOGQvuZuMUld8UrSHsU6pe1rzr1Vbot02yyPNy7VVLV

Putting this in equation (3), we obtain4hQV29uRXb ntd0 JgLosn0C6qOMtWPjXf1ngUmnQdXY4INdoM13EtsIyiJO vQO P934R3eVxrjVqbx5C5j1gUSj8APe0JdsKxqr310okg 5cKy wpNPyc7NgDOir5eILnizCw

Hence, fixed charge = Rs 5 And per km charge = Rs 10 Charge for 25 km = x + 25y

= 5 + 250 = Rs 255

  1. Let the fraction be  According to the given information,

From equation (1), we obtain

…..(3)

SBUxTRBwAuoGsVOzH EGf9vEzfK9ZRO1lCKpBSPpdcLlI4S4pz2fjTulZtr3CbnPQ2tv81cD j7qYfJ7Jy2PYPwunp iSCRjQKtZ32gc2Eww0Wc8WeO9RFZl Q7M2 IGmCXtw3Y

……(4)

Putting value of y in (3)

aHJRz76xZTVM6CI4DSP 0 a92gqwAmkwY9YaM5GvN2kAOdackILqvO64kthUPliYqnkGp6HD7t5fJDkyWgZL3cnvOtjMqSVpW4pipcWl0 fxxJzUxuU21BDVq6 xs cUAMQQGQw

Hence fraction is

(vi) Let the sage of Jacob and his son be x and y respectinvely 

vda4gVGbnV0jsvdleVuQTWYDuyQa7NftXcC upFKzqlkoHmvPxjqA9GnpJCr fLy9A3CEF0zd1JIYf O870lZ3fn1O1NJ2 NXf0899TGBZk8Drm7cGQi8X2urbJ0se9beFORZ6Y

…..(1)

saIwT8m Qw1YBEm6nxJDwaxfUl8nVHoUbaS2y SKbDvBJlZRANTgSmqODkMB4VQjA8w3M2xosj AvmMm 17cn ErGXKEFYtT0LbOuZzbv bFY5TBLJ149G2DbWI60U1r5ACNB6w

…..(2)

From (1)-(2) we get

gq1xA 3ld9Tgtn4mtxE8o e KdmAnblk1BOBLdxIqlXvheFX8K4bTtXrS1cAi8aqOABygAQCyqEqd bY5WlxwyZiCx83V i0sl76Ae3nRknAZ0w7cvSHLQRGdSfM2oQtTg9b9GM

Putting

Current age of Jacob = 40 yrs his son = 10 years.

Exercise 3.4

Q1 :

Solve the following pair of linear equations by the elimination method and the substitution 

method:V0VFXJNPoD1A2qzvfb22qwrRtdMcBcum4HL2QJmxRn4TyJUzyllj2ycim8ris0ZaAyD36p95Z0gD0bY5EpM3mMtPM1TsqLKx9pP5ecQT0GvZcWAHHHw7qqnDMSXTtQ ZF8 ylPYAq Sky1dWeJDF8W9rHhzVq3cI
EMoJ1BobPfs6qyzgH70i969ZDkIS4KByFv5CmGNAWDLjZlIcOSLRGF0WZtT6QO0IpmPO6M0H0cVvZCJix37WyKRAFG0esz 8iecBRHVCbnxVY4 tlNan 9Wa7hSyPpbDDdUKTjUxvH4oKD jjLO67Hc4SqyCuO1O6YNTPOzGM SQUpP0C7hGIq3jWoMkvI9zgZjTRhb9az8zY76TvEb0x8rSD0sMKVDz 6XCIpu2RXIBw218NZQ2vdYFZLEBTf539 KBPkOku82kMo

Answer :

(i) By elimination methodtT844yZWLRK3BbwWEDsXKu7P5buih1SbDaqh1 Bj5sf5MVtThHZXnAATDaQiqsrJT 346SV9cXtH7Ic SfZ

jnEYkOvWTVaSI2TtA RT PQvmgbVb vJoTRoqvLOcEE6nIX73 N2DgkG asH1BDu4zgo7UjSka5rwRxJgKW2AOR5M0 u89JbyLCO t0OcuI99hibhsUXHC7wHq09G0EfACaMG1c

Multiplying equation (1) by 2, we obtainEWkF0zcRNEioZrKifFroe DKTvoyvrkwOQzBHP80Ku4rFvjiStHEs4 J1knTK7Re btEDya34bjiQ qjpY6buuu6O OcNRIurvDISChQ6sfuFHr1kW7Ern4tVqo

Subtracting equation (2) from equation (3), we obtain

RIMIUFkWaOqYR5y AWpLwfBQj rbzCSvSxa4nm8UNGFHQO2NTAnXg1T06DWVxWxsmN5ChSiHCbZ2U9gOhad2e82 n3Hm RmjAdntJmcBcCdoNdUI63lal8bOII3MxYvUn k0aw8

Substituting the value in equation (1), we obtainC6fiF6ICIRYOhdWiZeqJrp6XeWtxqsgy5pA4eKzeT256nwN012T3g 4rJwfmBzIL2RMvY7zE8gdoFu do0HZyTtsj1gSDm1Ilx86ytFJkY8qRBvxq 2j4cLgPsvI IovL6wPLUI

By substitution method

From equation (1), we obtain (5)

Putting this value in equation (2), we obtaine 0TD6xY1fq61MwSJVksrV73a6DrFZ5TLXtR6

– 5y = – 6

8ze

Substituting the value in equation (5), we obtainC6fiF6ICIRYOhdWiZeqJrp6XeWtxqsgy5pA4eKzeT256nwN012T3g 4rJwfmBzIL2RMvY7zE8gdoFu do0HZyTtsj1gSDm1Ilx86ytFJkY8qRBvxq 2j4cLgPsvI IovL6wPLUI

(ii) By elimination method

GV j4qiNbudoA0nvi9xtVfeR8 X LkezmHA7q3typEk30tkJlYedS5QvpdukIqFFj mnEigB Itx4VNEz1miZrRQEAhStoftFSggAnhW
L7rDEyidGHif6TS8EobDESvgGrxKosnGwY6O

Multiplying equation (2) by 2, we obtainnm5Gce7r03DOcUmONpph3ZMMFqMvoWm18P7asctvnsRpRx9Hzq9uWoHnAx2w7SR4z6ex8esDXUK2WIz

