# Solving a One-Step Linear Equation Problem Type 2 Online Quiz

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Following quiz provides Multiple Choice Questions (MCQs) related to Solving a One-Step Linear Equation Problem Type 2. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz. Q 1 - Solve the following linear equation.

16 – 4y = 12

A - 1

B - 2

C - 3

D - 4

### Explanation

Step 1:

16 – 4y = 12

Subtracting 16 from both sides

Step 2:

16 – 4y – 16 = 12 – 16 = − 4

− 4y = − 4

Step 3:

Dividing both sides by – 4

So, y = 1

Q 2 - Solve the following linear equation.

4w + 6 = 18

A - 1

B - 2

C - 3

D - 4

### Explanation

Step 1:

4w + 6 = 18

Subtracting 6 from both sides

Step 2:

4w + 6 – 6 = 18 – 6 = 12

4w = 12

Step 3:

Dividing both sides by 4;

$\frac{4w}{4}$ = $\frac{12}{4}$ = 3

So, w = 3

Q 3 - Solve the following linear equation.

1 + 6p = 13

A - 1

B - 2

C - 3

D - 4

### Explanation

Step 1:

1 + 6p = 13

Subtracting 1 from both sides

Step 2:

1 + 6p – 1 = 13 – 1 = 12

6p = 12

Step 3:

Dividing both sides by 6 we get

$\frac{6p}{6}$ = $\frac{12}{6}$ = 2

So, p = 2

Q 4 - Solve the following linear equation.

7 + 2z = 19

A - 3

B - 4

C - 5

D - 6

### Explanation

Step 1:

7 + 2z = 19

Subtracting 7 from both sides

Step 2:

7 + 2z – 7 = 19 – 7 = 12

2z = 12

Step 3:

Dividing both sides by 2;

$\frac{2z}{2}$ = $\frac{12}{2}$ = 6

So, z = 6

Q 5 - Solve the following linear equation.

20 – 5m = 5

A - 1

B - 2

C - 3

D - 4

### Explanation

Step 1:

20 – 5m = 5

Subtracting 20 from both sides

Step 2:

20 – 5m – 20 = 5 – 20 = − 15

− 5m = − 15;

Step 3:

Dividing both sides by −5,

$\frac{-5m}{-5}$ = $\frac{-15}{-5}$ = 3

So, m = 3

Q 6 - Solve the following linear equation.

3t + 1 = 16

A - 4

B - 5

C - 6

D - 7

### Explanation

Step 1:

3t + 1 = 16

Subtracting 1 from both sides

Step 2:

3t + 1 – 1 = 16 – 1 = 15

3t = 15;

Step 3:

Dividing both sides by 3

$\frac{3t}{3}$ = $\frac{15}{3}$ = 5

So, t = 5

Q 7 - Solve the following linear equation.

7 = 3k − 5

A - 4

B - 5

C - 6

D - 7

### Explanation

Step 1:

7 = 3k − 5

Adding 5 to both sides

Step 2:

7 + 5 = 3k – 5 + 5 = 3k

3k = 12

Step 3:

Dividing both sides with 3

$\frac{3k}{3}$ = $\frac{12}{3}$ = 4

So, k = 4

Q 8 - Solve the following linear equation.

3x + 4 = 13

A - 0

B - 1

C - 2

D - 3

### Explanation

Step 1:

3x + 4 = 13

Subtracting 4 from both sides

Step 2:

3x + 4 – 4 = 13 – 4 = 9

3x = 9;

Step 3:

Dividing both sides with 3

$\frac{3x}{3}$ = $\frac{9}{3}$ = 3

So, x = 3

Q 9 - Solve the following linear equation.

6 = 2q − 4

A - 3

B - 4

C - 5

D - 6

### Explanation

Step 1:

6 = 2q − 4

Adding 4 to both sides

Step 2:

6 + 4 = 2q – 4 + 4 = 2q

2q = 10

Step 3:

Dividing both sides by 2

$\frac{2q}{2}$ = $\frac{10}{2}$ = 5

So, q = 5

Q 10 - Solve the following linear equation.

8 + 2x = 12

A - 1

B - 2

C - 3

D - 4

### Explanation

Step 1:

8 + 2x = 12

Subtracting 8 from both sides

Step 2:

8 + 2x – 8 = 12 – 8 = 4

2x = 4;

Step 3:

Dividing both sides by 2

$\frac{2x}{2}$ = $\frac{4}{2}$ = 2

So, x = 2

solving_one_step_linear_equation_problem_type2.htm
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