# DAV Class 6 Maths Chapter 7 Worksheet 4 Solutions

The DAV Class 6 Maths Solutions and DAV Class 6 Maths Chapter 7 Worksheet 4 Solutions of Linear Equations offer comprehensive answers to textbook questions.

## DAV Class 6 Maths Ch 7 WS 4 Solutions

Solve the following equations:

Question 1.
2x + 1 = 5
2x + 1 = 5
2x + 1 – 1 = 5 – 1 (Subtracting 1 from both sides)
⇒ 2x = 4 ⇒ $$\frac{2 x}{2}=\frac{4}{2}$$
(Dividing both sides by 2)
x = 2

Question 2.
6y – 5 = 19
6y – 5 = 19
⇒ 6y – 5 + 5 = 19 + 5 (Adding 5 to both sides)
⇒ 6y = 24
⇒ $$\frac{6 y}{6}=\frac{24}{6}$$ (Dividing both sides by 6)
⇒ y = 4

Question 3.
3 + 4y = – 5
3 + 4y = -5
⇒ -3 + 3 + 4y = -5 – 3 (Adding -3 to both sides)
⇒ 4y = -8
⇒ $$\frac{4 y}{4}=\frac{-8}{4}$$ (Dividing both sides by 4)
y = -2

Question 4.
$$\frac{2}{3}$$p = 6
$$\frac{2}{3}$$p = 6
⇒ $$\frac{2}{3}$$p × 3 = 6 × 3
⇒ 2p = 18
⇒ $$\frac{2 p}{2}=\frac{18}{2}$$ (Dividing both sides by 2)
∴ p = 9

Question 5.
5y + 10 = 4y – 10
5y + 10 = 4y – 10
⇒ 5y + 10 – 10 = 4y – 10 – 10 (Adding -10 both sides)
⇒ 5y = 4y – 20
⇒ 5y – 4y = -20
⇒ y = -20

Question 6.
$$\frac{1}{2}$$x + 3 = 5
$$\frac{1}{2}$$x + 3 = 5
⇒ $$\frac{1}{2}$$x + 3 – 3 = 5 – 3 (Adding -3 to both sides)
⇒ $$\frac{1}{2}$$x = 2
⇒ $$\frac{1}{2}$$x × 2 = 2 × 2 (Multiply both sides by 2)
∴ x = 4

Question 7.
3(x + 1) = 6
3(x + 1) = 6
⇒ $$\frac{3(x+1)}{3}=\frac{6}{3}$$ (Dividing both sides by 3)
⇒ x + 1 = 2
⇒ x + 1 – 1 = 2 – 1 (Adding -1 to both sides)
⇒ x = 1

Question 8.
4(x – 2) = -8
4(x – 2) = -8
⇒ $$\frac{4(x-2)}{4}=\frac{-8}{4}$$
(Dividing both sides by 4)
⇒ x – 2 = -2
⇒ x – 2 + 2 = -2 + 2 (Adding 2 to both sides)
∴ x = 0

Question 9.
3x + 8 = 5x + 2
3x + 8 = 5x + 2
⇒ 3x + 8 – 8 = 5x + 2 – 8
(Adding -8 to both sides)
⇒ 3x = 5x – 6
⇒ 3x – 5x = -6 (Shifting 5x to LHS)
⇒ -2x = -6
⇒ $$\frac{-2 x}{-2}=\frac{-6}{-2}$$ (Dividing both sides by -2)
x = 3

Question 10.
$$\frac{1}{3}$$x + 11 = 14
$$\frac{1}{3}$$x + 11 = 14
⇒ $$\frac{1}{3}$$x + 11 – 11 = 14 – 11 (Adding -11 to both sides)
⇒ $$\frac{1}{3}$$x = 3
⇒ $$\frac{1}{3}$$x × 3 = 3 × 3 (Multiplying both sides by 3)
∴ x = 9

Question 11.
$$\frac{x-5}{4}$$ = 3
$$\frac{x-5}{4}$$ = 3
$$\frac{x-5}{4}$$ × 4 = 3 × 4 (Multiplying both sides by 4)
⇒ x – 5 = 12
⇒ x – 5 + 5 = 12 + 5 (Adding 5 to both sides)
∴ x = 17

Question 12.
$$\frac{z+3}{4}$$ = 17
$$\frac{z+3}{4}$$ = 17
⇒ $$\frac{z+3}{4}$$ × 4 = 17 × 4 (Multiplying both sides by 4)