004 Sample Final A, Problem 6

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

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

Solution:

ExpandStep 1:
If we factor the denominators, we have ${\frac {1}{3(x+2)}}-{\frac {x}{(x+2)(x-2)}}+{\frac {3}{x-2}}$.
So, the common denominator of these three fractions is $3(x-2)(x+2)$.

ExpandStep 2:
So, we have ${\frac {1}{3(x+2)}}-{\frac {x}{(x+2)(x-2)}}+{\frac {3}{x-2}}={\frac {x-2}{3(x-2)(x+2)}}-{\frac {3x}{3(x+2)(x-2)}}+{\frac {3(3)(x+2)}{3(x+2)(x-2)}}$.

ExpandStep 3:
Now, combining into one fraction, we have ${\frac {x-2-3x+3(3)(x+2)}{3(x-2)(x+2)}}={\frac {7x+16}{3(x-2)(x+2)}}$

ExpandFinal Answer:
${\frac {7x+16}{3(x-2)(x+2)}}$

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