How do you solve $4 + 5 (3 - x) = 4 (8 + 2 x)$ $4 + 5(3 - x) = 4(8 + 2x)$ ?

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Answer 1 · 2016-02-26T18:28:47

$x = - 1$ $x = -1$

$4 + 5 (3 - x) = 4 (8 + 2 x)$ $4 + 5(3 − x) = 4(8 + 2x)$

Distribute $5$ $5$ and $4$ $4$ to make the equation in its expanded form

$4 + 15 - 5 x = 32 + 8 x$ $4 + 15 - 5x = 32 + 8x$

Joining common factors or alike integers for each side:

$13 x = - 13$ $13x = -13$

Simplify by dividing both sides with $13$ $13$

$x = - 1$ $x = -1$

How do you solve 4+5(3−x)=4(8+2x)4 + 5(3 - x) = 4(8 + 2x)?