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Class 10 Maths Chapter 3 Exercise 3.2 Solutions
In this section of Chapter 3 “Pair of Linear Equations in Two Variables“, you have studied the following points ⚡:
- We shall now discuss some algebraic methods of finding the solution which has non-integral coordinates like (√3, 2√7 ), (−1.75, 3.3), (4/13, 1/19), etc.
- In this section, we will learn how to use “substitution method“
- About Substitution Method – We substitute the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations.
In class 10 maths Chapter 3 Exercise 3.2 solutions, let’s try to find the answers to these questions —
EXERCISE 3.2
Question 1: Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x − y = 4
Answer 1 (i): x + y = 14 …… equation (1)
x − y = 4 …… equation (2)
From equation (1), we have y = 14 − x
Putting the value of y in equation (2),
x − (14 − x) = 4
⇒ 2x = 18
⇒ x = 9
So, value of y = 14 − x
Or, y = 14 − 9 = 5
Hence, x = 9 and y = 5 —Ans.
(ii) s − t = 3
s/3 + t/2 = 6
Answer 1 (ii): s − t = 3 …… equation (1)
s/3 + t/2 = 6 …… equation (2)
From equation (1), we have s = 3 + t
Putting the value of s in equation (2),
(3 + t)/3 + t/2 = 6
⇒ (6 +2t)/6 + 3t/6 = 6
⇒ 6 + 5t = 36
⇒ 5t = 36 − 6 = 30
⇒ t = 30/5 = 6
So, value of s = 3 + t
Or, s = 3 + 6 = 9
Hence, s = 9 and t = 6 —Ans.
(iii) 3x − y = 3
9x − 3y = 9
Answer 1 (iii): 3x − y = 3 …… equation (1)
9x − 3y = 9 …… equation (2)
From equation (1), we have −y = 3 − 3x
⇒ y = 3x − 3
Putting the value of y in equation (2),
9x − 3.(3x − 3) = 9
⇒ 9x − 9x + 9 = 9
⇒ 9 = 9, which is true.
Hence, the pair of linear equations have infinite many solution. —Ans.
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
Answer 1 (iv): 0.2x + 0.3y = 1.3 …… equation (1)
0.4x + 0.5y = 2.3 …… equation (2)
From equation (1), we have
y = (1.3 − 0.2x)/0.3
Putting the value of y in equation (2),
0.4x + 0.5[(1.3 − 0.2x)/0.3] = 2.3
⇒ 0.12x + 0.65 − 0.10x = 0.69
⇒ 0.02x = 0.04
⇒ x = 2
So, value of y = (1.3 − 0.2x)/0.3
Or, y = (1.3 − 0.2 × 2)/0.3
⇒ y = 3
Hence, x = 2 and y = 3 —Ans.
(v) √2 x + √3 y = 0
√3 x − √8 y = 0
Answer 1 (v): √2 x + √3 y = 0 …… equation (1)
√3 x − √8 y = 0 …… equation (2)
From equation (1), we have y = − √2x/√3
Putting the value of y in equation (2),
√3x − √2 (−√2x/√3) = 0
⇒ 3x − 2x = 0
⇒ x = 0
Again, value of y = − √2x/√3
Putting the value of x
y = 0
Hence, x = 9 and y = 5 —Ans.
(vi) 3x/2 − 5y/3 = −2
x/3 + y/2 = 13/6
Answer 1 (vi): 3x/2 − 5y/3 = −2 …… equation (1)
x/3 + y/2 = 13/6 …… equation (2)
From equation (1), we have 5/3 y = 3x/2 + 2
⇒ y = (9x + 12)/10
Putting the value of y in equation (2),
x/3 + (9x + 12)/(2 × 10) = 13/6
⇒ 20x/60 + (27x + 36)/60 = 13/6
⇒ 20x + 27x + 36 = 60 × 13/6
⇒ 47x = 130 − 36 = 94
⇒ x = 94/47 = 2
So, value of y = (9x + 12)/10
Or, y = (9×2 + 12)/10
⇒ y = 30/10 = 3
Hence, x = 2 and y = 3 —Ans.
Question 2: Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Answer 2: 2x + 3y = 11 …… equation (1)
2x − 4y = −24 …… equation (2)
From the equation (1), we get
y = (11 − 2x)/3
Putting the value of y in equation (2),
2x − 4.[(11 − 2x)/3] = −24
⇒ 6x − 44 + 8x = −72
⇒ 14x = −28
⇒ x = −2
So, value of y = (11 − 2x)/3
Or, y = (11 − 2 × −2)/3
⇒ y = (11 + 4)/3 = 15/3 = 5
Hence, x = −2 and y = 5.
