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

Question

Isaiah Anthony · Accepted Answer

#(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#

Abigail Allen · Answer

undefined To simplify (14g^3h^2)/(42gh^3), we can cancel out common factors in the numerator and denominator.

First, we can divide both the numerator and denominator by the greatest common factor, which is 14g. This leaves us with:

(14g^3h^2)/(42gh^3) = (g^2h^2)/(3h^3)

Next, we can simplify further by canceling out a common factor of h^2. This gives us:

(g^2h^2)/(3h^3) = g^2/(3h)

The excluded values are the values of h that would make the denominator equal to zero. In this case, h cannot be equal to zero, as it would result in division by zero.