How do you solve #8- ( 4y - 7) = 1- 5y#?

2 Answers
Jul 15, 2018

See a solution process below:

Explanation:

First, remove the terms from the parenthesis on the left side of the equation being careful to handle the signs correctly and then group and combine like terms:

#8 - 4y + 7 = 1 - 5y#

#8 + 7 - 4y = 1 - 5y#

#15 - 4y = 1 - 5y#

Now, subtract #color(red)(15)# and add #color(blue)(5y)# to each side of the equation to solve for #y# while keeping the equation balanced:

#15 - color(red)(15) - 4y + color(blue)(5y) = 1 - color(red)(15) - 5y + color(blue)(5y)#

#0 + (-4 + color(blue)(5))y = -14 - 0#

#1y = -14#

#y = -14#

Jul 15, 2018

#y=-14#

Explanation:

We essentially have a #-1# out in front of the parenthesis. Let's distribute that to the inner terms to get

#8-4y+7=1-5y#

We can combine like terms on the left to get

#-4y+15=1-5y#

We can add #5y# to both sides to get

#y+15=1#

Remember, we only want a #y# on the left. Next, we can subtract #15# from both sides to get

#y=-14#

Hope this helps!