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) x2 + 1 = 5
(d) \(\frac{x}{3}\) = 7
(e) x3 – 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.