How do you solve #13-5y = 8#?

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Answer 1 · 2018-06-02T12:13:49

See a solution process below:

First, subtract #color(red)(13)# from each side of the equation to isolate the #y# term while keeping the equation balanced:

#13 - color(red)(13) - 5y = 8 - color(red)(13)#

#0 - 5y = -5#

#-5y = -5#

Now, divide each side of the equation by #color(red)(-5)# to solve for #y# while keeping the equation balanced:

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

#(color(red)(cancel(color(black)(-5)))y)/cancel(color(red)(-5)) = 1#

#y = 1#