Question

Solve the equation and check the solution:
\[\dfrac{a}{(a-6) (a+6)} + \dfrac{2}{(a-6)} = \dfrac{1}{(a+6)}\]

Answer 1

Two choices: put all the fractions over a common denominator, or clear the fractions by multiplying through by \((a-6)(a+6)\)

Answer 2

\[\dfrac{a}{a^2-36}+\dfrac{2}{a-6}= \dfrac{1}{a+6}\]
\[\dfrac{a}{(a-6) (a+6)} + \dfrac{2}{(a-6)} = \dfrac{1}{(a+6)}\]
\[\dfrac{a}{(a-6)(a+6)} + \dfrac{2(a+6)}{(a-6)(a+6)} = \dfrac{(a-6)}{(a+6)(a-6)}\]
\[\dfrac{a + 2(a+6) - (a-6)}{(a+6)(a-6)} = 0\]
\[\dfrac{2+2a+16-a+6}{(a+6)(a-6)}=0\]
\[\dfrac{a+24}{(a+6)(a-6)}= 0\]
\[a+24 = 0\]
\[a = -24 \]

My textbook says the answer should be -9 what did I do wrong?

Answer 3

In 4th last step, you have done a silly mistake my dear.The correct equation will be :
(a+2a+12-a+6)/(a-6)(a+6)=0
(2a+18)=0
2a=-18
a=-9
which is the answer suggested by your textbook

Answer 4

I would clear the denominators, giving me\[a + 2(a+6) = a-6\]\[3a + 12 = a-6\]\[2a=-18\]\[a=-9\]

Answer 5

Your mistake was taking \(2(a+6)\) and making \(2a+16\) out of it instead of \(2a+12\)