## Algebra 1: Common Core (15th Edition)

a. $\frac{-5}{z+3}$ b. $\frac{3n-4}{5n-2}$ c. $\frac{1}{q-2}$
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: $\frac{q+2}{(q+2)(q-2)} = \frac{1}{q-2}$