Adding equation (1) and (3), we obtain

siqwCyvQXM3j0oKa pEwHM te6wJmdzcH0SVlW7 9OQrEGP45xq4P 7wHIsi6WH9BaBhNxkuku2gHzAtN8 69 utevAWxtgK2OpjcGHvK8WleQV7ntiQYRk0v75jBPTD4q1 c

Substituting in equation (1), we obtainR nSV6CFwByOwelbb1XkWsClwehplCIw7XIqQyw352Myr1HKoS5dKto8NYru5pSr1vp4EYZekKtg1DzAqNFRUoxAoiwZ4GfmECIs1VLBWWc7XlAQ VNraRFn9WaI3BgBReJoSrM

Hence, x = 2, y = 1

By substitution method

From equation (2), we obtain

fzSk7RmVcvLxDi69m4XQPdK7Acpt EaldGDLk3SlIVSJj06Uh5MAah4zEmiLPfYGTkyuAyyvb8b2JaaU6LRg8rUeX2d3klpbIpwkmvZwqFV HOT1 IAWB2FgooEDgnscn6iKLE (5)

Putting this value in equation (1), we obtainIh1UMBNZ1MBd lZi hcmDNvtYzVz3 R5M5O1RYFOPP2HgDt4PYx1imZMuhEMShJYJiNWDn n T s3 fVvfSCK0uHQHxnqHDAzy82DRsHDhWZtRoKDJqP6RlI0vJ0v

7y = 7

Substituting the value in equation (5), we obtaincmLFGUiv8QX6S4PRy2qHJBlpB4b1 eIP5ANHe4icl IEDUJwdGrnAZlV7uZDcTp8ji0yw089oV4DcX2Zud7joek7RLoV2buqrFXUmC7BXqBq8DhnGSTSzWmPj5An8bMVTO Pmc

Q2 :

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

  1. If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction?
  2. Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
  3. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
  1. Meena went to bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
  2. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer :

(i) Let the fraction be According to the given information,

Subtracting equation (1) from equation (2), we obtain

x = 3 (3)

RyimbxYDTLYLmpTxWDcK5UzOLYve4ejrZ2arB0TUxObB4QbxnB4bCw

Substituting this value in equation (1), we obtain

u2w6fV17wWRrK7mUW62K41TlQG2NXfNSANAtlVWxa19wY6pwUtouAJvdl1z9QvegeIkmPkAdRRJbY ihhiUBCERCdbhJ3gw6qngX9X7n1tYzM3do9XEijvEBO5JZ3WyOAiD7URM

Hence, the fraction is  

(ii) Let present age of Nuri = x and present age of Sonu = y

According to the given information,

jPEttUjUh REnLZo2rVIiM6QzpPX3JC8MkfMorrImVDalXpLLtmNFwJoeR7VFodAszVbhsuiGCGY2bHf3swf2gZgRst

Subtracting equation (1) from equation (2), we obtain

y = 20 (3)

Substituting it in equation (1), we obtain

RFdNiUH4JJZyt uplB8J 2CLJ0v1kDs4NkjLke2gDBj80 m5 F86aANfNIm08yV Rnd6QFZ0vaYXkmhYqcGISpqk9pL

Hence, age of Nuri = 50 years And, age of Sonu = 20 years

(iii) Let the unit digit and tens digits of the number be x and y respectively. Then, number

Number after reversing the digits

According to the given information,

x + y = 9 (1)

o52WXT8Ea7e8fvLPB0VTkgdOL9 YaK 9V3LXZbv A3f3zfsg5wRTigu0Sc8tIgT0H8mMEUptfFAlez8dQ3StZvDP7KG
Q2GqMgSvZkIlQuuwLMEsF8epDlILNpPlJ4053h0zlH7 gQ0BvOo4jripXbqUjAqlyMh 5oTNVNhCzfZ61 Ye7aMOodvK ejcxXZnTDJcXAdY DTZb3mKyk8OmXzzTZ5hXWMNceM
D2VO aDnFfEGGF49mCE32TjgqM6U9F7P8mjiE 5tmVUksOq g60N1csmcfs9FCAkEmfK26F9

Adding equation (1) and (2), we obtain 9y = 9

y = 1 (3)

Substituting the value in equation (1), we obtain

x = 8

Hence, the number is

(iv) Let the number of Rs 50 notes and Rs 100 notes be x and y respectively.

According to the given information,XKk x9tDbxOa8MGOdl7dOmpK9ISaFd ac7dyHAG22IGBVbzSRsjS RCeoDCtxB4xCrzxHEM14 wVpmrWORNvsUU8C65TNiHm oAdXQ717 jIApqp6AQ9cfMtIPjBVR0c zfIXU4

Multiplying equation (1) by 50, we obtain

POn2Q94eZLt68oA2 D1PmLlRq pJ h4Qra8ab5TKh9BhOL6GEdSIuaaaFclaWfOwp qzR OZpmUrIa aufY 9NX89bq40Nkr1uNoSX0Q9tS cdhnKfVfa3A3bjwnsvr20zZrWEg

Subtracting equation (3) from equation (2), we obtainQNo8Uzu201d ZnxUufTTQnSYWt2g2aatMZRIgrN20iWOpWjmVB5bNld99iyJ63ZSHIc5 pU6t8Nj9Stzns

Substituting in equation (1), we have x = 10

Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.

