Okay so i dont understand how to do this problem...
x+6 + x+3
x2+x 12 x2-16
(x2=x squared)
Im supposed to simplyfy and state the restrictions on the variables
if it was $\displaystyle x^2 {\color{red}-}x - 12$ in the denominator of the first fraction, we would be in business. but since it is + x, nothing cancels. we get: $\displaystyle \frac {x + 6}{x^2 + x - 12} \cdot \frac {x + 3}{x^2 - 16} = \frac {(x + 6)(x + 3)}{(x + 4)^2 (x - 4)(x + 3)}$
the restrictions are, we cannot choose the $\displaystyle x$'s that make the denominator zero. what x's are those?
ok.
$\displaystyle \frac {x + 6}{x^2 + x - 12} + \frac {x + 3}{x^2 - 16} = \frac {x + 6}{(x + 4)(x - 3)} + \frac {x + 3}{(x + 4)(x - 4)}$
LCD here is: $\displaystyle (x + 4)(x - 4)(x - 3)$
so we have: $\displaystyle \frac {x + 6}{(x + 4)(x - 3)} + \frac {x + 3}{(x + 4)(x - 4)} = \frac {(x + 6)(x - 4) + (x + 3)(x - 3)}{(x + 4)(x - 4)(x - 3)}$
$\displaystyle = \frac {x^2 + 2x - 24 + x^2 - 9}{(x + 4)(x - 4)(x - 3)}$
$\displaystyle = \frac {2x^2 + 2x - 33}{(x + 4)(x - 4)(x - 3)}$
that cannot be factored any further.
our restrictions are: $\displaystyle x \ne \pm 4$ and $\displaystyle x \ne 3$ since those x's make the denominator zero