How do you solve for x in #x+3(x+4)=20#?

1 Answer
Jun 15, 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:

#x + color(red)(3)(x + 4) = 20#

#x + (color(red)(3) * x) + (color(red)(3) * 4) = 20#

#x + 3x + 12 = 20#

#1x + 3x + 12 = 20#

#(1 + 3)x + 12 = 20#

#4x + 12 = 20#

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

#4x + 12 - color(red)(12) = 20 - color(red)(12)#

#4x + 0 = 8#

#4x = 8#

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

#(4x)/color(red)(4) = 8/color(red)(4)#

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

#x = 2#