The DAV Class 6 Maths Solutions and **DAV Class 6 Maths Chapter 7 Worksheet 1 **Solutions of Linear Equations offer comprehensive answers to textbook questions.

## DAV Class 6 Maths Ch 7 WS 1 Solutions

Question 1.

What are the two sides of an equation called?

Answer:

Left Hand Side (LHS) and Right Hand Side (RHS).

Question 2.

In a linear equation, what is the power of the variable?

Answer:

One.

Question 3.

Encircle the linear equations in the following:

(a) p + 3 = 5

(b) 2x + 3 = 7

(c) x^{2} + 1 = 5

(d) \(\frac{x}{3}\) = 7

(e) x^{3} – 1 = 9

(f) 2x – 5 = x – 3

Answer:

(a) p + 3 = 5 is linear equation

(b) 2x + 3 = 7 is linear equation

(c) \(\frac{x}{3}\) = 7 is linear equation

(d) 2x – 5 = x – 3 is linear equation

Question 4.

Separate the LHS and RHS of the following equations.

Linear equation | LHS | RHS |

(a) y – 3 = 9 |
||

(b) 2x + 1 = 5 | ||

(c) \(\frac{y}{3}\) = 9 |
||

(d) 5y – 1 = 9 – 2y |

Answer:

Linear equation | LHS | RHS |

(a) y – 3 = 9 | y – 3 | 9 |

(b) 2x + 1 = 5 | 2x + 1 | 5 |

(c) \(\frac{y}{3}\) = 9 | \(\frac{y}{3}\) | 9 |

(d) 5y – 1 = 9 – 2y | 5y – 1 | 9 – 2y |

Question 5.

Convert the following statements into an equation:

(a) 3 added to a number x is 5.

Answer:

3 + x = 5

(b) A number y increased by 6 is 9

Answer:

y – 6 = 9

(c) 2 subtracted from a number x is equal to 10

Answer:

x – 2 = 10

(d) Twice a number p equals to 12

Answer:

2p = 12

(e) 15 less than a number a is 7

Answer:

a – 15 = 7

(f) Two-third a number x is 5

Answer:

\(\frac{2}{3}\)x = 5

(g) Three times a number x decreased by 1 is 4

Answer:

3x – 1 = 4

(h) Six times a number y is 3 more than the number itself

Answer:

6y = y + 3

(i) 5 less than one-third a number 2 is 5

Answer:

\(\frac{z}{3}\) – 5 = 5

(j) Two times the sum of a number x and 3 equals 14.

Answer:

2(x + 3) = 14

### DAV Class 6 Maths Chapter 7 Worksheet 1 Notes

An equation in which the power of the variable is one is called a Linear Equation.

Example:

- 2x + 3 = 7
- \(\frac{2}{3}\)y + 5 = 0
- x – 7 = 0
- y = 0

Equation has two sides LHS and RHS

2x + 5 = 3

2x + 5 is the term on LHS and 3 is the term on RHS

Rules for solving an equation:

1. Same quantity can be added or subtracted from both sides of equality.

Example:

(i) 2x + 5 = 3

⇒ 2x + 5 – 3 = 3 – 3

⇒ 2x + 2 = 0

(ii) 3x – 7 = 5

⇒ 3x – 7 + 7 = 5 + 7

⇒ 3x = 12

2. Both sides can be multiplied by the same quantity.

Example:

5x + 2 = \(\frac{3}{2}\)

2 × (5x + 2) = \(\frac{3}{2}\) × 2

⇒ 10x + 4 = 3

3. Both sides can be divided by the same quantity

Example:

4x + 8 = 10

⇒ \(\frac{4 x+8}{2}=\frac{10}{2}\)

⇒ 2x + 4 = 5

From the above examples we can conclude that whatever we do to one side, we should do the same to the other side also.

Example 1.

Solve 3x + 8 = 14.

Solution:

3x + 8 = 14

⇒ 3x + 8 – 8 = 14 – 8 (Subtracting 8 from both sides)

⇒ 3x = 6

⇒ \(\frac{3 x}{3}=\frac{6}{3}\) (Dividing both sides by 3)

⇒ x = 2

Hence x = 2 is the solution of the given equation.

Example 2.

Solve : \(\) = 7

Solution:

\(\) = 7

⇒ \(\) x 2 = 7 x 2 (Multiplying both sides by 2)

⇒ 2y – 4 = 14

⇒ 2y-4 + 4 = 14 + 4 (Adding 4 to both sides)

⇒ 2y = 18

⇒ \(\frac{2 y}{2}=\frac{18}{2}\) (Dividing both sides by 2)

⇒ y = 9

Hence, y = 9 is the solution of the given equation.