Page 139 - Start Up Mathematics_8 (Non CCE)

\ 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 x- ˆ 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 ¯
Ê 2 - xˆ
fi 12x – 3 x - 6 ˜ ¯ = 4(2x + 8) – 36
Á
Ë
Ê 2 - ˆ x
fi 12x – 3x + 3 Á Ë 6 ¯ ˜ = 8x + 32 – 36
2 - x
fi 9x + = 8x – 4
2

Ê 2 x- ˆ
fi 2 9x + 2 ˜ = 2(8x – 4) (Multiplying both sides by 2)
Á
Ë
¯
fi 18x + 2 – x = 16x – 8
fi 17x + 2 = 16x – 8
fi 17x – 16x = –8 – 2 (Transposing 16x to LHS and 2 to RHS)
fi x = –10

Example 6: Solve 0.25(4f – 3) = 0.05(10f – 9) and check the solution. (NCERT)
Solution: 0.25(4f – 3) = 0.05(10f – 9)
fi 0.25 × 4f – 0.25 × 3 = 0.05 × 10f – 0.05 × 9

fi 1.0f – 0.75 = 0.5f – 0.45
fi f – 0.75 = 0.5f – 0.45
fi f – 0.5f = –0.45 + 0.75 (Transposing 0.5f to LHS and –0.75 to RHS)

fi 0.5f = 0.30
05. f 030. 03.
fi = = (Dividing both sides by 0.5)
05. 05. 05.

3
fi f = = 0.6
5
2
2
Example 7: Solve (3x – 5) + (3x + 5) = (18x + 10)(x – 2) + 44 and check the solution.
2
2
Solution: (3x – 5) + (3x + 5) = (18x + 10)(x – 2) + 44
2
2
2
2
2
2
fi 2{(3x) + (5) } = (18x + 10)(x – 2) + 44 {Using (a – b) + (a + b) = 2(a + b )}
2
2
fi 2(9x + 25) = 18x – 36x + 10x – 20 + 44
2
2
fi 18x + 50 = 18x – 26x + 24
131