Well-designed AP Board Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.3 offers step-by-step explanations to help students understand problem-solving strategies.

## Linear Equations in One Variable Class 8 Exercise 2.3 Solutions – 8th Class Maths 2.3 Exercise Solutions

Solve the following equations and check your results.

Question 1.

3x = 2x + 18

Answer:

Subtracting ‘2x’ on both sides

∴ x = 18

Question 2.

5t – 3 = 3t – 5

Answer:

5t – 3 = 3t – 5

Subtracting (3t) on both sides

(5t – 3t) – 3 = -5

2t – 3 = -5 (adding +3 on both sides)

2t = -2

⇒ t = \(\frac{-2}{2}\) = -1

∴ t = -1

Question 3.

5x + 9 = 5 + 3x

Answer:

Adding -3x on both sides we get

5x + 9 – 3x = 5 + – X

⇒ (5x – 3x) + 9 = 5

⇒ 2x + 9 = 5 (Transposing + 9 to RHS)

We get 2x = 5 – 9 = 4

2x = – 4

(now dividing by 2 on both sides)

\(\frac{2 x}{2}=\frac{-4}{2}\)

⇒ x = -2

Question 4.

4z + 3 = 6 + 2z

Answer:

4z + 3 = 6 + 2z

Adding (- 2z) on both sides we get

4z + 3 – 2z = 6 + 2z + (- 2z)

2z + 3 = 6 (transposing + 3 to RHS)

2z = 6 – 3 = 3

(dividing by 2 on both sides, we get)

⇒ z = \(\frac{3}{2}\)

Question 5.

2x – 1 = 14 – x

Answer:

2x – 1 = 14 – x

Adding ‘x’ on both sides

⇒ 2x + x — 1 = 14

⇒ 3x – 1 = 14

(adding 1 on both sides, we get)

3x – 1 + 1 = 14 +1

⇒ 3x = 15 (dividing by ’31)

⇒ x = 5

Question 6.

8x + 43(x – 1) + 7

Answer:

8x + 4 = 3 (x – 1) + 7

8x -t- 4 = 3 (x) + 3 (- 1) + 7 (By distributive law)

8x + 4 = 3x – 3 + 7 (Transpose + 3x to LHS, we get)

8x + 4 – 3x = -3 + 7 (Transpose + 4 to RHS)

8x – 3x = -3 + 7 – 4 = -7 + 7

5x = 0

\(\frac{5 x}{5}=\frac{0}{5}\)

⇒ x = 0

Question 7.

x = \(\frac{4}{5}\)(x + 10)

Answer:

x = \(\frac{4}{5}\)(x + 10) (multiply both sides by 5)

⇒ 5(x) = (X + 10)

⇒ 5x = 4 (x + 10) (By distributive law)

⇒ 5x = 4(x) + 4(10) = 4x + 40

⇒ 5x = 4x + 40 (Transpose + 4x toLHS)

⇒ 5x – 4x = 40

∴ x = 40

Question 8.

\(\frac{2x}{3}\) + 1 = \(\frac{7x}{15}\) + 3

Answer:

\(\frac{2x}{3}\) + 1 = \(\frac{7x}{15}\) + 3 (now multiply by 3 and then by distributivity)

(now multiply by 5 on both sides)

⇒ 5(2x) + 5(3) = 7x + 45

⇒ 10x + 15 = 7x + 45 (now transposing + 7x to LHS, + 15 to RHS)

⇒ 10x – 7x = 45-15

⇒ 3x = 30

(now divide by 3 on both sides)

⇒ \(\frac{3 \mathrm{x}}{3}=\frac{30}{3}\)

⇒ x = 10

Verification: Put x = 10 in given

Question 9.

2y + \(\frac{5}{3}\) = \(\frac{26}{3}\) – y

Answer:

2y + \(\frac{5}{3}\) = \(\frac{26}{3}\) – y (transposing (-y) to LHS)

⇒ 2y + \(\frac{5}{3}\) + y = \(\frac{26}{3}\) (transposing (+\(\frac{5}{3}\)) to LHS)

(now dividing by 3 on both sides)

⇒ 3y = 7

⇒ y = \(\frac{7}{3}\)

Question 10.

3m = 5m – \(\frac{8}{5}\)

Answer:

3m = 5m – \(\frac{8}{5}\) (transposing + 5m to LHS)

⇒ 3m – 5m = \(\frac{-8}{5}\)

⇒ -2m = \(\frac{-8}{5}\) [multiply by (-1)]

⇒ (-2m)(-1) = (\(\frac{-8}{5}\)) (-1)

⇒ 2m = \(\frac{-8}{5}\) (divide by 2 on both sides)

Verification:

3m = 5m – \(\frac{-8}{5}\)

Put m = \(\frac{4}{5}\) on both sides