How do you solve #4(x − 1) − (x + 2) = 12#?

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Answer 1 · 2016-01-13T16:06:51

#x =6#

#4 (x-1) - (x+2) =12#

#4 xx (x ) + 4 xx(-1) - x - 2 =12#

#4x -4 - x - 2 =12#

#4x - x - 2 -4 =12#

#3x - 6 =12#

#3x =12+6#

#3x =18#

#x =18/3#

#x =6#

Answer 2 · 2018-08-06T02:59:32

#x=6#

We can distribute the #4# and the negative sign to their respective terms. We now have the equation

#4x-4-x-2=12#

We can combine our variables and constants respectively to get

#3x-6=12#

Next, add #6# to both sides to get

#3x=18#

Lastly, to completely isolate #x#, we divide both sides by #3# to get

#x=6#

Hope this helps!