How do you solve #6(x+3)+38=2#?

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Answer 1 · 2016-06-06T07:48:04

#color(green)(x = -9#

As we need to find the value of #x#, we try to transpose all the other terms to the right hand side.

Transposing 38 to the right hand side, we get:

#6(x+3) = 2-38#

#6(x+3) = -36#

To eliminate 6, we can divide both sides by 6

#(cancel(6)(x+3))/cancel6 = -36/6#

#x + 3 = -6#

Transposing 3 to the right hand side, we get

#x = -6 - 3#

#color(green)(x = -9#

Test:
Try substituting the value of #x# in the original equation and check if it's satisfied