Page 171 - Maths Class 06

Þ x + 8 – 8 = 3 – 8 [ \ 8 + x = x + 8]
Þ x = – 5
So, x = – 5 is the solution of the given equation.

Check : Substituting x = – 5 in the given equation,
We get LHS = 8 – 5 = 3 and RHS = 3

\ when x = – 5, we have LHS = RHS.
EX AM PLE 3. Solve the equa tion 8x = 40 and check the re sult.
SOLUTION : 8x = 40
8x 40
Þ = [Dividing both sides by 8]
8 8

Þ x = 5
So, x = 5 is the solution of the given equation.

Check : Substituting x = 3 in the given equa tion,
We get LHS = ´8 5 40 and RHS =40.
=
\ when x = 5, we have, LHS = RHS

Trans po si tion

You know that one can add or subtract a number from both sides of the equation. So, for the equation
x - 4 = 5, we can write
x - +4 4 = +5 4 Þ x = +5 4.
Similarly, for the equation x + 5 = 3, we can write

x + - = -5 5 3 5 Þ x = -3 5.
In both these cases you will notice that after this operation, the number appears on the other side of the
equation, but with the opposite sign. So, you can straightaway change the sign of a term and transfer it
from one side of an equation to the other side. This is called transposition.
EXAMPLE 4. Solve : 3x + 5 = 13 - x. Check the result.

SOLUTION : 3x + 5 = 13 - x
Þ 3x + x = 13 5- [transposing - x to LHS and +5 to RHS]
Þ 4x = 8
4x 8
Þ = [dividing both sides by 4]
4 4
Þ x = 2

\ x = 2 is the solution of the given equation.
Check : Substituting x = 2 in the given equation, we get
LHS = ´ +3 2 5 = 11

and RHS = 13 - =2 11.

\ LHS = RHS, when x = 2.

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