(v) Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively. According to the given information,

Subtracting equation (2) from equation (1), we obtainxknyHDM8G5qo6LqSiRSnGpjhvZWuhmhFwaKSJbKUq62fgqSQrwQI9PBzRo7uJsIY1iSzsaTDJBYs5I70ocdpSzafr2A5uxj2N2 VPUP50EhihdNcpKn1sYutoElmdMlw it3Gzk

Substituting in equation (1), we obtain

cvc1Oxkn4XExRDFPw6VylkoBr4hhCFoqWvPN65FZLFvjgmYf3serrSMKM9feacS5ULugtjD7ynUHZeX4Dbjrb1Bz h5SJIN9R1cD1Tmi9H4747U0 gpyUzS6YU7 PXv8jji6ngI

Hence, fixed charge = Rs 15 And Charge per day = Rs 3

Exercise 3.5

Q1 :

Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication 

method.
NU2veS0XIePKezDY1tXJFfzrj5Fj1qLRE8sB4AMYPTws3wnROdFEbHMKYpC1KGN1

Answer :

Voiox446URKICVdfakisX4N7zmIG0iDX5oraDTFiOcARxW9 Kli9EopyoGF39XwNVa n2o8GR35HWK Q1

Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.pKicRLWZ7SojR3akNH gg2fn688YO iPYo5XeOSKLSXCgWNCRzYPd8FguBU0Jl79EL6r4Wvq8 UqTqn1wa1iLVXDZQAaClC3ekzPFGEDrqTNt v uAG7irQg8R488Xkt6pzPlL8

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication method,

∴ x = 2, y = 1

zpbqnjX3b8mwFPJgDa5x0ghzLV71KFlzViNfpLQHJOVT08O RixYVXm5c rGfzE9NJOUo1eZizaGB78ReCcxQTsTWYES1sI8zefiSgdZf V4LUziGZgqeqk3uyhPtfyIwtMPJQI

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

Ad6oEqOErWkak80gyx9mz h04PJEvVQDhR1iKZVEgPKUggNUxkfDsctQ48 w3BQhd hHooQqrISVc9oAdHEZLabAttHbQLihvzPzJhb m7C1eGQqaWA6EnX v RvDRbNqHXAGo

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication,t5sOd7ZeB3Gv1sraii eYxTU1MP8U

JI5zyRGT7Nb0psjHdl8mRnyyQ0jIy8mIR7 i3 G5JDaY323vEXo0XdFKFYhHriMBo4BFadUl 3dhw05jMKjXwaFqV44IAY6zYy5BZQwuXkyfFg6F5sF1qJy7o5uJyGD93zCt9LQ

Q2 :

(i) For which values of a and b will the following pair of linear equations have an infinite number of solutions?

4WaPHVBzSAv0pTzLVykDqRHSc0e6iJrM 6X7ff80I05VxDB3DE5wKJ2 SY2FJT9tKPiKlFaX1Hq7KXAUonNXphev usNNKJItCWMWJ7yZuMzbOdalL1FpaAUATlMqzkDJRPDVlE

(ii) For which value of k will the following pair of linear equations have no solution?l6 t85jedFkTZ7K4MM

Answer :

For infinitely many solutions,L4nQH3 Vsz ID9EKHANyOvPg5P3EH2BMLjwlA6ZzPA nJ7JSa0GS84d1bew wgfjPeBHVKfW0EtugU3ytuyXmtjBdGBYkC3Ok9aYaIFKD2RdbDYBGjSnwNpfTTP9QDrG9f aaM

9s4DjVepyOEtuAHL9EJYrH7QIP5X9BZ5ipbdnAKEc6bVCZM63mbpqqLoFKyV4yXAK79hDowUFJlP03M0dK1sSr1fCUBkM3l557CqPvjoPNlH7p3SPV60LLglwdp3CQqn0YvhpaQ

Subtracting (1) from (2), we obtain

Substituting this in equation (2), we obtain

7uq50qOvoP6fj5WmpWHzsOXBlTXNt5ktAFeS0BZkCybvODrFq39AdLUbi wz2ZUlb9CBCr8BVTYbRDRY467fSDtLaTfR1sXYSNsEc wn2JJzpN6HUFFBEx j jcOjyXQ4FZ4K2E

Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions.eO0r1NTTr4HX5v 8JEGbCl634LHsO21OGhNR6xqYYPko3LuL7JuoA1IgruGymw

For no solution,

t2THEDT33qqzSibs7MgYbTwP9nRYLlbk pMdAxS7vvq3lsrxHzCdgWPZ79W6q6ufKbQktBqmhPM86WHBtnIgE BglBkvMKNEEn3SH1LlbfpdcKunMC71LryHyGzFmzGI4V2vS w
kU uc2u8dR X4vxuY5kYz2zJlB3P4pO0KSY7TdgkRsTtgQuApvarVvzXMQ531nDpXlZZvIQ3rEstgCDDbIJ h9cVoQgkROkbw83 GbwokN9rkddGtGNiv0R9weWZIIWVuizB1g4

Hence, for k = 2, the given equation has no solution.

Q3 :

Solve the following pair of linear equations by the substitution and cross-multiplication methods:

FpgWJvHDeulPgGMuWa1eiTA6xn80ie1Vr4KLAkatps9F9JKXw qcI28AWnj wCv9Rpm9ah2 uwOexmITnD4O0EV1gG

Answer:Mq5Htvyz3aLH1X4jjtuEwujG2wJ9 hsXj3XsqOMQpB x9E1up1qUMvZF0Oi51blBjQ6IcAgvw9z84XN6r4oNtPLOgsDU8QzZs9Eh0EXQjsTMzRC2sc7sZSzcZKzq9BWJ tpmGNI

From equation (ii), we obtain

Y5DJFpJaKJUMWWF5TJ9jkJuhmCQSybY2SxQNbi3EbJAXb1QhTp2bLiikX3Dg3wVT1u76Mr7brz9EEw c AIu4ZH9uH2cJ2hjSsZd2BnSFv9mmOlwBzhcAE2D 2lI3uo5AQXw2 U

Substituting this value in equation (i), we obtainPm6rKIQ OJTKO0Ft3beX4tBWrZevDTW8ZJqyelqU4ZIa8QCBcrxNBLWVkZJXdeXmKu V000Dd39AmZDTV8AHy9wrWhXjjZa5egpumfUTR812g3i zCGFCQttM6nLv6R1uVD2b2Y

Substituting this value in equation (ii), we obtain

x6CJFrIf ur4UYtEg1Er2YhcH2BUS7ePIED9Y5bf0QVHI4Ht7IKNHm7gNZJeK81LKSOSPcr3pmd3fDcKl ESqShE94RaGeO UiSAQvp41FejzfcsTrXvQYsXdod1TFtNCAdn8c

Hence, IrYZd2nIZrFs

Again, by cross-multiplication method, we obtainofrXpBQe7MFg7coHNz6R7CGJvoHWny1m6LSvptI mrPH67JHgKFeuH3OXTgJcuE8tBKNUgaJG0Zt2R AHc8mmykirdbtCoMk a9yKwLEagLvc8Z771elF3P1lQuasQrrsOggf9o

Q4 :

