# 004 Sample Final A, Problem 6

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Simplify.      ${\displaystyle {\frac {1}{3x+6}}-{\frac {x}{x^{2}-4}}+{\frac {3}{x-2}}}$

Foundations
How do you simplify ${\displaystyle {\frac {1}{x}}+{\frac {1}{x+2}}}$ into one fraction?
You need to get a common denominator. The common denominator is ${\displaystyle x(x+2)}$. So,
${\displaystyle {\frac {1}{x}}+{\frac {1}{x+2}}={\frac {x+2}{x(x+2)}}+{\frac {x}{x(x+2)}}={\frac {2x+2}{x(x+2)}}}$.

Solution:

Step 1:
If we factor the denominators, we have ${\displaystyle {\frac {1}{3(x+2)}}-{\frac {x}{(x+2)(x-2)}}+{\frac {3}{x-2}}}$.
So, the common denominator of these three fractions is ${\displaystyle 3(x-2)(x+2)}$.
Step 2:
So, we have ${\displaystyle {\frac {1}{3(x+2)}}-{\frac {x}{(x+2)(x-2)}}+{\frac {3}{x-2}}={\frac {x-2}{3(x-2)(x+2)}}-{\frac {3x}{3(x+2)(x-2)}}+{\frac {3(3)(x+2)}{3(x+2)(x-2)}}}$.
Step 3:
Now, combining into one fraction, we have ${\displaystyle {\frac {x-2-3x+3(3)(x+2)}{3(x-2)(x+2)}}={\frac {7x+16}{3(x-2)(x+2)}}}$
${\displaystyle {\frac {7x+16}{3(x-2)(x+2)}}}$