# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

## AP Board Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 Book Answers AP Board Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 Book Answers

## Andhra Pradesh State Board Class 10th Maths Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 Books Solutions

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## AP Board Class 10th Maths Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 Textbooks Solutions with Answer PDF Download

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### 10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.3 Textbook Questions and Answers

Question 1.
Solve each of the following pairs of equations by reducing them to a pair of linear equations.
i) 5𝑥−1 + 1𝑦−2 = 2
6𝑥−1 + 3𝑦−2 = 1
Given
5𝑥−1 + 1𝑦−2 = 2
6𝑥−1 + 3𝑦−2 = 1
Put 1𝑥−1 = a and 1𝑦−2 = b,
then the given equations reduce to
5a + b = 2 ……… (1)
6a – 3b = 1 ………. (2) ⇒ b = 721 = 13
Substituting b = 13 in equation (1) we get ⇒ (x – 1) . 1 = 3 × 1
⇒ x – 1 = 3
⇒ x = 3 + 1 = 4
b = 1𝑦−2 ⇒ 13 = 1𝑦−2
⇒ (y – 2) . 1 = 3 × 1
⇒ y – 2 = 3
⇒ y = 3 + 2 = 5
∴ Solution (x, y) = (4, 5)

ii) 𝑥+𝑦𝑥𝑦 = 2;
𝑥−𝑦𝑥𝑦 = 6
Given
𝑥+𝑦𝑥𝑦 = 2
⇒ 𝑥𝑥𝑦 + 𝑦𝑥𝑦 = 2
⇒ 1𝑦 + 1𝑥 = 2
𝑥−𝑦𝑥𝑦 = 6
⇒ 𝑥𝑥𝑦 – 𝑦𝑥𝑦 = 6
⇒ 1𝑦 – 1𝑥 = 6
Take 1𝑥 = a and 1𝑦 = b,
then the given equations reduces to ⇒ b = 82 = 4
Substituting b = 4 in equation (1) we get
a + 4 = 2 ⇒ a = 2 – 4 = -2
but a = 1𝑥 = -2 ⇒ x = −12
b = 1𝑦 = 4 ⇒ y = 14
∴ Solution (x, y) = (−12,14)

iii) 2𝑥√ + 3𝑦√ = 2;
4𝑥√ – 9𝑦√ = -1
Given
2𝑥√ + 3𝑦√ = 2 and 4𝑥√ – 9𝑦√ = -1
Take 1𝑥√ = a and 1𝑦√ = b,
then the given equations reduces to
2a + 3b = 2 …….. (1)
4a – 9b = – 1 …….. (2) ⇒ b = 515 = 13
Substituting b = 13 in equation (1) we get
2a + 3(13) = 2
⇒ 2a + 1 = 2 ⇒ 2a = 2 – 1 ⇒ a = 12 ∴ Solution (x, y) = (4, 9)

iv) 6x + 3y = 6xy
2x + 4y = 5xy
Given
6x + 3y = 6xy
⇒ 6𝑥+3𝑦𝑥𝑦 = 6
⇒ 6𝑥𝑥𝑦 + 3𝑦𝑥𝑦 = 6
⇒ 6𝑦 + 3𝑥 = 6
2x + 4y = 5xy
⇒ 2𝑥+4𝑦𝑥𝑦 = 5
⇒ 2𝑥𝑥𝑦 + 4𝑦𝑥𝑦 = 6
⇒ 2𝑦 + 4𝑥 = 6
Take 1𝑥 = a and 1𝑦 = b,
then the given equations reduces to
3a + 6b = 6 ……. (1)
4a + 2b = 5 ……. (2) ⇒ b = 918 = 12
Substituting b = 12 in equation (1) we get
3a +6(12) = 6
⇒ 3a = 6 – 3
⇒ a = 33 = 1
but a = 1𝑥 = 1 ⇒ x = 1
b = 1𝑦 = 12 ⇒ y = 2
∴ Solution (x, y) = (1, 2)

v) 5𝑥+𝑦 – 2𝑥−𝑦 = -1
15𝑥+𝑦 + 7𝑥−𝑦 = 10
where x ≠ 0, y ≠ 0
Given
5𝑥+𝑦 – 2𝑥−𝑦 = -1 and
15𝑥+𝑦 + 7𝑥−𝑦 = 10
Take 1𝑥+𝑦 = a and 1𝑥−𝑦 = b, then
the given equations reduce to
5a – 2b = – 1 ……… (1)
15a + 7b = 10 ……… (2) ⇒ b = −13−13 = 1
Substituting b = 1 in equation (1) we get
5a – 2(1) = -1
⇒ 5a = -1 + 2
⇒ 5a = 1
⇒ a = 15
but a = 1𝑥+𝑦 = 15 ⇒ x + y = 5
b = 1𝑥−𝑦 = 1 ⇒ x – y = 1
⇒ x = 62 = 3
Solving the above equations
Substituting x = 3 in x + y = 5 we get
3 + y = 5 ⇒ y = 5 – 3 = 2
∴ Solution (x, y) = (3, 2)