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

  1. A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.
  2. A fraction becomes when 1 is subtracted from the numerator and it becomes when 8 is added to its denominator. Find the fraction.
  3. Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
  4. Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
  5. The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Answer:

(i) Let x be the fixed charge of the food and y be the charge for food per day. According to the given information,

Subtracting equation (1) from equation (2), we obtain

H4ivigxmvroYKHfK7t yMCoPqKqg6Y xDCPuUakARsUNjhPGeonfY5 EQid2YbyO4uV1TBZTO6Cw5KSFpaQAo2A8Z V44zi

Substituting this value in equation (1), we obtain

HGfJRw24IQ82V27DLxDsEVEySZ6aG9f0AH1zvLZuRf2Aa4yHudPFQoFdccJcQwU9MgiyiOVQ 8SW9MObZ5vQUg CVZl6Q3El1zysA hQgskpKKgfr D2Eh6AGN5VhOr3LTQrhLw

Hence, fixed charge = Rs 400 And charge per day = Rs 30

(ii) Let the fraction be

According to the given information,

udvxyPD G6TTB8CPwelY Q 3 s3mg50VzaDIH J6Hrw5Z ZziiG gGjOkykU95 yPorCbLBF0KIbsMx3zONBx1qXyFAssrX2FS oA4BND8mdU4qlC7cnUPu8RwwMHywmHtNdYiU

Subtracting equation (1) from equation (2), we obtain

x5y4STXALPu0R0PWkbEKoa2v2HKAOq8C80KIGGjvTYy4BoXdnPAdDVnTtsyluneEqKbvMSg2eXUsE28z0cK4L3 FNfIFQG3fGAzUlkF9dK sWY1ZTPe30CZVZY6M26vKq8x0 0

Putting this value in equation (1), we obtain

r7AbIuNzOk3bAqDcE6CyCQmNSnjqHeRpWdCquwHRy RdsfUJqvvkA5KJUmtKXco4bXIBoOZwtDTVB9hgTOEDhznr6SDubsAbZU82kM5L F6TICZ2IWT30UlPOTd2Gh7dml8riSc

Hence, the fraction is

(iii) Let the number of right answers and wrong answers be x and y respectively.

According to the given information,kVQjr8EWHj0Pvh6jzCuevE4qeYhlp1 zgDseR9R 5U1v oPP9QokSoOzqKIxki4GWCAHMSD6qlmCvMyxglCQXkC I2VRVKLQ

Subtracting equation (2) from equation (1), we obtain

x = 15 (3)

Substituting this in equation (2), we obtainE46M7L WT9ITfRfG2QLOWCuusWApMYL8f02O2Dy07pa23ZBaacfolVA3 Z NwU6FQOk82r3S8RXkNm9kc7mbC4UsSB4oJlLlgoA46 N1KF0VdSQsSQpILwQPKhGBYnM ejLXAw

Therefore, number of right answers = 15 And number of wrong answers = 5

Total number of questions = 20

(iv) Let the speed of 1st car and 2nd car be u km/h and v km/h.

Respective speed of both cars while they are travelling in same direction = km/h

Respective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = km/h

According to the given information,

PobLZ1Ro4 xB2zhpw3 uPdBmv53wqYOR5otVlH8Y5opR V eJ69K9LvoThyFQbHO3S wTfVHFAMPOewpNuGpnEGg

Adding both the equations, we obtain

Substituting this value in equation (2), we obtain

v = 40 km/h

Hence, speed of one car = 60 km/h and speed of other car = 40 km/h

(v) Let length and breadth of rectangle be x unit and y unit respectively. Area = xy

According to the question,

JM6QIYa8dbjjlZ2kFr7sBzVVR rKg0siQ cdYvXDNGPwNOUo 6FWRUG9b6YXLR1PFhjdMAgwfirojZAPqdaVnhFnxzdzaAmA8AApGpwoKASio9CygZe2Ajw jWKnLMM qZ49Ozo

By cross-multiplication method, we obtain

hhm10hpex6RU MLUWH14GWuU5Vr07dSZysfjiWqcWjcDGf

 Exercise 3.6 

Q1 :

Solve the following pairs of equations by reducing them to a pair of linear equations:

uq3P3V AEpQs3COKnbblo9xvwu 1MThBkmQgYV2zURP0BchVDQ1BXIeZ6KU6ssDXxiA 2VVHVTAt7BDcJG5FkYp6nHSlm1yMBPHQ1NP Hsj8KXMz ud43T7QCQO2J8dD6 q7mNY
ll65wpxDMmzPHXaKB4tJIMhz5SuamVHpr3e 0AgD5 TfF1jnfpXb3NwPixgw EbavnuKRHM8PgYjd4z6 ftmSHF 7HgSAWBcZfZFw Fk3k yOEWgkZinUmTzAyLS5ekiKueTwAA

3WIe GBOWmvvvuahTMW0cICaTuL0HNyqCfaj jMTsI8suQar6j6pNUd92D4Y6gWJJswcYOrHZuFdTExPF 43 TQhT0qapKjjDYdnCZSMBajCqAonuZgqjtNKKBV 0 97uA zgnI
vFPjkE7u8zkoKbuxGGTCpSosr7yFxUWVQxr4ZUQgPOTkQB8DN4yhOxykBihBncBggXny 4JKPZtQAbRl5VOcuu1Di5yoC49 53DaDPGeaLOASC0pIS uImJCDTs4vSh9PzGTMEU

Answer:togHRNgVKEl8Jx2eeLO KSVfnf ux0g5YohixllkPyUq6ooDsNSKArKg9MPOmu K1ETBeF0p1Es8dOmraMB0r1gOD7

Let    and then the equations change as follows.

