 # Linear equations in One Variable Exercise 2.1|NCERT Solutions for Class 8 Maths

## Linear equations in One Variable Chapter 2 Exercise 2.1

In this page we have NCERT Solutions for Class 8 Maths Chapter 2 : Linear equations in One Variable for Exercise 2.1. Hope you like them and do not forget to like , social share and comment at the end of the page.
Solving Equations which have Linear Expressions on one Side and Numbers on the other Side Question 1:
Solve: x-2=7
x - 2 = 7
Transposing 2 to R.H.S, we obtain
x = 7 + 2 = 9

Question 2:
Solve: y+3=10
y + 3 = 10
Transposing 3 to R.H.S, we obtain
y = 10 - 3 = 7

Question 3:
Solve: 6=z+2
6=z+2
Or
z+2=6
Transposing 2 to R.H.S, we obtain
z=6 -2=4

Question 4
Solve:
$\frac {3}{7} + x =\frac {17}{7}$

$\frac {3}{7} + x =\frac {17}{7}$
Multiplying 7 on both sides
3+7x=17
Transposing 3 to R.H.S, we obtain
7x=17-3=14
7x=14
Dividing both the sides by 7
x=2

Question 5:
Solve:
6x=12

6x=12
Dividing both the sides by 6
x=2

Question 6:
Solve:
$\frac {t}{5}=10$
$\frac {t}{5}=10$
Multiplying 5 on both sides
t=50

Question 7:
Solve
$\frac {2x}{3}=18$
$\frac {2x}{3}=18$
Multiplying both the sides by 3
2x=54
Dividing both the sides by 2
x=27

Question 8:
Solve:
$1.6 = \frac {y}{1.5}$

$1.6 = \frac {y}{1.5}$
Multiplying both the sides by 1.5
1.6 × 1.5=y
y=2.4

Question 9:
Solve:
7x-9=16
7x-9=16
Transposing 9 to R.H.S, we obtain
7x=16+9
7x=25
Dividing by 7 on both the sides
x=25/7

Question 10:
Solve:
14y-8=13

14y-8=13
Transposing 8 to R.H.S, we obtain
14y=13+8
14y=21
Dividing 14 on both the sides
y=21/14=3/2

Question 11:
Solve:
17+6p=9
17+6p=9
Transposing 17 to R.H.S, we obtain
6p=9-17
6p=-8
Dividing 6 on both the sides
p=-8/6=-4/3

Question 12:
Solve:
$\frac {x}{3}+1=\frac {7}{15}$

$\frac {x}{3}+1=\frac {7}{15}$
The LCM of denominator is 15, So multiplying both sides by 15
5x +15=7
Transposing 15 to R.H.S, we obtain
5x=-8
Dividing 5 on both the sides
x=-8/5