How do you solve #43= - 5( y + 11) + 7y#?

1 Answer
Sep 21, 2017

#y=49#

Explanation:

First we switch sides from
#43=-5(y+11)+7y# to #-5(y+11)+7y=#.
We will remove the #7y# and the #43=# and focus on the rest.
We will distribute parentheses/brackets using:
#a(b+c)=ab+ac#.

#a# would be #-5#, #b# would be #y# and #c# would be #11# in this case.

Using #a(b+c)=ab+ac#, we know that #-5*y-5*11#. Multiply #-5*y=-5y# and #5*11=55#. Therefore the equation would be (with the #7y#) #-5y-55+7y#.

Next, we group like terms: #-5y+7y-55#, and then add #-5y+7y=2y#.
The whole equation will then be #2y-55=43#. Add #55# to both sides:

#2y-55color(blue)+color(blue)55=43color(blue)+color(blue)55# and simplify it to be #2y=98#.
Divide both sides by #2# and you get #y=49#.