Question

Simplify, thank you in advance

Answer 1

$\bf \cfrac{3}{x+4}+\cfrac{2}{x^2-16}\implies \cfrac{3}{x+4}+\cfrac{2}{\stackrel{\textit{difference of squares}}{x^2-4^2}}\implies \cfrac{3}{x+4}+\cfrac{2}{(x-4)(x+4)}\\\\\\\stackrel{\textit{so our LCD will be }(x-4)(x+4)}{\cfrac{(x-4)3+(1)2}{(x-4)(x+4)}}\implies \cfrac{3x-12+2}{(x-4)(x+4)}\implies \cfrac{3x-10}{(x-4)(x+4)}$

Answer 2

So firstly, we will be using the difference of squares with the second fraction's denominator. The difference of squares is $x^2+y^2=(x+y)(x-y)$ . Apply this rule to this expression: $\frac{3}{x+4}+\frac{2}{(x+4)(x-4)}$

Next, we have to find the LCM, or least common multiple, of both denominators. In this case, the LCM is (x + 4)(x - 4). Multiply the denominators with the quantity that gets the LCM as the denominator, and then multiply by that same amount on their numerators:

$\frac{3}{x+4}*\frac{(x-4)}{(x-4)}=\frac{3(x-4)}{(x+4)(x-4)}\\ \\ \frac{2}{(x+4)(x-4)}*\frac{1}{1}=\frac{2}{(x+4)(x-4)}\\ \\ \frac{3(x-4)}{(x+4)(x-4)}+\frac{2}{(x+4)(x-4)}$

Now foil 3(x - 4): $\frac{3x-12}{(x+4)(x-4)}+\frac{2}{(x+4)(x-4)}$

And lastly, add the numerators up and your final answer will be $\frac{3x-10}{(x+4)(x-4)}$