OaXiAusIunFmvGWpQXtVxyI5e6fMjFxnlljVVgUbDYX dFEbtOlOzYdUq5fHPuVOvxcJJeX4b6oSYFhvVRS Cy IkPbMHmQpQc7z3F3vH FYjFZJbL41uWZnALTFwg fNkQK oE

Using cross-multiplication method, we obtainz

MZm5OsxHU1ELHig3rO mNZvP4ENpO2NyChYsAAjCivt8354FxxvpujZNeQ4kIC3vFH6h051 WxR0Wawm0RW j7mizIscPL80OuajDNnzHAjPJueBYo4qjaxXiFAPJGX WT Ll7M

Putting and in the given equations, we obtain
4ESzjVcJTtGEU2lwWktQj37Ba6jqf2xPXv0Ei8EDpas1UJfJmfzKGaumuslNcThaOrZ6Ex7jwyrATGi xIHF

Multiplying equation (1) by 3, we obtain     …..(3)

Adding equation (2) and (3), we obtain7XPDTHtOgiIuPXLKKIQFts4g3oG3fm3BiKOQU kZuM2HYaMeEAB5FU7pAMUn0Dhih

Putting in equation (1), we obtain

rm4 rvyEen9NU kreXNmdQEmCnO9mhDSx9Z9nygpCUjx7bvRu9n hJ5I4Ma6ny XOfzBTr9Vyt1zH8kWsqHIFSVnQ8MOpvk34gbSd1jZ9SRpBozDz eeHbVro3fjFKbNzS ASPQ
OvtfX3sTyAcn2AQLvQxiQwf8vOUwLGzCTbbBsWO3F53w 72OTrKJt coB1nXEXMvSi5aGA02EVs2TYXj0kERULIutmb2BOeb1uPhNvGt6birCKcui5P5zqpzkka2K4soc25fjcE

Hence, GjulQ9 HjKiHZdiN0KXa20LcanwCg385

uq3P3V AEpQs3COKnbblo9xvwu 1MThBkmQgYV2zURP0BchVDQ1BXIeZ6KU6ssDXxiA 2VVHVTAt7BDcJG5FkYp6nHSlm1yMBPHQ1NP Hsj8KXMz ud43T7QCQO2J8dD6 q7mNY

R89f62LrPgSrzlMc7cnqmcRL3PFxjX44ghquY38Q0X2Dgo2kkLRVMG6Ytw2S3DUdGcfnjOGQISCHzOsy SH0g8xrqm B6MCjquk76OPP0kbSYH5foFKJD5uEbgt l6NsNU1izHI

Substituting in the given equations, we obtain

By cross-multiplication, we obtain

diXkYhfrfhd0Tp8imcXJZ2amZIGlXEZGtmioJKAAe1W4sTncm7UeqrjnUQ4I779a3exblDkC4vot8AJLfDMDiVsurmfvaLLclPqdlvmFVaYYK3CAzG00k DKAXP uAGf0UdemM
ll65wpxDMmzPHXaKB4tJIMhz5SuamVHpr3e 0AgD5 TfF1jnfpXb3NwPixgw EbavnuKRHM8PgYjd4z6 ftmSHF 7HgSAWBcZfZFw Fk3k yOEWgkZinUmTzAyLS5ekiKueTwAA

Putting  and  in the given equation, we obtain

Multiplying equation (1) by 3, we obtain

*Q2 :

Formulate the following problems as a pair of equations, and hence find their solutions:

  1. Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
  2. 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
  3. Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Answer :

(i) Let the speed of Ritu in still water and the speed of stream be x km/h and y km/h respectively.

Speed of Ritu while rowing

Upstream km/h Downstream km/h According to question,

Adding equation (1) and (2), we obtain

H9bVlFTY3UavfvInI5uAH RCDxjs kdEKHce6KTHUfoxxMb6IsqkW OemB3utqOm7 enmYFvqJSYjzz7zWcSBy2TBbH3ya2oaQpIF 5Pw5lE 5MLV6ogqhVVjLIrFRIIbByuT g

Putting this in equation (1), we obtain

y = 4

Hence, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h. 

(ii) Let the number of days taken by a woman and a man be x and y respectively.

Therefore, work done by a woman in 1 day

Work done by a man in 1 day According to the question,

SAM Vlzg4zuL4u715nmkbj 5Moq2wnDzOV071VvXnBfkLMmVUVI3mEkK2Jbefdu5PjkCSB HEjEHPTMVyRsK8OvmXJFMaa6ANToaZ5os9PMf06gJNrqVsqEKnaSKHU scU0ZHK0

Putting in these equations, we obtain
WhqZPZBbSMxhYpaol9qt89Tj2EcTfvHKO083zvpT8r3EkwN2DL SjqmeKVEBMaeeIyCf9eLG5DS artZ6sMfjsFUVJGh8mjTM0VHDqe4yuJ6LoOzkMk4NqfTyfTtowKOAkd1vo4

By cross-multiplication, we obtain

elRZUg99BMz0ugV7BlKu4AySXpftw49qo3luzv
lbPZuV0DwlmaZC6o5Pa7Ba6B359ZMmVVMO1rrytRsvmSJQF9yDS6wgo6h 8NpxAGagJnIZUdZMvJECNJumVR2EmD2 E8E0wPscibnsidqnbY XsfBrV4earfWL9M7qM2OwZjrQ0

Hence, number of days taken by a woman = 18 Number of days taken by a man = 36

(iii) Let the speed of train and bus be u km/h and v km/h respectively. According to the given information,

q8pZkUUJMSVEdDgCB37pLKHZE3YpG9UAz3AUBfs5QHvIe4QuxPTyGMBWMdV

Putting and in these equations, we obtain

LVcdaf3gfJJigNuLpEAGjV1 QXTC1LuBf1fegbZuT3yMWhBNgXM7neUBBl

Multiplying equation (3) by 10, we obtain

Subtracting equation (4) from (5), we obtain

L68qG2kPm1tVT9961YzOo0KB ino2RdwReuYojnuoX89rHa6HJlgrx4qK0xqs1 jpILeWivKvlAyIM3RyBwtaCxvbUZug0IZReZsPXjgmzZmec0 Rb fTKdkU0gG1FC8VUFKmjc

Substituting in equation (3), we obtain

AN0l 88rOkzc y0FWK54nCG7ghULLA0qO7hSUAT2J7XnScnGioIeR4vQIoeKS INhm95QCv6ZcAzOG9ho3f1bXDYnAf0NcK8RrkYlvLcam7WSImlPH5UuS0LFKuQIaMnT Rtgow

Hence, speed of train = 60 km/h Speed of bus = 80 km/h

Exercise 3.7 

Q1 :

The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differs by 30 years. Find the ages of Ani and Biju.

Answer :

The difference between the ages of Biju and Ani is 3 years. Either Biju is 3 years older than Ani or Ani is 3 years older than Biju. However, it is obvious that in both cases, Ani’s father’s age will be 30 years more than that of Cathy’s age.

