Excluded Values

Division by 0 has no answer, so a rational expression is not defined where its bottom is 0. Those inputs are its excluded values. For $\frac{x-2}{x\left(x+6\right)}$ the bottom is 0 at x = 0 and at x = -6, so both are excluded. The top being 0 is fine: at x = 2 the fraction is simply 0.

Cancelling can hide an excluded value. $\frac{{x}^{2}-9}{x-3}$ factors to $\frac{\left(x+3\right)\left(x-3\right)}{x-3}$, and cancelling leaves $x+3$. For almost every x the two agree: at x = 1, the fraction is -8 over -2, which is 4, and 1 + 3 is 4. But at x = 3 the original is 0 over 0, which has no value, while x + 3 gives 6. So the honest answer is x + 3 for every x other than 3.

Read the excluded values from the original bottom, before anything cancels. On a graph, this expression is the line y = x + 3 with a single hole at (3, 6). Later in this section you will solve equations with fractions, and an answer that lands on an excluded value is a false answer: it has to be thrown out.