How do you solve $3 - 5 (x + 2) = 3 (x + 1) + 6$ $3 - 5(x + 2) = 3(x + 1) + 6$ ?

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Answer 1 · 2016-07-13T08:58:23

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

Use the distributive law to multiply out the brackets first so we have separate terms.

$3 - 5 (x + 2) = 3 (x + 1) + 6$ $3 - 5(x + 2) = 3(x + 1) + 6$

$3 - 5 x - 10 = 3 x + 3 + 6$ $3-5x-10 = 3x +3+6$

$cancel3-5x-10 = 3x +cancel3+6$

$-10-6 = 3x+5x " x terms on right to keep them positive"$
$-16 = 8x$

$x = -2$

How do you solve 3 - 5(x + 2) = 3(x + 1) + 63−5(x+2)=3(x+1)+63 - 5(x + 2) = 3(x + 1) + 6?