Question 1201042
<font color=black size=3>
LHS = left hand side
RHS = right hand side


Let n be the unknown numerator expression in the RHS fraction.
{{{14/(z^2-3z) = n/(z(z-3)(z-2))}}}
n is some expression in terms of z.


Let's factor the denominator of the LHS.
{{{z^2-3z = z(z-3)}}}


This means
{{{14/(z^2-3z) = n/(z(z-3)(z-2))}}}
is the same as
{{{14/(z(z-3)) = n/(z(z-3)(z-2))}}}


The factors for the LHS denominator are z and (z-3). Both of which are present in the RHS denominator {{{z(z-3)(z-2)}}}. The LHS denominator is missing the factor (z-2).


We'll multiply top and bottom of the LHS fraction by (z-2) to fill in that missing gap.


{{{14/(z(z-3)) = n/(z(z-3)(z-2))}}}


{{{(14*highlight((z-2)))/(z(z-3)*highlight((z-2))) = n/(z(z-3)(z-2))}}}


{{{(14*(z-2))/(z(z-3)*(z-2)) = n/(z(z-3)(z-2))}}}


{{{(14z-14*2)/(z(z-3)(z-2)) = n/(z(z-3)(z-2))}}}


{{{(14z-28)/(z(z-3)(z-2)) = n/(z(z-3)(z-2))}}}


Both denominators of the LHS and RHS have the same exact factorization. The fractions are only equal if the numerators were the same. Therefore, we must have n = 14(z-2) = 14z-28


In other words,
{{{14/(z^2-3z) = (14z-28)/(z(z-3)(z-2))}}}
after multiplying top and bottom by (z-2).
</font>