Page 166 - ICSE Math 8

Solving Linear Equations with Variables on Both Sides (Transposition Method)

In case where an equation consists of variables and numbers on both sides of equality, the method of solving
involves transposition or shifting of variable terms on one side and numerals on the other.
Step 1: Identify the variable terms and the numerals.
Step 2: Simplify the LHS and the RHS.
Step 3: Transpose variable terms to the LHS and numerals to the RHS following the rules of transposition
given below.
Step 4: Simplify the transposed LHS and RHS and reduce each side to a single term.
Step 5: Divide both the sides by the coefficient of the variable on the LHS.

Rules for transposition
(i) A term with plus (+) or minus (–) sign changes to minus (–) or plus (+) sign respectively when transposed
from one side to the other.
(ii) A multiplier changes to a divisor and vice versa when transposed.
6 3
Example 4: Solve 3x + = 6x + and check the solution.
5 25
6 3 3 6 6
Solution: 3x + = 6x + fi 3x – 6x = − (Transposing 6x to LHS and to RHS)
5 25 25 5 5
3 6 330−
fi –3x = − fi –3x = (LCM of 25 and 5 = 25)
25 5 25
−27 −3x −27
(
fi –3x = fi = ÷−3) (Dividing both sides by –3)
25 −3 25
−27 1 9
fi x = × =
25 −3 25
Check:
6 3
LHS = 3x + RHS = 6x +
5 25
Ê 9 ˆ 6 27 6 Ê 9 ˆ 3 54 3
= 3 Á 25¯ + 5 = 25 + = 6 Á Ë 25¯ ˜ + 25 = 25 + 25
˜
Ë
5
27 30+ 57 54 3+ 57
= = = =
25 25 25 25
\ LHS = RHS

9
So, x = is the solution of the given equation.
25
1 Ê 2 x- ˆ 2x + 8
Example 5: Solve x – Á x - ˜ = - 3.
4 Ë 6 ¯ 3

-
1 Ê 2 xˆ 2x + 8
Solution: x – Á x - ˜ = - 3
4 Ë 6 ¯ 3
LCM of 4, 6, 3 = 12. On multiplying both sides by 12, we get

Ï 1 Ê 2 xˆ ¸ Ê 2x + 8 ˆ
-
12 x - Á x - ˜ ˝ = 12 Á - 3 ˜
Ì
Ó 4 Ë 6 ¯ ˛ Ë 3 ¯
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