# PSEB 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

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) √2x + √3y = 0
√3x – √8y = 0

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

Solution:
(i) Given pair of linear equati3ns
x + y = 14 …………(1)
and x – y = 4
From (2) x = 4 ………….(3)
Substitute this value of x in Equation we get: .
4 + y + y = 14
2y = 14 – 4
2y = 10
y = $$\frac{10}{2}$$ = 5
Substitute this value of y in equation (3), we get:
x = 4 + 5 = 9
Hence x = 9 and y = 5

(ii) Given pair of linear equations
s – t = 3 …………….(1)
and
$$\frac{s}{3}+\frac{t}{2}$$ = 6
$$\frac{2 s+3 t}{6}$$ = 6
2s + 3t = 36 …………….(2)
From (1), s = t + 3 ……………….(3)
Substitute this value of s in equation (1), we get:
2(3 + t) + 3t = 36
Or 6 + 2t + 3t = 36
Or 6 + 5t = 36
Or 5t = 36 – 6
Or 5t = 30
Or t = $$\frac{30}{5}$$ = 6
Substitute this value of r in equation (3), we get:
s = 3 + 6 = 9
Hence, s = 9 and t = 6

(iii) Given pair of linear equation is:
3x – y = 3 .
and 9x – 3y = 9
From (1),
3x – 3 = y
Or y = 3x – 3 ………….(3)
Substitute this value of y in equation (2), we get :
9x – 3(3x – 3) = 9
Or 9x – 9x + 9 = 9
Or 9 = 9
This statement is true for all values of x. However, we do not get a specific value of x as a solution. Therefore we cannot obtain a specific value of y. This situation has arises because both the given equations are same. Therefore, equations (1) and (2) have infinitely many solutions.

(iv) Given pair of linear equation is:
0.2x + 0.3y = 1.3
or $$\frac{2}{10} x+\frac{3}{10} y=\frac{13}{10}$$
or 2x + 3y = 13 ……………..(1)
0.4x + 0.5y = 2.3
or $$\frac{4}{10} x+\frac{5}{10} y=\frac{23}{10}$$
Or 4x + 5y = 23 ……………(2)
From (1),
2x = 13 – 3y
x = $$\frac{13-3 y}{2}$$ …………..(2)
Substitrne this value of x in (2), we get:
4$\frac{13-3 y}{2}$ + 5y = 23
26 – 6y + 5y = 23
-y = 23 – 26 = -3
y = 3
Substitute this value of y in (3), we get:
x = $$\frac{13-3 \times 3}{2}$$
= $$\frac{13-9}{2}=\frac{4}{2}$$ = 2
Hence, x = 2 and y = 3.

(v) Given pair of linear equation is:
√2x + √3y = 0 ……………..(1)
√3x – √8y = 0 …………..(2)
From (2), √3x = √8y
or x = $$\frac{\sqrt{8}}{\sqrt{3}}$$y …………..(3)
Substitute this value of x in (1), we get
√2($$\frac{\sqrt{8}}{\sqrt{3}}$$y) + √3y = 0
or [$$\frac{4}{\sqrt{3}}$$ + √3]y = 0
y = 0
Substitute this value of y in (3), we get:
x = $$\frac{\sqrt{8}}{\sqrt{3}}$$ × 0 = 0
Hence x = 0 and y = 0

(vi) Given pair of linear equation is:
$$\frac{3 x}{2}-\frac{5 y}{3}=-2$$

or $$\frac{9 x-10 y}{6}$$ = -2
or 9x – 10y = -12 …………..(1)
$$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$$
$$\frac{2 x+3 y}{6}=\frac{13}{6}$$
or 2x + 3y = $$\frac{13}{6}$$ × 6
or 2x + 3y = 13 …………..(2)
From (1), 9x = 10y – 12
or x = $$\frac{10 y-12}{9}$$ …………….(3)
2$\frac{10 y-12}{9}$ + 3y = 13

or $$\frac{20 y-24}{9}$$ + 3y = 13

or $$\frac{20 y-24+27 y}{9}$$ = 13

or 47y – 24 = 13 × 9 = 117
47y = 117 + 24 = 141
or y = $$\frac{141}{47}$$ = 3
substitute this value of y in (3), we get
x = $$\frac{10 \times 3-12}{9}=\frac{30-12}{9}$$

= $$\frac{18}{9}$$ = 2
Hence, x = 2 and y = 3

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + 3.
Solution:
Given pair of linear equations is:
2x + 3y = 11
and 2x – 4y = -24 …………(2)
From (2),
2x = 4y – 24
2x = 2 [2y – 12]
Or x = 2y – 12 …………..(3)
Substitute this value of x in (1), we get:
2 (2y – 12) + 3y = 11
Or 4y – 24 + 3y = 11
Or 7y = 11 + 24
Or 7y = 35
y = $$\frac{35}{7}$$ = 5
Substitute this value of y in (3), we get:
x = 2(5) – 12 = 10 – 12 = -2
Now, consider y = mx + 3
Substitute the value of x = -2, y = 5, we get:
5 = m(-2) + 3
Or 5 – 3 = -2m
Or 2 = – 2m
Or -2m = 2
Or m = -1
Hence, x = -2, y = 5 and 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 kilometre? 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 ¡s added to both the numerator and the denominator it becomes $$\frac{5}{6}$$. 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?
Solution:
(I) Let two number be x and y.
According to 1st condition,
x – y=26 ……….(1)
According to 2nd condition,
x = 3y …………(2)
Substitute this value of x in (1), we get :
3y – y = 26
Or 2y = 26
y = $$\frac{26}{2}$$ = 13
Substitute this value of y in (2), we get:
x = 3 × 13 = 39
Hence, two numbers are 39, 13.