Putting the value of x and y in y = mx + 3,
5 = −2m + 3
⇒ m = −1 —Ans.
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.
Answer 3 (i): Suppose that the two numbers are x and y.
According to the conditions given in the question —
x − y = 26 …… equation (1)
x = 3y …… equation (2)
Putting the value of x in equation (1),
3y − y = 26
⇒ 2y = 26
⇒ y = 13
Putting the value of y in equation (2),
x = 3(13) = 39
Hence, the two numbers are 13 and 39. —Ans.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Answer 3 (ii): Suppose that the measurements of two supplementary angles are x and y.
According to the conditions given in the question —
x + y = 180 …… equation (1)
x − y = 18 …… equation (2)
⇒ x = 18 + y
Putting this value of x in equation (1),
(18 + y) + y = 180
⇒ 18 + 2y = 180
⇒ y = 81
Putting the value of y in equation (2),
x = 18 + y = 18 + 81 = 99
Hence, the two supplementary angles are 99° and 81°. —Ans.
(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.
Answer 3 (iii): Suppose that the cost of each bat is x and the cost of each ball is y.
According to the conditions given in the question —
7x + 6y = 3800 …… equation (1)
3x + 5y = 1750 …… equation (2)
⇒ 3x = 1750 − 5y
⇒ x = (1750 − 5y)/3 …… equation (3)
Putting this value of x in equation (1),
7 × (1750 − 5y)/3 + 6y = 3800
⇒ (7 × 1750) − (7 × 5y) + (3 × 6y) = 3 × 3800
⇒ −17y = (3 × 3800) − (7 × 1750)
⇒ −17y = −850
⇒ y = 50
Putting the value of y in equation (3),
x = (1750 − 5.50)/3
⇒ x = 1500/3 = 500
Hence, the cost of each bat is ₹ 500 and the cost of each ball is ₹ 50. —Ans.
(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?
Answer 3 (iv): Suppose that the fixed charge is x and the charge per km is y.
For a distance of 10 km, the charge paid is ₹ 105 —
x + 10y = 105 …… equation (1)
For a distance of 15 km, the charge paid is ₹ 155 —
x + 15y = 155 …… equation (2)
⇒ x = 155 − 15y
Putting the value of x in equation (1),
(155 − 15y) + 10y = 105
⇒ −5y = 105 − 155
⇒ y = 50/5 = 10
Putting the value of y in equation (1),
x + 10.10 = 105
⇒ x = 105 − 100 = 5
So, the taxi charges consist of a fixed charge of ₹ 5 together with the charge ₹ 10 per km for the distance covered.
Therefore, the charge for 25 km distance = ₹ 5 + ₹ 10.25
= ₹ 255 —Ans.
(v) A fraction becomes 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 5/6. Find the fraction.
Answer 3 (v): Suppose that the numerator is x and the denominator is y.
According to the first condition given in the question,
(x + 2)/(y + 2) = 9/11
⇒ 11x + 22 = 9y + 18
⇒ x = (9y − 4)/11 …… equation (1)
According to the second condition given in the question,
(x + 3)/(y + 3) = 5/6
⇒ 6x + 18 = 5y + 15
⇒ 6x − 5y = −3 …… equation (2)
Putting the value of x (see first equation) in equation (2),
6.[(9y − 4)/11] − 5y = −3
⇒ 54y − 24 − 55y = −33
⇒ y = 9
Putting this value of y in equation (1),
x = (9.9 − 4)/11 = 7
Hence, the fraction is x/y = 7/9 —Ans.
(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 3 (vi): Let the age of Jacob = x years
Let the age of son = y years
After 5 years —
Jacob’s age is (x + 5) years
and son’s age is (y + 5) years
According to the given condition,
x + 5 = 3.(y + 5)
⇒ x = 3y + 10 …… equation (1)
5 years ago —
Jacob’s age was (x − 5) years
and son’s age is (y − 5) years
x − 5 = 7(y − 5)
⇒ x − 5 = 7y − 35
⇒ x − 7y = −30 …… equation (2)
Putting the value of x in equation (2),
3y + 10 − 7y = −30
⇒ −4y = −40
⇒ y = 10
Putting the value of y in equation (1),
x = 3.10 + 10 = 40
Hence, Jacob is 40 years old, and his son is 10 years old. —Ans.