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

Answer:

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

Answer:

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

Answer:

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

Answer:

\(\frac{2}{3}\)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

Answer:

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

Answer:

\(\frac{1}{2}\)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

Answer:

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

Answer:

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

Answer:

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

Answer:

\(\frac{1}{3}\)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

Answer:

\(\frac{x-5}{4}\) = 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

Answer:

\(\frac{z+3}{4}\) = 17

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

⇒ z + 3 = 68

⇒ z + 3 – 3 = 68 – 3 (Adding -3 to both sides)

⇒ z = 6