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) \(\frac{1}{4}\)

View Answer

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

View Answer

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

View Answer

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

View Answer

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 = \(\frac{5}{3}\)

View Answer

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

View Answer

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

View Answer

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

View Answer

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) \(\frac{-7}{2}\)

View Answer

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

View Answer

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**.

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