vi) 2𝑥 + 3𝑦 = 13
5𝑥 – 4𝑦 = -2
where x ≠ 0, y ≠ 0
Given
2𝑥 + 3𝑦 = 13 and
5𝑥 – 4𝑦 = -2
Take 1𝑥 = a and 1𝑦 = b, then
the given equations reduce to
2a + 3b = 13 ……… (1)
5a – 4b = -2 ……… (2) ⇒ b = 6923 = 3
Substituting b = 3 in equation (1) we get
2a + 3 (3) = 13
⇒ 2a = 13 – 9
⇒ a = 42 = 2
but a = 1𝑥 = 2 ⇒ x = 12
b = 1𝑦 = 3 ⇒ y = 13
∴ Solution (x, y) = (12, 13)

vii) 10𝑥+𝑦 + 2𝑥−𝑦 = 4
15𝑥+𝑦 – 5𝑥−𝑦 = -2
Given
10𝑥+𝑦 + 2𝑥−𝑦 = 4 and
15𝑥+𝑦 – 5𝑥−𝑦 = -2
Take 1𝑥+𝑦 = a and 1𝑥−𝑦 = b, then
the given equations reduce to
10a + 2b = 4 ……… (1)
15a – 5b = – 2 ……… (2) ⇒ b = 1616 = 1
Substituting b = 1 in equation (1) we get
10a + 2(1) = 4
⇒ 10a = 4 – 2
⇒ a = 210 = 15
but a = 1𝑥+𝑦 = 15 ⇒ x + y = 5 ……. (3)
b = 1𝑥−𝑦 = 1 ⇒ x – y = 1 …….. (4) ⇒ x = 62 = 3
Substituting x = 3 in x + y = 5 we get
3 + y = 5 ⇒ y = 5 – 3 = 2
∴ Solution (x, y) = (3, 2)

viii) 13𝑥+𝑦 + 13𝑥−𝑦 = 34
12(3𝑥+𝑦) – 12(3𝑥−𝑦) = −18
Given
13𝑥+𝑦 + 13𝑥−𝑦 = 34 and
12(3𝑥+𝑦) – 12(3𝑥−𝑦) = −18
Take 13𝑥+𝑦 = a and 13𝑥−𝑦 = b, then
the given equations reduce to ⇒ a = 28 = 14
Substituting a = 14 in equation (1) we get Solving (3) and (4) ⇒ x = 66 = 1
Substituting x = 1 in 3x + y = 4
⇒ 3(1) + y = 4
⇒ y = 4 – 3 = 1
∴ The solution (x, y) = (1, 1)

Question 2.
Formulate the following problems as a pair of equations and then find their solutions.
i) A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Let the speed of the boat in still water = x kmph
and the speed of the stream = y kmph
then speed in downstream = x + y
Speed in upstream = x – y
and time =  distance  speed
By problem,
30𝑥−𝑦 + 44𝑥+𝑦 = 10
40𝑥−𝑦 + 55𝑥+𝑦 = 13
Take 1𝑥−𝑦 = a and 1𝑥+𝑦 = b, then
the given equations reduce to
30a + 44b = 10 ……… (1)
40a + 55b = 13 ……… (2) ⇒ b = 111
Substituting b = 111 in equation (1) we get ⇒ x = 8
Substituting x = 8 in x – y = 5 we get
8 – y = 5
⇒ y = 8 – 5 = 3
∴ The solution (x, y) = (8, 3)
Speed of the boat in still water = 8 kmph
Speed of the stream = 3 kmph.

ii) Rahim travels 600 km to his home partly by train and partly by car. He takes 8 hours if he travels 120 km by train and rest by car. He takes 20 minutes more if he travels 200 km by train and rest by car. Find the speed of the train and the car.
Let the speed of the train be x kmph
and the speed of the car = y kmph
By problem, Take 1𝑥 = a and 1𝑦 = b, then
the given equations reduce to
15a + 60b = 1 ……… (1)
8a + 16b = 13 ⇒ 24a + 48b = 1 ……… (2) ⇒ a = −1−60 = 160
Substituting a = 160 in equation (1) we get but a = 1𝑥 = 160 ⇒ x = 60 kmph
b = 1𝑦 = 180 ⇒ y = 80 kmph
Speed of the train = 60 kmph and
speed of the car = 80 kmph

iii) 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 and 1 man alone to finish the work.
Let the time taken by 1 woman to complete the work = x days
and time taken by 1 man to complete the work = y days
∴ Work done by 1 woman in 1 day = 1𝑥
Work done by 1 man in 1 day = 1𝑦
By problem, Take 1𝑥 = a and 1𝑦 = b,
then the above equations reduce to
2a + 5b = 14 and 3a + 6b = 13
⇒ 8a + 20b = 1 …….. (1) and
9a + 18b = 1 ……… (2) ⇒ b = 136
Substituting b = 136 in equation (1) we get but a = 1𝑥 = 118 ⇒ x = 18 and
b = 1𝑦 = 136 ⇒ y = 36
∴ Time taken by 1 woman = 18 days
1 man = 36 days

## Andhra Pradesh Board Class 10th Maths Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 Textbooks for Exam Preparations

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