# How do you find the restrictions and simplify (3x-2)/(x+3)+7/(x^2-x-12)?

May 17, 2016

Let's first find the restrictions:

This can be done by factoring the denominator of the expression.

$\frac{3 x - 2}{x + 3} + \frac{7}{\left(x - 4\right) \left(x + 3\right)}$

A restriction in a rational expression occurs when the denominator equals 0, since division by 0 in mathematics is undefined.

Therefore, we must now set the factors in the denominator to 0 and solve for x. These will be our restrictions.

$x + 3 = 0 \mathmr{and} x - 4 = 0$

$x = - 3 \mathmr{and} x = 4$

Therefore, $x \ne - 3 , 4$
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Now, let's simplify:

This can be done by placing everything on a common denominator.

Since both expressions already has $\left(x + 3\right)$ as a factor in the denominator, and only one has $\left(x - 4\right)$, we must multiply the expression to the left by $\left(x - 4\right)$ to make it equivalent to the one on the right.

$\frac{\left(3 x - 2\right) \left(x - 4\right)}{\left(x + 3\right) \left(x - 4\right)} + \frac{7}{\left(x + 3\right) \left(x - 4\right)}$

$= \frac{3 {x}^{2} - 14 x + 8 + 7}{\left(x + 3\right) \left(x - 4\right)}$

$= \frac{3 {x}^{2} - 14 x + 15}{\left(x + 3\right) \left(x - 4\right)}$

The trinomial in the numerator is factorable. Always factor it when possible to see if anything can be simplified. You will be docked marks if you don't simplify fully. I factored this one and nothing needs to be eliminated. This is in simplest form.

Hopefully this helps!