Questions and Solutions for NCERT Class 8th Mathematics Chapter 2 -Exercise 2.1

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NCERT Class 8th Mathematics Chapter 2 – Linear Equations in One Variable Solutions Exercise 2.1 Solutions
                                                                       Exercise 2.1

Q1.x-2=7

Sol: x-2=7

[Adding 2 both sides]

x-2+2=7+2

x=9

Q2.y+3=10

Sol:y+3=10

[Subtracting 3from both sides]

y+3-3=10-3

y=7

Q3.6=z+2

Sol:6=z+2

[Subtracting 2 from both sides]
6-2=z+2-2

z=4

$\mathit{Q}\mathbf{4}\mathbf{.}\frac{\mathbf{3}}{\mathbf{7}}\mathbf{+}\mathit{X}\mathbf{=}\frac{\mathbf{17}}{\mathbf{7}}$

Sol:

$\frac{3}{7}+X=\frac{17}{7}$

[Subtracting 3/7 from both sides]

$\frac{3}{7}+X–\frac{3}{7}=\frac{17}{7}–\frac{3}{7}$

$X=\frac{17–3}{7}$

$X=\frac{14}{7}$

X=2

Q5.6x=12

Sol:6x=12

[Dividing both sides by 6]

$\frac{6x}{6}=\frac{12}{6}$

x=2

$\mathit{Q}\mathbf{6}\mathbf{.}\frac{\mathbf{t}}{\mathbf{5}}\mathbf{=}\mathbf{10}$

Sol:

$\frac{t}{5}=10$

[Multiplying both sides by 5]

$\frac{t}{5}\times 5=10\times 5$

t=50

$\mathit{Q}\mathbf{7}\mathbf{.}\frac{\mathbf{2}\mathbf{x}}{\mathbf{3}}\mathbf{=}\mathbf{18}$

Sol:

$\frac{2x}{3}=18$

[Multiplying both sides by 3 ]

$\frac{2x}{3}\times 3=18\times 3$

2x=54

[Dividing both sides by 2]

$\frac{2x}{2}=\frac{54}{2}$

x=27

$\mathit{Q}\mathbf{8}\mathbf{.}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{=}\frac{\mathbf{y}}{\mathbf{1}\mathbf{.}\mathbf{5}}$

Sol:

$1.6=\frac{y}{1.5}$

[Multiplying both sides by 1.5]

$1.6\times 1.5=\frac{y}{1.5}\times 1.5$

y=2.40

Q9.7x-9=16

Sol:7x-9=16

[Adding 9 both sides]
7x-9+9=16+9

7x=25
[Dividing both sides by 7]

$\frac{7x}{7}=\frac{25}{7}$

$x=\frac{25}{7}$

Q10.14y-8=13

Sol:14y-8=13

[Adding 8 both sides]

14y-8+8=13+8
14y=21

[Dividing both sides by 14]

$\frac{14y}{21}=\frac{21}{14}$ $y=\frac{3}{2}$

Q11.17+6p=9

Sol:17+6p=9

[Subtracting 17from both sides]

17+6p-17=9-17

6p=-8
[Dividing both sidesby 6]

$\frac{6p}{6}=\frac{–8}{6}$

$p=\frac{–4}{3}$

$\mathit{Q}\mathbf{12}\mathbf{.}\frac{\mathbf{X}}{\mathbf{3}}\mathbf{+}\mathbf{1}\mathbf{=}\frac{\mathbf{7}}{\mathbf{15}}$

Sol:

$\frac{X}{3}+1=\frac{7}{15}$

[Subtracting 1 from both sides]

$\frac{X}{3}+1–1=\frac{7}{15}–1$

$\frac{X}{3}=\frac{–8}{15}$

[Multiplying both sides by 3]

$\frac{X}{3}\times 3=\frac{–8}{15}\times 3$

$X=\frac{–8}{5}$

Chapter 2-Linear Equations in One Variable Solutions Exercise 2.2 Solutions