Let the age of Ani and Biju be x and y years respectively. Therefore, age of Ani’s father, Dharam years

And age of Biju’s sister Cathy years

By using the information given in the question, Case (I) When Ani is older than Biju by 3 years, x – y = 3 (i)
S 7jm2p1x0XCN b8CfrtEtu0HSUxSa72rkd UJQxotp21Urf5tGMXQcSBlXc5eayA55VB5b9

4x – y = 60 (ii)

Subtracting (i) from (ii), we obtain 3x = 60 – 3 = 57
c47rnuRhALh1Elacf HYiCG560VBDvunYAmknAVoq3XMkWXHd5zxE

Therefore, age of Ani = 19 years And age of Biju = 19 – 3 = 16 years

Case (II) When Biju is older than Ani,

y – x = 3 (i)S 7jm2p1x0XCN b8CfrtEtu0HSUxSa72rkd UJQxotp21Urf5tGMXQcSBlXc5eayA55VB5b9

4x – y = 60 (ii)

Adding (i) and (ii), we obtain 3x = 63

x = 21

Therefore, age of Ani = 21 years And age of Biju = 21 + 3 = 24 years

Q2 :

One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II)

[Hint: x + 100 = 2 (y – 100), y + 10 = 6(x – 10)]

Answer :

Let those friends were having Rs x and y with them. Using the information given in the question, we obtain x + 100 = 2(y – 100)

x + 100 = 2y – 200

x – 2y = -300 (i)

And, 6(x – 10) = (y + 10) 6x – 60 = y + 10

6x – y = 70 (ii)

Multiplying equation (ii) by 2, we obtain 12x – 2y = 140 (iii)

Subtracting equation (i) from equation (iii), we obtain 11x = 140 + 300

11x = 440

x = 40

Using this in equation (i), we obtain 40 – 2y = -300

40 + 300 = 2y

2y = 340

y = 170

Therefore, those friends had Rs 40 and Rs 170 with them respectively.

Q3 :

A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Answer :

Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel was d km. We know that,FTF4w0CWJz7dmFVsHLlhBMGFcL7nOei6FCkCdjCpumd38JaSsps0wvdHGHxsaGs

TRf 9m1l s3RRG Ooubdy5Dq8b73wYWXm Q7mlg

Or, (i)

Using the information given in the question, we obtain

bXfwHnqj onfaw7PAOYl4ark97TIBGPyf N9OpeivVV21ljmxnxGrYdzZraagQY

By using equation (i), we obtain

(ii)PEtHF7gDLJdL91yQiPQLdXo6GHq A3x6n9mg85TUyVHw66e6Vk5ZmuNUueB9CpLjLZY0OranRbo9SWh62QCXU6UjCB0

By using equation (i), we obtain 3x – 10t = 30 (iii)

Adding equations (ii) and (iii), we obtain

x = 50

Using equation (ii), we obtain

zTrw5W 13tpRzFMaSgeES88zNn 0lRFA6JF7s YOZRzmHJQQ xmnxJwW7DKHmcLm8ruxfZYnO29vVp2ro lqVJQyC7cWe9 smzwE62BerzWdwJbdxRRvmCyFIjgBM6LA oIdtrI
XIPCOaaV4gU0nbnq4Txbx204M6wGL4gyQoeGrKLsdQlkfq4PL7jNB5vsXhSeobFxb9sHFVCh Oh1JNJcyVpADS8iP3BS4yGPRL9P0ok PdUUIv4qsjr6HfUxw97CY asGJfFS0Q

t = 12 hours

From equation (i), we obtain Distance to travel

ICkmfyOruVpYuW9gcn29ucqFChCp BqHcz L6frnTOXgjdbRV7AMMjCOZwI 0Macjm45S6nTWsBwV3 BlyMkFBnMuohGS9Hckv7TZtHVWY7UowoRIGs1uEQXecKHwkOZDeigBdY
Ey6JPcDGRGmc7IS3JEFFR0u0zqjIOO uaY6OYFewMJ3 zmFk3mlwhldesiwhyJgTqegYIU1dvY8TtMvyXe79aOHhUnw KHuED4czT7E 1vZPp kvCJ0YlvUmArnAEIndejYJZPY

Hence, the distance covered by the train is 600 km.

Q4 :

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Answer :

Let the number of rows be x and number of students in a row be y. Total students of the class

= Number of rows x Number of students in a row

= xy

Using the information given in the question,

Condition 1

Total number of students

LmPXivl5PZYuyzlp6CvR58zxzTFfDpmy7gWitvq pOscUZGOhHcBEqeVkrV14

(i)

Condition 2

Total number of students

(ii)

Subtracting equation (ii) from (i),

9HFh3Kr9IFgYj1RF0TaJEHZjMKku9etvthFztp6ne9yIlE4cDWtgPqpcrwxS uIIj4UsNDZhRuePUbLflRNUuHrmjWUvtOrkbJK6P7eCcgrty01yN1
fwnWTdL4rdJiiadgzKxAweqleCQcEiPksKqlkAp U Q kkK1Q3Ht2m265LZ45qcz2eN4e5Z1lQjnnUW 0XV8YiplmHePaNyOBpIzJM8zidPRdhYl4KDI1iINMhBU164 unbymHQ

By using equation (i), we obtain

1gGnFyZUssm YvLLQuUtRVp acpthEo0TtlfMvot5oaY4LjrskT2ofVqC9orSU25azjwsPSBzmypzeFlCmsoSe0Mpnf429E3PSuKCpPRZoJ7j z2zQkvpac8l4q A1d9yXHnv8g
MYFaL TgsD6hTI5tOkR DhXIb7cVVwgAzFRrBPHwBVuz6McKOloQN5Bbh2Mrg1fITyrYErMznUVc52bvGHqWwBOOP4qD0y5hrB3UbpA7Nr5onEJB1hiSkesKfD7LOL fIj 7wc

Number of rows

Number of students in a row

Number of total students in a class

Q5 :

In a ΔABC, Find the three angles.

