How do you solve #4(y-1)=3(2+y)#?

1 Answer
Jan 5, 2016

Answer:

#y=10#
(see below for solution method)

Explanation:

Given
#color(white)("XXX")4(y-1)=3(2+y)#

Expanding both sides:
#color(white)("XXX")4y-4=6+3y#

Subtract #(3y-4)# from both sides to isolate the variable on one side and the constant on the other
#color(white)("XXX")(4y-4)-(3y-4) = (6+3y)-(3y-4)#

Simplify
#color(white)("XXX")y = 10#