How do you solve #z+4(2z+3)=15#?

1 Answer
Feb 5, 2017

Answer:

See the entire solution process below:

Explanation:

First, expand the terms in parenthesis on the left side of the equation:

#z + (4 xx 2z) + (4 xx 3) = 15#

#z + 8z + 12 = 15#

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

#1z + 8z + 12 = 15#

#(1 + 8)z + 12 = 15#

#9z + 12 = 15#

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

#9z + 12 - color(red)(12) = 15 - color(red)(12)#

#9z + 0 = 3#

#9z = 3#

Now, divide each side of the equation by #color(red)(9)# to solve for #z# while keeping the equation balanced:

#(9z)/color(red)(9) = 3/color(red)(9)#

#(color(red)(cancel(color(black)(9)))z)/cancel(color(red)(9)) = 1/3#

#z = 1/3#