How do you solve #2( 3x + 8) + 2( x - 1) = 4( x + 8)#?

1 Answer
Sep 4, 2017

See a solution process below:

Explanation:

First, expand the terms within parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

#color(red)(2)(3x + 8) + color(blue)(2)(x - 1) = color(orange)(4)(x + 8)#

#(color(red)(2) xx 3x) + (color(red)(2) xx 8) + (color(blue)(2) xx x) - (color(blue)(2) xx 1) = (color(orange)(4) xx x) + (color(orange)(4) xx 8)#

#6x + 16 + 2x - 2 = 4x + 32#

Next, group and combine like terms on the left side of the equation:

#6x + 2x + 16 - 2 = 4x + 32#

#(6 + 2)x + (16 - 2) = 4x + 32#

#8x + 14 = 4x + 32#

Then, subtract #color(red)(14)# and #color(blue)(4x)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#8x - color(blue)(4x) + 14 - color(red)(14) = 4x - color(blue)(4x) + 32 - color(red)(14)#

#(8 - color(blue)(4))x + 0 = 0 + 18#

#4x = 18#

Now, divide each side of the equation by #color(red)(4)# to solve for #x# while keeping the equation balanced:

#(4x)/color(red)(4) = 18/color(red)(4)#

#(color(red)(cancel(color(black)(4)))x)/cancel(color(red)(4)) = 9/2#

#x = 9/2#