NCERT Solutions for Class 10 Maths Exercise 3.3 Chapter 3 Pair of Linear Equations in Two Variables – FREE PDF Download
NCERT Class 10 Maths Ch 3 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter
3 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Pair of Linear Equations in Two Variables solutions will help you understand the chapter thoroughly.
NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables
1. Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 4
(ii) s – t = 3
(iii) 3x – y = 3
9x − 3y = 9
(iv)0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v) 
(vi) 
x – y = 4 … (2)
x = 4 + y from equation (2)
Putting this in equation (1), we get
4 + y + y = 14
⇒ 2y = 10⇒ y = 5
Putting value of y in equation (1), we get
x + 5 = 14
⇒ x = 14 – 5 = 9
Therefore, x = 9 and y = 5
(ii) s – t = 3 … (1)
…(2)
Using equation (1), we can say that s = 3 + t
Putting this in equation (2), we get
⇒ 6+2t+3t6=66+2t+3t6=6
⇒ 5t + 6 = 36
⇒ 5t = 30⇒ t = 6
Putting value of t in equation (1), we get
s – 6 = 3⇒ s = 3 + 6 = 9
Therefore, t = 6 and s = 9
(iii) 3x – y = 3 … (1)
9x − 3y = 9 … (2)
Comparing equation 3x – y = 3 with 
and equation 9x − 3y = 9 with 
,
We get 
Here 
Therefore, we have infinite many solutions for x and y
(iv) 0.2x + 0.3y = 1.3 … (1)
0.4x + 0.5y = 2.3 … (2)
Using equation (1), we can say that
0.2x = 1.3 − 0.3y
⇒ x = 
Putting this in equation (2), we get
0.4 
+ 0.5y = 2.3
⇒ 2.6 − 0.6y + 0.5y = 2.3
⇒ −0.1y = −0.3 ⇒ y = 3
Putting value of y in (1), we get
0.2x + 0.3 (3) = 1.3
⇒ 0.2x + 0.9 = 1.3
⇒ 0.2x = 0.4 ⇒ x = 2
Therefore, x = 2 and y = 3
(v) ……….(1)
……….(2)
Using equation (1), we can say that
x = 
Putting this in equation (2), we get
⇒ 
⇒ 
⇒ y = 0
Putting value of y in (1), we get x = 0
Therefore, x = 0 and y = 0
(vi) … (1)
… (2)
Using equation (2), we can say that
x = 
⇒ x = 
Putting this in equation (1), we get
⇒ 
⇒ 
⇒ 
⇒ 
⇒ y = 3
Putting value of y in equation (2), we get
⇒ 
⇒ 
⇒ x = 2
Therefore, x = 2 and y = 3
2. Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which
y = mx + 3.
2x − 4y = −24 … (2)
Using equation (2), we can say that
2x = −24 + 4y
⇒ x = −12 + 2y
Putting this in equation (1), we get
2 (−12 + 2y) + 3y = 11
⇒ −24 + 4y + 3y = 11
⇒ 7y = 35 ⇒ y = 5
Putting value of y in equation (1), we get
2x + 3 (5) = 11
⇒ 2x + 15 = 11
⇒ 2x = 11 – 15 = −4⇒ x = −2
Therefore, x = −2 and y = 5
Putting values of x and y in y = mx + 3, we get
5 = m (−2) + 3
⇒ 5 = −2m + 3
⇒ −2m = 2 ⇒ m = −1
3. Form a 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 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?
According to given conditions, we have
x – y = 26 (assuming x > y) … (1)
x = 3y(x > y)… (2)
Putting equation (2) in (1), we get
3y – y = 26
⇒ 2y = 26
⇒ y = 13
Putting value of y in equation (2), we get
x = 3y = 
Therefore, two numbers are 13 and 39.
(ii) Let smaller angle =x and let larger angle =y
According to given conditions, we have
y = x + 18 … (1)
Also, 
(Sum of supplementary angles) … (2)
Putting (1) in equation (2), we get
x + x + 18 = 180
⇒ 2x = 180 – 18 = 162
⇒ 
Putting value of x in equation (1), we get
y = x + 18 = 81 + 18 = 
Therefore, two angles are 
.
(iii) Let cost of each bat = Rs x and let cost of each ball = Rs y
According to given conditions, we have
7x + 6y = 3800 … (1)
And,3x + 5y = 1750 … (2)
Using equation (1), we can say that
7x = 3800 − 6y ⇒ x = 
Putting this in equation (2), we get
3 
+ 5y = 1750
⇒ 
+ 5y = 1750
⇒ 
⇒ 
⇒ 17y = 850 ⇒ y = 50
Putting value of y in (2), we get
3x + 250 = 1750
⇒ 3x = 1500 ⇒ x = 500
Therefore, cost of each bat = Rs 500 and cost of each ball = Rs 50
(iv) Let fixed charge = Rs x and let charge for every km = Rs y
According to given conditions, we have
x + 10y = 105… (1)
x + 15y = 155… (2)
Using equation (1), we can say that
x = 105 − 10y
Putting this in equation (2), we get
105 − 10y + 15y = 155
⇒ 5y = 50 ⇒ y = 10
Putting value of y in equation (1), we get
x + 10 (10) = 105
⇒ x = 105 – 100 = 5
Therefore, fixed charge = Rs 5 and charge per km = Rs 10
To travel distance of 25 Km, person will have to pay = Rs (x + 25y)
= Rs (5 + 25 × 10)
= Rs (5 + 250) = Rs 255
(v) Let numerator = x and let denominator = y
According to given conditions, we have
… (1)
… (2)
Using equation (1), we can say that
11 (x + 2) = 9y + 18
⇒ 11x + 22 = 9y + 18
⇒ 11x = 9y – 4
⇒ x = 
Putting value of x in equation (2), we get
6 
= 5 (y + 3)
⇒ 
⇒ 
⇒ −24+3311=55y−54y11−24+3311=55y−54y11
⇒ y = 9
Putting value of y in (1), we get
⇒ x + 2 = 9 ⇒ x = 7
Therefore, fraction =
(vi) Let present age of Jacob = x years
Let present age of Jacob’s son = y years
According to given conditions, we have
(x + 5) = 3 (y + 5) … (1)
And, (x − 5) = 7 (y − 5) … (2)
From equation (1), we can say that
x + 5 = 3y + 15
⇒ x = 10 + 3y
Putting value of x in equation (2) we get
10 + 3y – 5 = 7y − 35
⇒ −4y = −40
⇒ y = 10 years
Putting value of y in equation (1), we get
x + 5 = 3 (10 + 5) = 
⇒ x = 45 – 5 = 40 years
Therefore, present age of Jacob = 40 years and, present age of Jacob’s son = 10 years
