Answer:
[tex]\frac{r^2-10}{(r+4)(r+1)}[/tex]
Step-by-step explanation:
We want to evaluate:
[tex]\frac{r+2}{r+4}-\frac{3}{r+1}[/tex]
We collect the least common multiple of the denominators to get:
[tex]\frac{(r+2)(r+1)-3(r+4)}{(r+4)(r+1)}[/tex]
Expand the numerator:
[tex]\frac{r^2+3r+2-3r-12}{(r+4)(r+1)}[/tex]
[tex]\frac{r^2-3r+3r+2-12}{(r+4)(r+1)}[/tex]
[tex]\frac{r^2-10}{(r+4)(r+1)}[/tex]
Answer:
[tex]\frac{ r^2-10}{(r+4)(r+1)}[/tex]
Step-by-step explanation:
[tex]\frac{r+2}{r+4} - \frac{3}{r+1}[/tex]
To simplify the fractions we take common denominator
LCD is (r+4)(r+1)
we try to get same denominator on both fraction
multiply the first fraction by r+1 and second fraction by r+4
[tex]\frac{(r+2)(r+1)}{(r+4)(r+1)} - \frac{3(r+4)}{(r+1)(r+4)}[/tex]
(r+2)(r+1) becomes r^2 +3r+2
3(r+4) becomes 3r+12
[tex]\frac{ r^2 +3r+2}{(r+4)(r+1)} - \frac{3r+12)}{(r+1)(r+4)}[/tex]
[tex]\frac{ r^2 +3r+2-3r-12}{(r+4)(r+1)}[/tex]
combine like terms
[tex]\frac{ r^2-10}{(r+4)(r+1)}[/tex]
