How do you solve $2 - 3 [7 - 2 (4 - x)] = 2 x - 1$ $2-3[7-2(4-x)]= 2x-1$ ?

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Answer 1 · 2016-01-08T14:35:37

$x = \frac{3}{4}$ $x = 3/4$

this is a multi-step equation.

the first step is to simplify the contents of the 'square bracket'

$2 - 3 (7 - 8 + 2 x) = 2 x - 1$ $2 -3 ( 7- 8 + 2x ) = 2x - 1$

now simplify contents of 'new bracket'

$2 - 3 (- 1 + 2 x) = 2 x - 1$ $2 - 3 ( - 1 + 2x ) = 2x - 1$

now multiply the bracket by - 3

$2 + 3 - 6 x = 2 x - 1$ $2 + 3 - 6x = 2x - 1$

now collect like terms to separate sides of the = sign

$2 + 3 + 1 = 2 x + 6 x$ $2 + 3 + 1 = 2x + 6x$

$6 = 8 x$ $6 = 8x$ ( divide both sides by 8 to obtain x )

$\frac{6}{8} = \frac{8 x}{8} \Rightarrow x = \frac{3}{4}$ $6/8 = (8x)/8 rArr x = 3/4$

How do you solve 2-3[7-2(4-x)]= 2x-12−3[7−2(4−x)]=2x−1 2-3[7-2(4-x)]= 2x-1?