Question

add or subtract the rational expressions below as indicated, and express the result in simplest form. (\frac{c - 1}{c + 1} + \frac{c + 2}{5c + 3}) (provide your answer in expanded form.)

Explanation:

Step1: Find common denominator

The denominators are $c+1$ and $5c+3$, so the common denominator is $(c+1)(5c+3)$.
Rewrite each fraction:
$\frac{(c-1)(5c+3)}{(c+1)(5c+3)} + \frac{(c+2)(c+1)}{(c+1)(5c+3)}$

Step2: Expand numerators

Expand $(c-1)(5c+3)$:
$c(5c+3)-1(5c+3) = 5c^2+3c-5c-3 = 5c^2-2c-3$
Expand $(c+2)(c+1)$:
$c(c+1)+2(c+1) = c^2+c+2c+2 = c^2+3c+2$

Step3: Add numerators

Combine the expanded numerators over the common denominator:
$\frac{(5c^2-2c-3)+(c^2+3c+2)}{(c+1)(5c+3)}$
$\frac{5c^2-2c-3+c^2+3c+2}{(c+1)(5c+3)}$
$\frac{6c^2+c-1}{(c+1)(5c+3)}$

Step4: Factor numerator (simplify)

Factor $6c^2+c-1$:
$6c^2+3c-2c-1 = 3c(2c+1)-1(2c+1) = (3c-1)(2c+1)$
Factor denominator: $(c+1)(5c+3)$ (already factored, no common terms with numerator)
Expand the denominator to meet expanded form requirement:
$(c+1)(5c+3)=5c^2+3c+5c+3=5c^2+8c+3$

Answer:

$\frac{6c^2+c-1}{5c^2+8c+3}$