# AP 8th Class Maths 2nd Chapter Linear Equations in One Variable Exercise 2.1 Solutions

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
x – 2 + 2 = 7 + 2
∴ x = 9

IInd Method:
x – 2 = 7
x = 7 + 2 (Transposing -2 to R.H.S)
∴ x = 9

Question 2.
y + 3 = 10
Subtracting ‘3’ from both sides
y + 3 – 3 = 10 – 3
∴ y = 7

IInd Method:
y + 3 = 10
y = 10 – 3 (Transposing + 3 to RHS)
∴ y = 7

Question 3.
6 = z + 2
Subtracting ‘2’ from both sides 2
6 – 2 = z + 2 – 2
4 = z
So z = 4

IInd Method:
6 = z + 2 (Transposing + 2 to LHS)
6 – 2 = z
4 = z [re-shaping]
z = 4

Question 4.
$$\frac{3}{7}$$ + x = $$\frac{17}{7}$$
Subtracting $$\frac{3}{7}$$ from both sides

x = $$\frac{17-3}{7}=\frac{14}{7}$$ = 2
∴ x = 2

IInd 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
Dividing by ‘6’ on both sides
$$\frac{6 x}{6}=\frac{12}{6}$$
∴ x = 2

IInd Method:
6x = 12
x = $$\frac{12}{6}$$ (Shifting 6 to RHS)
∴ x = 2

Question 6.
$$\frac{t}{5}$$ = 10
Multiplying both sides by ‘5’
we get

∴ t = 50

Question 7.
$$\frac{2x}{3}$$ = 18
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 = $$\frac{y}{1.5}$$
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
7x – 9 + 9 = 16 + 9
7x = 25
∴ x = $$\frac{25}{7}$$

Question 10.
14y – 8 = 13
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
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 = $$\frac{7}{15}$$
= $$\frac{7-15}{15}=\frac{-8}{15}$$
$$\frac{x}{3}=\frac{-8}{15}$$
⇒ x = $$\frac{-8}{5}$$