How do you solve #1/2(4x-6)=11# using the distributive property?

1 Answer
Jun 27, 2017

See a solution process below:

Explanation:

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

#color(red)(1/2)(4x - 6) = 11#

#(color(red)(1/2) xx 4x) - (color(red)(1/2) xx 6) = 11#

#4/2x - 6/2 = 11#

#2x - 3 = 11#

Next, add #color(red)(3)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#2x - 3 + color(red)(3) = 11 + color(red)(3)#

#2x - 0 = 14#

#2x = 14#

Now, divide each side of the equation by #color(red)(2)# to solve for #x# while keeping the equation balanced:

#(2x)/color(red)(2) = 14/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = 7#

#x = 7#