How do you solve #4(3-y)+y = 22+2y#?

2 Answers
Jun 23, 2015

Answer:

I found: #y=-2#

Explanation:

First multiply the #4#:
#12-4y+y=22+2y#
now collect all the #y# on one side (left for example):
#-4y+y-2y=-12+22#
Add:
#-5y=10#
#y=10/-5=-2#

Jun 23, 2015

Answer:

#y=(-2)#

Explanation:

#4(3-y)+y = 22+2y#
multiply out the first term:
#color(white)("XXXX")##12-4y + y = 22+2y#
combine the #y# terms on the left side
#color(white)("XXXX")##12-3y = 22+2y#
add #3y# and subtract #22# from both sides
#color(white)("XXXX")##-10 = 5y#
divide both sides by #5# (and reverse equation for more normal form)
#color(white)("XXXX")##y = -2#