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

Simplified expression: $\frac{x+3}{5}$ Excluded value: $-3$
$\frac{x^2+6x+9}{5x+15}$ In a rational expression, a value would be excluded if it makes a denominator of any fraction equal to zero. In this case, all solutions of $5x+15=0$ would be excluded values. $5x+15=0$ $5x=-15$ $x=-3$ Then, we can simplify the expression. $\frac{x^2+6x+9}{5x+15}$ $=\frac{(x+3)^2}{5(x+3)}$ $=\frac{x+3}{5}$