Solution Bank Rational Expressions & Equations Target B
Answers to the Practice Problems are below in random order.
\(\begin{align}&-\frac{5}{2}\ & \ \ \ &6x(x-6)\ & \ \ \ &-\frac{1}{3}\\\\
&\frac{x^2-15}{(x+3)(x-3)}\ & \ \ \ &2x(x-2)(x+2)\ & \ \ \ &0, 9\\\\
&-16-15i\ & \ \ \ &-7\ & \ \ \ &x=-\frac{8}{3}\\\\
&\frac{1}{2}, 1, 9\ & \ \ \ &2\ & \ \ \ &\text{Since a denominator is} (x-2), x=2\ \text{is in the restricted domain and}\\
&\ & \ \ \ &\ & \ \ \ &\text{ cannot be a solution}\\\\
&-9, -2\ & \ \ \ &-\frac{9}{2}, 1\ & \ \ \ &-1\\\\
&-\frac{11}{2}, -1\ & \ \ \ &\frac{5}{4}, 2 \ & \ \ \ &3, 8\\\\
&-2\ & \ \ \ &1, \frac{11}{3}\end{align}\)
Answers to the Practice Problems are below in random order.
\(\begin{align}&-\frac{5}{2}\ & \ \ \ &6x(x-6)\ & \ \ \ &-\frac{1}{3}\\\\
&\frac{x^2-15}{(x+3)(x-3)}\ & \ \ \ &2x(x-2)(x+2)\ & \ \ \ &0, 9\\\\
&-16-15i\ & \ \ \ &-7\ & \ \ \ &x=-\frac{8}{3}\\\\
&\frac{1}{2}, 1, 9\ & \ \ \ &2\ & \ \ \ &\text{Since a denominator is} (x-2), x=2\ \text{is in the restricted domain and}\\
&\ & \ \ \ &\ & \ \ \ &\text{ cannot be a solution}\\\\
&-9, -2\ & \ \ \ &-\frac{9}{2}, 1\ & \ \ \ &-1\\\\
&-\frac{11}{2}, -1\ & \ \ \ &\frac{5}{4}, 2 \ & \ \ \ &3, 8\\\\
&-2\ & \ \ \ &1, \frac{11}{3}\end{align}\)