How do you solve #4x + 2( 8- 3x ) = 12x#?

2 Answers
May 16, 2018

#x = 8/7#

Explanation:

Ok first you need to get the #x#'s on to one side and the numbers on their own on the over, so first expanding the bracket you get

#4x + 16 - 6x =12x#

Once we collect like terms we end up with

#-2x +16 = 12x#

Now what we do to one side, we do to the other, so if we add #2x# to get rid of the #-2x#, we have to also add #2x# to the #12# so we end up with

#16=14x#

And supposedly you want just #x#, so we have to divide both sides by #14#, which is

#x = 16/14 = 8/7 ~~ 1.4286#

May 16, 2018

Simplify the expression, rearrange terms, and then solve to find #x=8/7#

Explanation:

First, we'll simplify. We could multiply both the terms that are in the parentheses by the coefficient (2), but note that the other terms outside of the parentheses are divisible by 2. For this, I elected to divide both sides by 2, which eliminates the need for the parentheses:

#(4x+2(8-3x))/2=(12x)/2#

#2x+(8-3x)=6x#

#2x+8-3x=6x#

Next, we'll combine like terms:

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

#8-x=6x#

Next, we rearrange by adding #x# to both sides:

#8cancel(-x)color(red)(cancel(+x))=6xcolor(red)(+x)#

#8=x(6+color(red)(1))#

#8=7x#

Finally, we divide through by #x#'s coefficient:

#8/color(red)(7)=(cancel(7)x)/color(red)(cancel(7))#

#color(green)(x=8/7~=1.1429#