Page 172 - Maths Class 06

x 1 x
EX AM PLE 5. Solve: – = – 2.
8 2 6
SOLUTION : Multiplying each term by 24, the LCM of 8, 2 and 6, the given equation becomes
3x – 12 = 4x – 48

Þ 3x – 4x = – 48 + 12 [Transposing 4x to LHS and – 12 to RHS]

Þ – x = – 36

Þ x = 36
\ x = 36 is the solution of the given equation.
Check : Substituting x = 36 in the given equation, we get

æ 36 1 ö æ 36 – 4 ö 32
LHS = ç – ÷ = ç ÷ = = 4
è 8 2 ø è 8 ø 8
æ 36 ö
and RHS = ç – 2÷ = (6 – 2) = 4
è 6 ø

\ LHS = RHS, when x = 36.

Exercise11.5

1. Solve the following equation and check the result in each case:
(a) x + 2 = 7 (b) 3x - 3 = 12 (c) 3 - x = 1
3x
(d) x - 2 = -5 (e) x + 5 = -7 (f) = 18
5
x x
4
(g) 6x - 5 = 2x + 11 (h) 4x - = 16 (i) = + 5
2 3
2. Solve the follow ing:
m 1 m 3y
(a) - = + 1 (b) - 4 = 11 (c) 3(x + 2) - 2(x - 3) = 5
4 2 3 10
m 2x x
(d) 3(x + 6) + 2(x + 3) = 54 (e) + 8 = 12 (f) + 8 = - 1
4 3 2
(g) 12m - 3 = 5 2m +( 1) (h) 2(x - 2)- 3(x - 3) = 5(x - 5)
(i) 6x + 5 = 3x + 20

3. Solve each of following equation and verify answer:
x x x n n 1
(a) + 17 – + = 0 (b) 3 2( - 5 ) -x 2 1 6( - x ) = 1 (c) - 5 = +
5 6 4 4 6 2
3x 2x 3 x x - 3 2 x
(d) - 4 = 14 (e) - = + 1 (f) - 2 =
10 5 2 2 5 5

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