How do you solve #-4(3+x)+5=4(x+3)#?

1 Answer
Jul 24, 2017

See a solution process below:

Explanation:

First, expand the terms in parenthesis on each side of the equation by multiplying all the terms within the parenthesis by the term outside the parenthesis:

#color(red)(-4)(3 + x) + 5 = color(blue)(4)(x + 3)#

#(color(red)(-4) xx 3) + (color(red)(-4) xx x) + 5 = (color(blue)(4) xx x) + (color(blue)(4) xx 3)#

#-12 + (-4x) + 5 = 4x + 12#

#-12 - 4x + 5 = 4x + 12#

#-12 + 5 - 4x = 4x + 12#

#-7 - 4x = 4x + 12#

Next, add #color(red)(4x)# and subtract #color(blue)(12)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#-7 - color(blue)(12) - 4x + color(red)(4x) = 4x + color(red)(4x) + 12 - color(blue)(12)#

#-19 - 0 = (4 + color(red)(4))x + 0#

#-19 = 8x#

Now, divide each side of the equation by #color(red)(8)# to solve for #x# while keeping the equation balanced:

#-19/color(red)(8) = (8x)/color(red)(8)#

#-19/8 = (color(red)(cancel(color(black)(8)))x)/cancel(color(red)(8))#

#-19/8 = x#

#x = -19/8#