Answer :

Given that,

Lxj74WFGmAMcxqlq1AIW1lU6dGHr9otE5kCMcGmxqQ09Z0c0AIwbyQnh
DiHedsmHGut0oelmxIJyccNbNIjOsTEvw4veWLy6lLQ8SMWEPqf6XjdMpPbS8apv56I66q1 joCt8uN3DBwT pGciOlJaroPhuL9 1temaCAwfKk2b6g9Umg61OCW 2ECTlOV2I

v9eVG5CSWJtBpuxPPgNCPc5Hlx KBymTg90W hNhsB GM5x2hohaxU6PLNI 3ulmHW07lzwbSeRfRtMVmOPo … (i)

We know that the sum of the measures of all angles of a triangle is 180°. Therefore,

vj3hTQSIyCxxoTyJQfduzy57Xt0iZy1MP9LC vxaWs2LXQxt5aW0QkWANZu1jPNJjyIAIvCz4iBh7EIgiMObu3wvkuWiDL867Y5

∠ A + ∠ B + 3 ∠ B = 180°

∠ A + 4 ∠ B = 180° … (ii)

Multiplying equation (i) by 4, we obtain 8 ∠ A – 4 ∠ B = 0 … (iii)

Adding equations (ii) and (iii), we obtain 9 ∠ A = 180°

∠ A = 20°

From equation (ii), we obtain 20° + 4 ∠ B = 180°

4 ∠ B = 160°

∠ B = 40°

∠ C = 3 ∠ B

= 3 x 40° = 120°

Therefore, ∠ A, ∠ B, ∠ C are 20°, 40°, and 120° respectively.

Q6 :

Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.

Answer:

Or,

The solution table will be as follows.

x012
y– 505
eJbKSV82tYdO0RrYsJIZlsAN9iZ

Or,

The solution table will be as follows.

x012
y– 303

The graphical representation of these lines will be as follows.

J6NRa9avMdjB6zyvqpbvRdnDLYeLa z 1m1eLzT9fRBt1CHAOzHcpOZaHxVnH2lbZvXr2m6hqAufFIxPyyxWG1KVMu h171p2UAFeV aRqPRu1o00dLFDiygIR 9C1sg0NY2JrE

It can be observed that the required triangle is ΔABC formed by these lines and y-axis. The coordinates of vertices are A (1, 0), B (0, – 3), C (0, – 5).

Q7:

Solve the following pair of linear equations.

(i) px + qy = p – q, qx – py = p + q

(ii) ax + by = c, bx + ay = 1 + c

(iii)

(iv)

cF9oHVXYv4bEVCB0NDFZhcjbnpexcqS1P
c68sG8isn41SjELNl18xINxIZaUAEh56v t IHGO NQQj1NgOWSTfD9JeOIsdnNY0W18cJ6mOIJq06JzV0JJhFepA9NxLziBCQf5k2ZFKj3vVBOxnY BRE OJpbE8qnYUsDPy7o

(v)

O9laI1UVeb9hCKqhg3Bp MPBMh6qvS0hyUAPZvhpBXyRPvX6B99OuMv0tJznCExCsZ4K2rDSTjuAWuKbL5p CAee2NRTHGAy22r AHCCBxfIDQ0yZvOSX fRB

Answer :

(i) … (1)

… (2)

Multiplying equation (1) by p and equation (2) by q, we obtain

… (3)

… (4)

Adding equations (3) and (4), we obtain

l519v3wHJEPIzGaTQmJZlw9N2ICXQtTwEmuQbpiDjhSokiK PeWHRc9nWrLZnMFSwDURYByluceZI7eG41w ygp9wVBLxtKOecY u5EPgXhQP G0nxG3bIC PY
Qb78h1HnqBtLWVsyhizaAzmEapQn JvuYEGdW

From equation (1), we obtain

2bp2Ozx1IZ lHblFX9 dpMYwCemBDc7hPtIUgE6YAzSvSByvRzSMGpuPZ90b7oGSwhkDUJ0NvwJQ7eZpLjTRx2A5kio9 E R5S1kQku9VnLYh YNOliAg tMGnADlAK0EwjzI74

(ii) … (1)

… (2)

Multiplying equation (1) by a and equation (2) by b, we obtain

… (3)

… (4)

Subtracting equation (4) from equation (3), (a2 – b2) x = ac – bc – b
ESHl8pil X OfPMnpYDuk sz xtBpmy34rMGv77MpcllOiNfuc7yLhhJnz vsqvv0YJjBaEszaS2lWSjDmW8WMH4Ia4UYm0KqtKpOtMHESXcoOeLuFlOrAqQVfLqswI4jPqjVNk

From equation (1), we obtain

qHcvw1aG4EyXltF1ufFlDbBkzo3zxdkjyE 3XopjyuEiwJu58q9g HL6puzyWtWraV 2dLtGh1AiSKYoKp01v2f6akTgloOrYKjUOcE247lmN CaN8ocIuw1w

1ycSniYiqEjmGDePNTL0Q2FjxlS1iWz32RhWRjAGGlZOYC2LZT2ULzhe ci6ZMRDZiae9CcecOMt2NNsQcZizyO7MqUqfm2QFZ0GSOClUp4mjkSHeap ZJ

(iii)

Or, … (1)

… (2)

Multiplying equation (1) and (2) by b and a respectively, we obtain

… (3)

… (4)

Adding equations (3) and (4), we obtain

51X 4vQstAoYdLFCUBfZE2E0wg7ju0TB fYKI3YNzRkxAlFzXCSTa YiKesSiNRgBSRhhB1xYQKht9LvKw6lnwluGj0DvaC143NG lp5a2cYNp2Te 8OGYMg2gZttRLpBSwJbOQ
bgmwoTtwYppSXJHBPu3YsYlcSETz3hjgK crwAoHzasCgR2U877jKKJN4qu6CzY 6gfMXpMPTXzRFtGBQB 1CWgfVOKcQ 8JBJW4 fg tnUj EGYMzQnB5fZaMeM4l64oaGc7I

By using (1), we obtain

lBDT3mA h2zYpGIjr5YkOGN HWT1VMfr646G0jFV81FJaDZTF iJV6alNsXLZQLT7mkhrlGCMAn8HZSrs6k87CBrX2gAilPpkqD1gSC9yuoI4swdPytROvjhwNLr gHBDD6ye E
4Aqs5E75b6m3wx09vhi16paw1M2y8LGkoanw5r2Q7Szhw4xwtbxgbPA6MP0oEFigs2SmG00qjBI8XMaS6Hd33VTu0av 7EXhyfMQnVachT6dtV xG0jbgz7QgtlN w15OUNtOP0

