Answer:
[tex] \frac{2(2x+3)}{(x+2)(x+1)}= \frac{4x+6}{(x+2)(x+1)}[/tex]
And the numerator for the simplied sum is [tex] 4x+6[/tex]
Step-by-step explanation:
For this case we have the following expression :
[tex] \frac{x}{x^2 +3x +2} +\frac{3}{x+1}[/tex]
And we can begin doing this:
[tex] \frac{x(x+1) + 3(x^2 +3x+2)}{(x^2+3x+2)(x+1)}[/tex]
Now we can distribute the terms on the numerator, and factorize the [tex] x^2 +3x+ +2 = (x+1)(x+2)[/tex] on the denominator and we got:
[tex] \frac{x^2 +x + 3x^2 +9x +6}{(x+2) (x+1)^2}[/tex]
[tex] \frac{4x^2 +10x +6}{(x+2) (x+1)^2}[/tex]
We can now factorize the nuemrator like this:
[tex] 4x^2 +10x + 6 = (2x+2) (2x+3)[/tex]
And if we replace this we got:
[tex] \frac{(2x+2) (2x+3)}{(x+2) (x+1)^2}[/tex]
Now we can take common factor 2 on the numerator like this:
[tex] \frac{2(x+1) (2x+3)}{(x+2) (x+1)^2}[/tex]
And we can cancel the x+1 on the numerator and denominator like this:
[tex] \frac{2(2x+3)}{(x+2)(x+1)}= \frac{4x+6}{(x+2)(x+1)}[/tex]
And the numerator for the simplied sum is [tex] 4x+6[/tex]
