# AP 8th Class Maths 2nd Chapter Linear Equations in One Variable Exercise 2.3 Solutions

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
Subtracting ‘2x’ on both sides

∴ x = 18

Question 2.
5t – 3 = 3t – 5
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
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
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
2x – 1 = 14 – x

⇒ 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
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)
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
$$\frac{2x}{3}$$ + 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
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 – $$\frac{8}{5}$$
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)
3m = 5m – $$\frac{-8}{5}$$
Put m = $$\frac{4}{5}$$ on both sides