Well-designed AP Board Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2 offers step-by-step explanations to help students understand problem-solving strategies.

## Pair of Linear Equations in Two Variables Class 10 Exercise 3.2 Solutions – 10th Class Maths 3.2 Exercise Solutions

Question 1.

Solve the following pair of linear equations by the substitution method.

i) x + y = 14, x – y = 4

ii) s – t = 3, \(\frac{s}{3}+\frac{t}{2}\) = 6

iii) 3x – y = 3, 9x – 3y = 9

iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3

v) \(\sqrt{2}\)x + \(\sqrt{3}\)y = 0, \(\sqrt{3}\)x – \(\sqrt{y}\)y = 0 \(\sqrt{3}\)x – \(\sqrt{8}\)y = 0

vi) \(\frac{3 x}{2}-\frac{5 y}{3}\) = -2, \(\frac{x}{3}+\frac{y}{2}\) = \(\frac{13}{6}\)

Solution:

i) Given linear equations are

x + y = 14 ……..(1)

x – y = 4

x = 4 + y …..(2)

Put (2) in (1)

4 + y + y = 14

4 + 2y = 14

2y = 14 – 4

y = \(\frac{10}{2}\) = 5

Put y = 5 in (2)

x = 4 + 5 = 9

Solution is x = 9 and y = 5

Check:

Put x = 9 and y = 5 in x + y = 14

9 + 5 = 14

14 = 14

LHS = RHS

So, our solution is correct.

ii) Given linear equations are

s – t = 3 …….. (1)

\(\frac{\mathrm{s}}{3}\) + \(\frac{\mathrm{t}}{2}\) = 6

\(\frac{2 s+3 t}{6}\) = 6

2s + 3t = 36 …….. (2)

Put (1) in (2)

2(3 + t) + 3t = 36

6 + 2t + 3t = 36

5t = 36 – 6 = 30

t = \(\frac{30}{5}\) = 6

Therefore solution is s = 9, t = 6

Put t = 6 in (1)

∴ s = 3 + 6 = 9

∴ s = 9 and t = 6

Check: Put s = 9 and t = 6 in 2s + 3t = 3

2(9) + 3(6) = 36

18 + 18 = 36

36 = 36

LHS = RHS

So, our solution is correct.

iii) Given linear equations are

3x – y = 3 and 9x – 3y – 9

3x – y = 3

3x = 3 + y

y = 3x – 3

9x – 3y = 9

Put (1) in (2)

9x – 3(3x – 3) = 9

9x – 9x – 9 = 9

9x – 9x = 9 – 9

0 = 0

So, these linear equations can represent coinsident lines.

Therefore, they have infinitely many solutions.

iv) Given linear equations are

0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3

Therefore solution is x = 2 and y = 3

Check:

Put x = 2 and y = 3 in (2)

4(2) + 5(3) = 23

8 + 15 = 23

23 = 23

LHS = RHS

So, our solution is correct.

v) Given linear equations are

\(\sqrt{2}\)x + \(\sqrt{3}\)y = 0 and \(\sqrt{3}\)x – \(\sqrt{8}\)y = 0 ……… (2)

∴ x = 0 and y = 0

Solution is x = 0 and y = 0

vi) Given linear equations

Therefore solution is x = 2 and y = 3

Check:

Put x = 2 and y = 3 in (2)

2(2) + 3(3) = 13

4 + 9 = 13

13 = 13

So, our solution is correct.

Question 2.

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

Solution:

Given linear equations are

2x + 3y = 11 and 2x – 4y = -24 …… (2)

2x = 11 – 3y

x = \(\frac{11-3 y}{2}\) ……. (1)

Put (1) in (2)

To get m value put x = -2 and y = 5 in

y = mx + 3

5 = m(-2) + 3

-2m = 5 – 3 = 2

Therefore, m = -1.

Question 3.

Form 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 ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 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 ₹ 105 and for a journey of 15 km, the charge paid is ₹ 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 \(\frac{9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.

Solution:

i) Let the two numbers are x and y (x>y)

difference of two numbers = 26

x – y = 26 ……. (1)

and x = 3y ……. (2)

Put (2) in (1) 3y – y = 26

2y = 26

y = \(\frac{26}{2}\) = 13

Put y = 13 in (2) x = 3(13) = 39

Therefore, numbers are 39 and 13.

ii) Let the x and y are supplementary angles (x>y)

i.e., x + y = 180 ……. (1)

and x – y = 18

x = 18 + y ……..(2)

Put (2) in (1) 18 + y + y = 180

2y = 180 – 18 = 162

y = \(\frac{162}{2}\) = 81°

Put y = 81 in (2) x = 18 + y

x = 18 + 81 = 99°

Therefore, angles are 99° and 81°

iii) Let the cost of bat is ₹ x and ball is ₹ y.

7 bats cost 6 balls cost = ₹ 3800

i.e. 7x + 6y = 3800 …….(1)

3 bats cost + 5 balls cost = ₹ 1750

i.e. 3x + 5y = 1750

3x = 1750 – 5y

x = \(\frac{1750-5 y}{3}\) …. (2)

Put (2) in (1)

7\(\left(\frac{1750-5 y}{3}\right)\) + 6y = 3800

\(\frac{7(1750-5 y)+18 y}{3}\) = 3800

12250 – 35y + 18y = 3800 × 3

12250 – 17y = 11400

-17y = 11400 – 12250

iv) Let the fixed charges of taxi be ₹ x and the running charges be ₹ y per km.

then x + 10y = 105 and

x = 105 – 10y…(1)x + 15y = 155…….(2)

Put (1) in (2) 105 – 10y + 15y = 155 + 5y = 155 – 105

5y = 50

y = \(\frac{50}{5}\) = 10

Put y = 10 in (1)

x = 105 – 10(10) = 105 – 100 = 5

Therefore. x = 5 and y 10

So, fixed charges of taxi = ₹ 5

and the running charges ₹ 10 per km.

Charges to pay for traveliling a distance of 2km = x + 25y

= 5 + 25(10)

= 5 + 250 = ₹ 255.

v) Let the numerator be x and denominator be y

then fraction = \(\frac{x}{y}\)

If 2 added to numerator and denominator

11(x + 2) = 9(y + 2)

11x + 22 = 9y + 18

11x – 9y = 18 – 22

11x – 9y = -4

11x = 9y – 4

x = \(\frac{9 y-4}{11}\) …….(1)

If 3 is added to numerator and deminator

\(\frac{x+3}{y+3}\) = \(\frac{5}{6}\)

6(x + 3) = 5(y + 3)

6x + 18 = 5y + 15

6x – 5yv = 15 – 18

6x – 5y = -3 ……. (2)

Put (1) in (2)

3y + 10 – 7y = -30

-4y = -30 – 10

-4y = -40

y = 10

Put y = 10 in (1)

x = 3(10) + 10 = 30 + 10 = 40

Therefore, Jacobs present age (x) = 40 years

and his sons age (y) = 10 years.