What is #(-5y^5 + 7y^4) -: (-y)#?

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1 Answer
Michael · Stefan V.
May 24, 2018

#5y^4-7y^3#

Explanation:

When dividing exponents, it is just subtraction. The #-y# is actually #-y^1#. Since the numerator has the same variable and nothing is being multiplied or divided in it, we can just subtract #1# from each exponent.

#-5y^5# would just be #-5y^4# and #7y^4# would become #7y^3#.

Now the denominator is not gone. The #y# is gone, but there is still a #-1# there. All this means is you need to switch the signs up top. The final simplified solution would be

#5y^4-7y^3#

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