Checkpoint: Adding and Subtracting Rational Expressions Solution

Checkpoint: Adding and Subtracting Rational Expressions Solutions

Simplify. Determine the restrictions on the denominator.

$\frac{x}{x + 3} - \frac{6}{x - 8}$

$\begin{array}{rcl} \begin{matrix} x + 3 = 0 & x - 8 = 0 \\ x = - 3 & x = 8 \end{matrix} \\ LCD : (x + 3) (x - 8) \end{array}$

$\frac{x (x - 8)}{(x - 8) (x + 3)} - \frac{6 (x + 3)}{(x + 3) (x - 8)} x^{2} - 8 x - 6 x - 18 \frac{x^{2} - 14 x - 18}{(x + 3) (x - 8)}, x \neq - 3, 8$

Note

Q: How do you know that the restrictions cannot be part of the denominator?
A: They make the expression undefined (OR the denominator contains a zero).

Q: Is your answer a rational expression? Explain.
A: Yes, when you add or subtract rational expressions, the result is a rational expression.

Recall that rational expressions are closed under the 4 basic operations.