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

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Answer 1 · 2018-05-13T14:46:09

$x = - 2$ $x = -2$

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

Let's start with the left side. First we distribute the $- 3$ $-3$ .

$2 (1 - 3 x - 6) = - x$ $2(1- 3x -6 )= -x$

We combine the integers.

$2 (- 3 x - 5) = - x$ $2(-3x - 5)= -x$

Let's distribute the two.

$- 6 x - 10 = - x$ $-6x -10 = -x$

Now we add $6 x$ $6x$ to both sides to move the $- 6 x$ $-6x$ to the right.

$- 10 = 5 x$ $-10 = 5x$

Dividing by five,

$x = - 2$ $x = -2$

Check:

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

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