## Algebra 1

$\frac{3t -5}{(t-3)(t-2)}$
To solve: $\frac{4}{t-3} - \frac{1}{t-2}$, you will first need to make sure both fractions have common denominators. To do this, we must find the LCD, or the Least Common Multiple. Examining both denominators, you can see that they do not share any common factors. Therefore, the LCD will be $(t-3)(t-2)$. Now\times we can rewrite the fractions using the new LCD we found. $\frac{4}{t-3} - \frac{1}{t-2}$ = $\frac{4\times (t-2)}{(t-3) \times (t-2)} - \frac{1 \times (t-3)}{(t-2) \times (t-3)}$ =$\frac{4t - 8}{(t-3)(t-2)} - \frac{t-3}{(t-3)(t-2)}$ Now that we have common denominators, we can add the numerators: $\frac{4t - 8}{(t-3)(t-2)} - \frac{t-3}{(t-3)(t-2)}$ = $\frac{4t - 8 - (t-3)}{(t-3)(t-2)}$ = $\frac{4t - 8 - t+3}{(t-3)(t-2)}$ = $\frac{4t - t - 8 +3}{(t-3)(t-2)}$ = $\frac{3t -5}{(t-3)(t-2)}$