Question

module 7 pretest: rational expressio equations
module 7 pretest: rational expressions and equations
score: 4/9 answered: 4/9
question 5
simplify completely.
\\(\frac{r - 1}{r - 3} + \frac{r + 9}{r + 4} = \\)
the domain is \\(r \
eq \\)
question help: video

Explanation:

Step1: Find common denominator

The common denominator of \(\frac{r - 1}{r - 3}\) and \(\frac{r + 9}{r + 4}\) is \((r - 3)(r + 4)\). Rewrite each fraction:
\(\frac{(r - 1)(r + 4)}{(r - 3)(r + 4)}+\frac{(r + 9)(r - 3)}{(r - 3)(r + 4)}\)

Step2: Expand numerators

Expand \((r - 1)(r + 4)\) and \((r + 9)(r - 3)\):
\((r - 1)(r + 4)=r^2 + 4r - r - 4=r^2 + 3r - 4\)
\((r + 9)(r - 3)=r^2 - 3r + 9r - 27=r^2 + 6r - 27\)

Step3: Add numerators

Add the expanded numerators:
\(\frac{r^2 + 3r - 4 + r^2 + 6r - 27}{(r - 3)(r + 4)}\)
Combine like terms in the numerator: \(r^2+r^2=2r^2\), \(3r + 6r = 9r\), \(-4-27=-31\). So numerator is \(2r^2 + 9r - 31\).
Denominator is \(r^2 + 4r - 3r - 12=r^2 + r - 12\).
So the simplified expression is \(\frac{2r^2 + 9r - 31}{r^2 + r - 12}\).

Step4: Find domain restrictions

For the original fractions, the denominators \(r - 3\) and \(r + 4\) cannot be zero. So \(r
eq3\) and \(r
eq - 4\).

Answer:

Simplified: \(\frac{2r^2 + 9r - 31}{r^2 + r - 12}\); Domain: \(r
eq3,-4\)