(iv)

… (2) Subtracting equation (2) from (1), we obtain

0kuy8rzthvFGn34snZ0bsq01sVIjKbn9hYDicHVQaGZ0M4Mf KD8i4bn2wFVvMg6KVuK3LCOXYTDhppAbH5QZi2qDWuuf9OFD9mDIXDQJygSFp4OfzQhllsquhPYeq5dmDNelpE

Using equation (1), we obtain

arC41Pjq0xaZw5526wdxmdO MJVaB7k3dTK3HUtUlAi WZIrhkr2MH3PYr1dU3bk16Wbh8PEMPd 7t OEJNMWqvxJDKmZB6V4oA4qQEYAeJLSC mADfitLAL5EUWsvOvbVpGgs
7lMw gEIegOqIRxUZ0vazps3Km7yQnhX

(v)

Q8 :

ABCD is a cyclic quadrilateral finds the angles of the cyclic quadrilateral.

Answer :

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°. Therefore, ∠ A + ∠ C = 180

4y + 20 – 4x = 180

-4x + 4y = 160

x – y = – 40 (i)

Also, ∠ B + ∠ D = 180 3y – 5 – 7x + 5 = 180

– 7x + 3y = 180 (ii)

Multiplying equation (i) by 3, we obtain 3x – 3y = – 120 (iii)

Adding equations (ii) and (iii), we obtain

– 7x + 3x = 180 – 120

– 4x = 60

x = -15

By using equation (i), we obtain

x – y = – 40

-15 – y = – 40

y = -15 + 40 = 25

∠ A = 4y + 20 = 4(25) + 20 = 120°

∠ B = 3y – 5 = 3(25) – 5 = 70°

∠ C = – 4x = – 4(- 15) = 60°

∠ D = – 7x + 5 = – 7(-15) + 5 = 110°

+ 5 = – 7(-15) + 5 = 110°

NCERT Solutions for Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables

Chapter 3- Pair of Linear Equations in Two Variables holds a weightage of 11 marks in the examinations. This chapter gives an account of the various topics related to the Pair of Linear Equations in Two Variables .The topics discussed in the chapter are mentioned below:

3.1 Introduction
In earlier classes, you have studied Linear Equations in Two Variables. You have also studied that a Linear Equation in Two Variables has infinitely many solutions. In this chapter, the knowledge of Linear Equations in Two Variables shall be recalled and extended to that of Pair of Linear Equations In Two Variables. 
3.2 Pair of Linear Equations in Two Variables
An equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not zeros, is called a linear equation in two variables x and y. The solution of such a problem is a pair of values, one for x and the other for y, which makes the two sides of the equation equal. The topic also discusses the geometrical representation of Pair of Linear Equations in Two Variables along with suitable examples.
3.3 Graphical Method of Solution of a Pair of Linear Equations
In the previous section, you have seen how we can graphically represent a pair of linear equations as two lines. You have also seen that the lines may intersect, or may be parallel, or may coincide. In this section, you will know how to solve it in each case from the geometrical point of view.
3.4 Algebraic Methods of Solving a Pair of Linear Equations
In the previous section, we discussed how to solve a pair of linear equations graphically. In some of the cases, the graphical method is not convenient. In this topic, we shall discuss various algebraic methods such as the Substitution Method, Elimination Method, and Cross – Multiplication Method. Each subtopic is explained elaborately with suitable examples, for better understanding.
3.5 Equations Reducible to a Pair of Linear Equation in Two Variables
In this section, we shall discuss the solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitutions. The process is explained through some examples related to the subtopic.
3.6 Summary
The Summary section consists of the overall important points that need to be memorized while solving the exercise questions of the chapter Pair of Linear Equations in Two Variables. The points mentioned in this section will help you to revise all the concepts mentioned in the chapter.
Two linear equations in the same two variables are called pair of linear equations in two variables. The pair of linear equations in two variables can be represented graphically and algebraically. The graph can be represented by two lines:

  • If the lines intersect at a point, the pair of equations is said to be consistent.
  • If the lines coincide, the pair of equations is dependent.
  • If the lines are parallel, the pair of equations is inconsistent.

Algebraically the following methods can be used to solve the pair of linear equations in two variables:

  • Substitution method
  • Elimination method
  • Cross-multiplication method

Key Features of NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables

  • NCERT Solutions are created by subject experts.
  • The answers are provided after a lot of brainstorming and are accurate.
  • These contain questions related to all the important topics.
  • NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables includes the solutions to the exercises given in the textbook as well.

Conclusion The NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables provided by Swastik Classes are an excellent resource for students who want to understand the fundamental concepts of linear equations in two variables. Our expert team has created the solutions in a step-by-step approach to enable students to understand the concepts thoroughly.
Through our solutions, students can improve their problem-solving skills and prepare well for their board exams. These solutions are prepared based on the latest CBSE guidelines and are strictly based on the NCERT textbook.
We believe that our solutions will help students achieve their academic goals and pave the way for their success. We encourage students to use our solutions and strengthen their concepts in this chapter.

What are the important topics from an exam perspective present in NCERT Solutions for Class 10 Maths Chapter 3?

The important topics present in NCERT Solutions for Class 10 Maths Chapter 3 are substitution method, elimination method and cross-multiplication method of pair of linear equations in two variables. By solving problems based on these concepts students can score well in Class 10 board exams.

Is it necessary to learn all three methods to solve pairs of linear equations in two variables of NCERT Solutions for Class 10 Maths Chapter 3?

Yes, it is compulsory to learn all three methods to solve a pair of linear equations in two variables of NCERT Solutions for Class 10 Maths Chapter 3. These topics will continue in higher studies as well as it may come in their finals of Class 10. These concepts are explained by SWC’S flawlessly. Hence, the main aim of these solutions formulated SWC’S experts is to provide knowledge on the fundamental aspects of Maths, which in turn, helps students to understand every concept clearly.

How to start reading NCERT Solutions for Class 10 Maths Chapter 3?

Very first learn and understand the definition of linear equations by the help of SWC’s website. Then go through the solutions provided by SWC’S experts at their site. By solving these exercises one can be thorough in all the concepts present in NCERT Solutions for Class 10 Maths Chapter 3.


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