How do you solve #20y + 2y + 3y + 4y - 25y = 20#?

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Answer 1 · 2017-03-30T02:16:50

See the entire solution process below:

First, combine like terms on the left side of the equation:

#(20 + 2 + 3 + 4 - 25)y = 20#

#(29 - 25)y = 20#

#4y = 20#

Now, divide each side of the equation by #color(red)(4)# to solve for #y# while keeping the equation balanced:

#(4y)/color(red)(4) = 20/color(red)(4)#

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

#y = 5#