# Mathematics Questions and Answers – Solving Linear Equations with Variables on One Side and Number on the Other

«
»

This set of Mathematics MCQs for Class 8 focuses on “Solving Linear Equations with Variables on One Side and Number on the Other”.

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.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

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 = $$\frac{5}{7}$$.

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.

Sanfoundry Global Education & Learning Series – Mathematics – Class 8.

To practice Mathematics MCQs for Class 8, here is complete set of 1000+ Multiple Choice Questions and Answers.