Answer:
4x + 6
Step-by-step explanation:
Given
[tex]\frac{x}{x^2+3x+2}[/tex] + [tex]\frac{3}{x+1}[/tex]
Before we can add the fractions we require them to have a common denominator.
Factor the denominator of the first fraction
[tex]\frac{x}{(x+1)(x+2)}[/tex] + [tex]\frac{3}{x+1}[/tex]
Multiply the numerator / denominator of the second fraction by (x + 2)
= [tex]\frac{x}{(x+1)(x+2)}[/tex] + [tex]\frac{3(x+2)}{(x+1)(x+2)}[/tex] ← fractions now have a common denominator
Add the numerators leaving the denominators
= [tex]\frac{x+3(x+2)}{(x+1)(x+2)}[/tex]
= [tex]\frac{x+3x+6}{(x+1)(x+2)}[/tex]
= [tex]\frac{4x+6}{(x+1)(x+2)}[/tex] ← simplified sum with numerator 4x + 6