Answer:
[tex]\frac{3x+6}{x^{2}-x-6 } +\frac{2x}{x^{2} +x-12}=\frac{5x+12}{x^{2} +x-12}[/tex]
Step-by-step explanation:
The given expression is
[tex]\frac{3x+6}{x^{2}-x-6 } +\frac{2x}{x^{2} +x-12}[/tex]
First, we need to factor each part of the expression
[tex]\frac{3(x+2)}{(x+2)(x-3)} +\frac{2x}{(x-3)(x+4)}[/tex]
Remember that quadractic expression are factored in two binomial factors. The first quadratic expression factors are about two numbers which product is 6 and which difference is one. The second quadratic expression is about two numbers which product is 12 and which difference is 1.
Now, we simplify equal expression at each fraction.
[tex]\frac{3}{(x-3)} +\frac{2x}{(x-3)(x+4)}[/tex]
Then, we use the least common factor about the denominators to sum those fractions. In this case, the least common factor is [tex](x-3)(x+4)[/tex], because those are the factors present in the denominators.
Now, we divide each fraction by the least common factor, and then multiply the numeratos by its result.
[tex]\frac{3(x+4)+2x}{(x-3)(x+4)}[/tex]
Finally, we multiply all products and sum like terms.
[tex]\frac{3x+12+2x}{x^{2} +4x-3x-12}=\frac{5x+12}{x^{2} +x-12}[/tex]
Therefore, the sum of the initial expression is equal to
[tex]\frac{5x+12}{x^{2} +x-12}[/tex]
3(x+2)/(x+2)(x-3) + 2x/(x-3)(x+4)
↓
3/(x-3) + 2x/(x-3)(x+4)
↓
3(x+4)/(x-3)(x+4) + 2x/(x-3)(x+4)
↓
(3x+12)+2x/(x-3)(x+4)
↓
5x+12/(x-3)(x+4)
(Apologies! I got it incorrect the first time in the image below, but I edited it-so, now it's correct)
