How do you simplify #(14g^3h^2)/(42gh^3)# and find the excluded values?

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Answer 1 · 2017-01-23T20:05:02

#(14g^3h^2)/(42gh^3)=g^2/(3h), h !=0#

In order to simplify, we must remove the indices from the variables where they appear in the numerator and denominator.

#14/42=1/3#

We know that #a^3/a^2=a^(3-2)=a^1=a#, so we can apply this to #g# and #h# in the fraction.

#g^3/g=g^(3-1)=g^2#

#h^2/h^3=h^(2-3)=h^-1=1/h#

We can now simplify the fraction:

#(1cancel(14)g^(2cancel(3))h^(-1cancel(2)))/(3cancel(42)cancel(g)cancel(h^3))=1/3g^2h^-1=g^2/(3h)#

Since this is a fraction, the denominator can't equal #0#, so the only excluded value is #h=0#