How do you solve #4y - 5= - y + 15#?

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Answer 1 · 2017-07-20T01:08:41

See a solution process below:

Step 1) Add #color(red)(5)# and #color(blue)(y)# to each side of the equation to isolate the #y# term while keeping the equation balanced:

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

#color(blue)(1y) + 4y - 0 = 0 + 20#

#(color(blue)(1) + 4)y = 20#

#5y = 20#

Step 2) Divide each side of the equation by #color(red)(5)# to solve for #y# while keeping the equation balanced:

#(5y)/color(red)(5) = 20/color(red)(5)#

#(color(red)(cancel(color(black)(5)))y)/cancel(color(red)(5)) = 4#

#y = 4#