(ii) Let, required two supplementary angles are x, y and x > y
According to 1st condition,
x + y = 180 ………..(1)
According to 2nd condition,
x = y + 18 …………..(2)
Substitute this value of x in (1), we get:
y + 18 + y = 180
Or 2y = 180 – 18
or 2y =162
Or y = $$\frac{162}{2}$$ = 81
Substitute this value of y in (2), we get:
x = 81 + 18 = 99
Hence, required angles are 99, 81.

(iii) Let cost of one bat = ₹ x
and cost of one ball = ₹ y
According to 1st condition,
7x + 6y = ₹ 3800 ……………(1)
According to 2nd condition,
3x + 5y = ₹ 1750 …………….(2)
From(1), 7x = 3800 – 6y
Or x =$$\frac{3800-6 y}{7}$$ ………….(3)
substitute this value of x in (2), we get:
3$\frac{3800-6 y}{7}$ + 5y = 1750
Or $$\frac{11400-18 y+35 y}{7}$$ = 1750
Or 11400 + 17y = 1750 × 7
0r 11400 + 17y = 12250
Or 17y = 12250 – 11400
Or 17y = 850
or y = $$\frac{850}{17}$$ = 50
Substitute this value of y in (3), we get:
x = $$\frac{3800-6 \times 50}{7}$$
= $$\frac{3800-300}{7}=\frac{3500}{7}$$
x = 500
Hence cost of one bat = ₹ 500
and cost of one ball = ₹ 50.

(iv) Let the fixed charges for the taxi = ₹ x
and charges for travelling one km = ₹ y
According to 1st condition,
x + 10y = 105 …………..(1)
According to 2nd condition,
x + 15y = 155 ……………(2)
From (1),
x = 105 – 10y …………..(3)
Substitute the value of x in (2), we get
105 – 10y + 15y = 155
Or 5y = 155 – 105
Or 5y = 50
Or y = $$\frac{50}{5}$$ = 10
Substitute the value of y in (3), we get:
x = 105 – 10 × 10
= 105 – 100 = 5
Hence, fixed charges for the taxi = ₹ 5
and charges for travelling one km = ₹ 10
Also, charges for travelling 25 km = ₹(10 × 25) + ₹ 5
= ₹[250 + 5] = ₹ 255

(v) Let numerator of given fraction = x
Denominator of given fraction = y
∴ Required fraction = $$\frac{x}{y}$$
Acccrding to 1st condition,
$$\frac{x+2}{y+2}=\frac{9}{11}$$
Or 11(x + 2) = 9(y + 2)
Or 11x + 22 = 9y + 18
Or 11x = 9y + 18 – 22
Or 11x = 9y – 4
Or x = $$\frac{9 y-4}{11}$$ ……………..(1)
Acccrding to 2nd equation
$$\frac{x+3}{y+3}=\frac{5}{6}$$
Or 6 (x + 3) = 5(y + 3)
Or 6x + 18 = 5y + 15
Or 6x – 5y = 15 – 18
Or 6x – 5y = -3
Putting the value of x from (1). we get:
6$\left[\frac{9 y-4}{11}\right]$ – 5y = -3
Or $$\frac{54 y-24}{11}$$ – 5y = -3
Or $$\frac{54 y-24-55 y}{11}$$ = -3
Or -y – 24 = -3 × 11
Or -y = -33 + 24
Or -y = -9
Or y = 9
Substitute the value of y in (1), we get:
x = $$\frac{9 \times 9-4}{11}=\frac{81-4}{11}$$
= $$\frac{77}{11}$$ = 7
Hence, required fraction is $$\frac{7}{9}$$.

(vi) Let Jacob’s present age = x years
and Jacob son’s present age = y years
Five years hence
Jacob’s age = (x + 5) years
His son’s age = (y + 5)years
According to 1st condition.
x + 5 = 3(y + 5)
Or x + 5 = 3y + 15
Or x = 3y + 15 – 5
Or x = 3y + 10 ……………(1)
Five years ago
Jacobs age = (x – 5) years
His son’s age = (y – 5) years
According to 2nd condition.
x – 5 = 7(y – 5)
Or x – 5 = 7y – 35
Or x – 7y = -35 + 5
Or x – 7y = -30
Substitute the value of x from (1), we get:
3y + 10 – 7y = -30
– 4y = – 30 – 10
-4y = -40
y = 10
Substitute tins value of y in (1), we get:
x = 3 (10) + 10
= 30 + 10 = 40
Hence. Jacob and his son’s ages are 40 years and 10 years respecùveiy.