Well-designed AP Board Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.1 offers step-by-step explanations to help students understand problem-solving strategies.

## Linear Equations in One Variable Class 8 Exercise 2.1 Solutions – 8th Class Maths 2.1 Exercise Solutions

Solve the following equations.

Question 1.

x – 2 = 7

Answer:

Adding ‘2’ on both sides

x – 2 + 2 = 7 + 2

∴ x = 9

II_{nd} Method:

x – 2 = 7

x = 7 + 2 (Transposing -2 to R.H.S)

∴ x = 9

Question 2.

y + 3 = 10

Answer:

Subtracting ‘3’ from both sides

y + 3 – 3 = 10 – 3

∴ y = 7

II_{nd} Method:

y + 3 = 10

y = 10 – 3 (Transposing + 3 to RHS)

∴ y = 7

Question 3.

6 = z + 2

Answer:

Subtracting ‘2’ from both sides 2

6 – 2 = z + 2 – 2

4 = z

So z = 4

II_{nd} Method:

6 = z + 2 (Transposing + 2 to LHS)

6 – 2 = z

4 = z [re-shaping]

z = 4

Question 4.

\(\frac{3}{7}\) + x = \(\frac{17}{7}\)

Answer:

Subtracting \(\frac{3}{7}\) from both sides

x = \(\frac{17-3}{7}=\frac{14}{7}\) = 2

∴ x = 2

II_{nd} Method:

\(\frac{3}{7}\) + x = \(\frac{17}{7}\)

x = \(\frac{17}{7}-\frac{3}{7}\) (Transposing \(\frac{3}{7}\) to RHS)

x = \(\frac{17-3}{7}=\frac{14}{7}\) = 2

∴ x = 2 is the solution.

Question 5.

6x = 12

Answer:

Dividing by ‘6’ on both sides

\(\frac{6 x}{6}=\frac{12}{6}\)

∴ x = 2

II_{nd} Method:

6x = 12

x = \(\frac{12}{6}\) (Shifting 6 to RHS)

∴ x = 2

Question 6.

\(\frac{t}{5}\) = 10

Answer:

Multiplying both sides by ‘5’

we get

∴ t = 50

Question 7.

\(\frac{2x}{3}\) = 18

Answer:

Dividing by ‘3’ on both sides

⇒ 2x = 54

Divided by ‘2’ on both sides

\(\frac{2 x}{2}=\frac{54}{2}\)

⇒ x = 27

Question 8.

1.6 = \(\frac{y}{1.5}\)

Answer:

Re-write it \(\frac{\mathrm{y}}{1.5}\) = 1.6

Multiply by 1.5 on both sides

\(\frac{\mathrm{y}}{1.5}\)(1.5) = (1.6)(1.5) = 2.40

∴ y = 24

Question 9.

7x – 9 = 16

Answer:

Add (9) on both sides

7x – 9 + 9 = 16 + 9

7x = 25

∴ x = \(\frac{25}{7}\)

Question 10.

14y – 8 = 13

Answer:

Transposing – 8 to RHS

14y = 13 + 8 = 21

Dividing by 14 on both sides

y = \(\frac{21}{14}=\frac{3}{2}\)

∴ y = \(\frac{3}{2}\)

Question 11.

17 + 6p = 9

Answer:

Transposing +17 to RHS

6p = 9 – 17 = -8

6p = -8

Now dividing with ‘6’ on both sides

p = \(\frac{-8}{6}=\frac{-4}{3}\)

∴ p = \(\frac{-4}{3}\) is the solution

Question 12.

\(\frac{x}{3}\) + 1 = \(\frac{7}{15}\)

Answer:

Subtracting ‘1’ on both sides we get

= \(\frac{7-15}{15}=\frac{-8}{15}\)

\(\frac{x}{3}=\frac{-8}{15}\)

Multiplying by ‘3’ on both sides

⇒ x = \(\frac{-8}{5}\)