Question 198040
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Once you have determined a common denominator, adding fractions is simply the sum of the numerators over that common denominator.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y + 2}{y^2 + 5y  + 6} + \frac{2 - y}{y^2 + y - 6}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y + 2}{(y + 3)(y + 2)} + \frac{2 - y}{(y + 3)(y - 2)}]


So far, so good.  But realize that *[tex \LARGE 2 - y = - (y - 2)], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{y + 3} + \frac{-1}{y + 3}]


Now just add the numerators -- <b><i>do not</i></b> add the denominators


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1 - 1}{y + 3}\ =\ 0]


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You started with the correct strategy, i.e. you multiplied the left hand fraction by 3 over 3, but then you again added the denominators when you were adding the fractions.  Your result should have been:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{10}{3x}]


And, by the way, you also made an error when you added the denominators:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x + 3x \neq 6x^2] rather *[tex \LARGE 3x + 3x = 6x]


Remember adding fractions in elementary school:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}]


<b><i>not</i></b>


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{4} + \frac{1}{4} \neq \frac{2}{8} = \frac{1}{4}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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