Answer
a. $\frac{-5}{z+3}$
b. $\frac{3n-4}{5n-2}$
c. $\frac{1}{q-2}$
Work Step by Step
a. Since both fractions are of the same denominator, we can directly perform subtraction on the numerators and leave the result as a single fraction.
$\frac{2}{z+3} - \frac{7}{z+3} = \frac{(2-7)}{z+3}$
$ = \frac{-5}{z+3}$
b. Again, directly perform subtraction on the numerators.
$\frac{9n-3}{10n-4} - \frac{3n+5}{10n-4} = \frac{9n-3-3n-5}{10n-4}$
$\frac{6n-8}{10n-4}$
All of the numbers in the fraction can be divided by two. Factor out two on both the top and bottom, and then cancel them out.
$\frac{6n-8}{10n-4} = \frac{2(3n-4)}{2(5n-2)}$
$ = \frac{3n-4}{5n-2}$
c. Again directly perform subtraction on the numerator, but do take care to convert two minus signs to a plus ($ a-(-b) = a+b$):
$\frac{7q-3}{q^2-4} - \frac{6q-5}{q^2-4} = \frac{(7q-3)-(6q-5)}{q^2-4}$
$ = \frac{7q-3-6q+5}{q^2-4}$
$ = \frac{q+2}{q^2-4}$
Next, factor: $(q^2 -4)$:
$\frac{q+2}{q^2-4} = \frac{q+2}{(q+2)(q-2)}$
Since $(q +2)$ is on both the top and the bottom, we can cancel them out:
$\frac{q+2}{(q+2)(q-2)} = \frac{1}{q-2}$
