Section A.4 – Rational Expressions
- Definition and properties of rational expressions
- Simplifying rational expressions
- Operations with rational expressions
- Simplifying compound fractions
Exercises
Directions: You should try to solve each problem first, and then click "Reveal Answer" to check your answer. You can click "Watch Video" if you need help with a problem.
1. Perform the operations and simplify: \(\dfrac{x^2-3x-10}{2x^2-9x-5}\div \dfrac{x^2-2x-8}{2x^2-9x+4}\)
Answer: \(\dfrac{2x-1}{2x+1}\)
Solution: \[
\begin{align}
\dfrac{x^2-3x-10}{2x^2-9x-5}\div \dfrac{x^2-2x-8}{2x^2-9x+4}
&=\dfrac{(x-5)(x+2)}{(2x+1)(x-5)}\div\dfrac{(x-4)(x+2)}{(2x-1)(x-4)}\\
&=\dfrac{x+2}{2x+1}\cdot\dfrac{2x-1}{x+2}\\
&=\dfrac{2x-1}{2x+1}
\end{align}\]
Solution: \[
\begin{align}
\dfrac{x^2-3x-10}{2x^2-9x-5}\div \dfrac{x^2-2x-8}{2x^2-9x+4}
&=\dfrac{(x-5)(x+2)}{(2x+1)(x-5)}\div\dfrac{(x-4)(x+2)}{(2x-1)(x-4)}\\
&=\dfrac{x+2}{2x+1}\cdot\dfrac{2x-1}{x+2}\\
&=\dfrac{2x-1}{2x+1}
\end{align}\]
2. Perform the operations and simplify: \(\dfrac{x+1}{x^2-4x+4}-\dfrac{x-3}{x^2-4}\)
Answer: \(\dfrac{4(2x-1)}{(x-2)^2(x+2)}\)
Solution: \[\begin{align}
\dfrac{x+1}{x^2-4x+4}-\dfrac{x-3}{x^2-4}
&=\dfrac{x+1}{(x-2)(x-2)}-\dfrac{x-3}{(x-2)(x+2)}\\
&=\dfrac{x+1}{(x-2)(x-2)}\cdot\dfrac{x+2}{x+2}-\dfrac{x-3}{(x-2)(x+2)}\cdot\dfrac{x-2}{x-2}\\
&=\dfrac{x^2+3x+2}{(x-2)(x-2)(x+2}-\dfrac{x^2-5x+6}{(x-2)(x-2)(x+2)}\\
&=\dfrac{8x-4}{(x-2)(x-2)(x+2)}\\
&=\dfrac{4(2x-1)}{(x-2)^2(x+2)}
\end{align}\]
Solution: \[\begin{align}
\dfrac{x+1}{x^2-4x+4}-\dfrac{x-3}{x^2-4}
&=\dfrac{x+1}{(x-2)(x-2)}-\dfrac{x-3}{(x-2)(x+2)}\\
&=\dfrac{x+1}{(x-2)(x-2)}\cdot\dfrac{x+2}{x+2}-\dfrac{x-3}{(x-2)(x+2)}\cdot\dfrac{x-2}{x-2}\\
&=\dfrac{x^2+3x+2}{(x-2)(x-2)(x+2}-\dfrac{x^2-5x+6}{(x-2)(x-2)(x+2)}\\
&=\dfrac{8x-4}{(x-2)(x-2)(x+2)}\\
&=\dfrac{4(2x-1)}{(x-2)^2(x+2)}
\end{align}\]
3. Perform the operations and simplify: \(\dfrac{\displaystyle\frac{2}{x}+\displaystyle\frac{1}{3x^2}}{\displaystyle\frac{4}{x}-1}\)
Answer: \(\dfrac{6x+1}{3x(4-x)}\)
Solution: We can clear out the complex fraction by multiplying top and bottom by the ENTIRE fraction's common denominator.
\[\begin{align}
\dfrac{\displaystyle\frac{2}{x}+\displaystyle\frac{1}{3x^2}}{\displaystyle\frac{4}{x}-1}
&=\dfrac{\dfrac{2}{x}+\dfrac{1}{3x^2}}{\dfrac{4}{x}-1}\cdot\dfrac{3x^2}{3x^2}\\
&=\dfrac{6x+1}{12x-3x^2}\\
&=\dfrac{6x+1}{3x(4-x)}
\end{align}\]
Solution: We can clear out the complex fraction by multiplying top and bottom by the ENTIRE fraction's common denominator.
\[\begin{align}
\dfrac{\displaystyle\frac{2}{x}+\displaystyle\frac{1}{3x^2}}{\displaystyle\frac{4}{x}-1}
&=\dfrac{\dfrac{2}{x}+\dfrac{1}{3x^2}}{\dfrac{4}{x}-1}\cdot\dfrac{3x^2}{3x^2}\\
&=\dfrac{6x+1}{12x-3x^2}\\
&=\dfrac{6x+1}{3x(4-x)}
\end{align}\]
4. Perform the multiplication and simplify: \(\displaystyle \frac{t^2-t-6}{t^2+6t+9}\cdot\frac{t+3}{t^2-4}\)
\(\dfrac{t-3}{(t+3)(t-2)}\)