Solve the following- If 8x-1 - 4x-1-x-2 -3x-1-2- then x-

The correct option is A 2 Given, (8x 1) #160;(4x 1 x 2 ) = 3x + 1 2 (8x 1) ( 8x (1 x) 2 ) = 6x + 1 2 2(8x 1) (8x 1 +x ...

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Standard VIII

Mathematics

Variable on Both Sides

Solve the fol...

Question

Solve the following: If $(8 x - 1) - (4 x - \frac{1 - x}{2}) = 3 x + \frac{1}{2}$ , then $x =$

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Solution

The correct option is A $2$
Given, $(8 x - 1) - (4 x - \frac{1 - x}{2}) = 3 x + \frac{1}{2}$
$(8 x - 1) - (\frac{8 x - (1 - x)}{2}) = \frac{6 x + 1}{2}$
$2 (8 x - 1) - (8 x - 1 + x) = 6 x + 1$
$16 x - 2 - 8 x + 1 - x = 6 x + 1$
$x - 1 = 1$
$∴ x = 2$

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Solve the following: If (8x−1)−(4x−1−x2)=3x+12, then x=

The correct option is A 2Given, (8x−1)−(4x−1−x2)=3x+12(8x−1)−(8x−(1−x)2)=6x+122(8x−1)−(8x−1+x)=6x+116x−2−8x+1−x=6x+1x−1=1∴x=2

Solve the following: If $(8 x - 1) - (4 x - \frac{1 - x}{2}) = 3 x + \frac{1}{2}$ , then $x =$

The correct option is A $2$
Given, $(8 x - 1) - (4 x - \frac{1 - x}{2}) = 3 x + \frac{1}{2}$
$(8 x - 1) - (\frac{8 x - (1 - x)}{2}) = \frac{6 x + 1}{2}$
$2 (8 x - 1) - (8 x - 1 + x) = 6 x + 1$
$16 x - 2 - 8 x + 1 - x = 6 x + 1$
$x - 1 = 1$
$∴ x = 2$