# Class 8 Maths MCQ – Linear Equations in One Variable

This set of Class 8 Maths Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Linear Equations in One Variable”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.

1. Find the solution for the equation 3x + 3 = 4.
a) 3
b) $$\frac{1}{3}$$
c) 4
d) $$\frac{1}{4}$$

Explanation: 3x + 3 = 4
Step 1: Subtract 3 from both the sides,
3x + 3 – 3 = 4 – 3
3x = 1
Step 2: Divided both the sides by 3,
x = $$\frac{1}{3}$$
So, we have solution for the equation x = $$\frac{1}{3}$$

2. In equation 3x + 4 = 10, by transposing the variable on RHS we get ________
a) -4 = 10 – 3x
b) 4 = 3x + 10
c) 4 = -3x + 10
d) -4 = – 3x – 10

Explanation: While shifting the variable from LHS to RHS the sign changes from positive to negative and vice versa. Here, the given equation is 3x + 4 = 10.

3. Solve equation 7x + 14 = 21 to find value of x.
a) x = 1
b) x =-1
c) x = 2
d) x = -2

Explanation: 7x + 14 = 21
Step 1: Solving the equation, we take constant 14 on the RHS making 7x = 7
Step 2: Dividing LHS and RHS with 7 we get x = 1.

4. Solve: $$\frac{5}{2}x + \frac{3}{2} = \frac{6}{4}$$.
a) x = 1
b) x = 2
c) x = 3
d) x = 0

Explanation: $$\frac{5}{2}x + \frac{3}{2} = \frac{6}{4}$$
Step 1: Reducing the fraction we get,
$$\frac{5}{2}x + \frac{3}{2} = \frac{3}{2}$$
Step 2: Multiplying LHS and RHS by 2 we get,
5x + 3 = 3
i.e. 5x = 0
Step 3: Dividing both sides by 5 we get,
x = 0.

5. Solve: 7x – 2 = 3.
a) x = $$\frac{5}{7}$$
b) x = $$\frac{7}{5}$$
c) x = $$\frac{3}{5}$$
d) x = $$\frac{5}{3}$$

Explanation: 7x – 2 = 3
Step 1: Transposing -2 on RHS we get,
7x = 5
Step 2: Dividing both sides by 7 we get,
x = $$\frac{5}{7}$$.
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6. Pick the equation from the given one’s which have solution as z = 2.
a) 2z -2 = 3
b) 3z -2 = -2
c) 3z -3 = 3
d) 4z + 3 = 3

Explanation: Solving the 3z – 3 = 3 we get,
Step 1: Transposing -3 on RHS we get,
3z = 6
Step 2: Dividing both sides by 3 we get,
z = 2
Hence, this is the solution of our question.

7. Pick the equation which has the solution in the form of prime number.
a) 2x = 3
b) 3z = -6
c) 4y – 3 = 2
d) 2z – 2 = 2

Explanation: A prime number is the number which can be formed by multiplying only two numbers which are 1 and the number itself.
Solving equation, 2z – 2 = 2
Step 1: Transposing – 2 on RHS we get,
2z = 4
Step 2: Dividing equation by 2 we get,
z = 2
Since 2 is greater than 0 and can be formed only by multiplying 1 and 2 it is a prime number.

8. Solve: 16 = 3m – 2.
a) m = -5
b) m = 5
c) m = 6
d) m = -6

Explanation: Step 1: Transposing – 2 to LHS we get,
18 = 3m
Step 2: Dividing both the sides by 3 we get,
m = 6.

9. What is the solution of the equation, 7 + 2x = 0?
a) $$\frac{2}{7}$$
b) $$\frac{-2}{7}$$
c) $$\frac{7}{2}$$
d) $$\frac{-7}{2}$$

Explanation: 7 + 2x = 0
Step 1: Transposing 7 to RHS we get,
2x = -7
Step 2: Dividing both sides by 2 we get,
x = $$\frac{-7}{2}$$.

10. The solution of equation 3x + 6 = -3 is x = -3.
a) True
b) False

Explanation: 3x + 6 = -3
Step 1: Transposing 6 to RHS we get,
3x = -9
Step 2: Dividing both sides by 3 we get
x = -3.

More MCQs on Class 8 Maths Chapter 2:

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