NCERT Class 10 Maths – Chapter 3
Pair of Linear Equations in Two Variables
Exercise 3.2 – Solutions
Class: 10
Subject: Mathematics
Chapter: Pair of Linear Equations in Two Variables
Exercise: 3.2
Question 1: Solve by Substitution Method
(i) x + y = 14 ; x − y = 4
Add both equations:
2x = 18
x = 9
Substitute in x + y = 14:
9 + y = 14
y = 5
Solution: x = 9, y = 5
(ii) s − t = 3 ; s/3 + t/2 = 6
From first: s = t + 3
Substitute:
(t + 3)/3 + t/2 = 6
Multiply by 6:
2(t + 3) + 3t = 36
2t + 6 + 3t = 36
5t = 30
t = 6
s = 6 + 3 = 9
Solution: s = 9, t = 6
(iii) 3x − y = 3 ; 9x − 3y = 9
Second equation is 3 × first equation.
Both represent the same line.
Infinitely many solutions.
(iv) 0.2x + 0.3y = 1.3 ; 0.4x + 0.5y = 2.3
Multiply first by 10:
2x + 3y = 13
Multiply second by 10:
4x + 5y = 23
Multiply first by 2:
4x + 6y = 26
Subtract:
y = 3
Substitute:
2x + 9 = 13
2x = 4
x = 2
Solution: x = 2, y = 3
(v) √2x + √3y = 0 ; √3x − √8y = 0
Solve simultaneously → x = 0, y = 0
Solution: x = 0, y = 0
(vi) 3x/2 − 5y/3 = −2 ; x/3 + y/2 = 13/6
Multiply first by 6:
9x − 10y = −12
Multiply second by 6:
2x + 3y = 13
Solve simultaneously:
x = 1, y = 3
Solution: x = 1, y = 3
Question 2
2x + 3y = 11
2x − 4y = −24
Subtract equations:
7y = 35
y = 5
Substitute:
2x + 15 = 11
2x = −4
x = −2
Given y = mx + 3
5 = m(−2) + 3
5 = −2m + 3
−2m = 2
m = −1
Value of m = −1
Question 3
(i) Difference is 26 and one is three times the other
Let numbers be x and y.
x − y = 26
x = 3y
Substitute:
3y − y = 26
2y = 26
y = 13
x = 39
Numbers: 39 and 13
(ii) Supplementary angles differ by 18°
Let angles be x and y.
x + y = 180
x − y = 18
Solving:
x = 99°, y = 81°
Angles: 99° and 81°
(iii) Cost of bats and balls
Let cost of bat = x
Cost of ball = y
7x + 6y = 3800
3x + 5y = 1750
Solving:
x = 500
y = 50
Bat = ₹500, Ball = ₹50
(iv) Taxi charges
Let fixed charge = x
Charge per km = y
x + 10y = 105
x + 15y = 155
Subtract:
5y = 50
y = 10
x + 100 = 105
x = 5
For 25 km:
5 + 25(10) = 255
Fixed charge = ₹5, Per km = ₹10, Fare for 25 km = ₹255
(v) Fraction problem
Let numerator = x, denominator = y
(x + 2)/(y + 2) = 9/11
(x + 3)/(y + 3) = 5/6
Solving simultaneously:
x = 1, y = 2
Required fraction = 1/2
Important Concepts Used
- Substitution Method
- Forming Linear Equations from Word Problems
- Solving Simultaneous Equations
Keywords:
NCERT Class 10 Maths Exercise 3.2, Substitution Method, Pair of Linear Equations Solutions, Chapter 3 Maths Solutions
Category:
Class 10 Maths