Question

Solve the eqation:
a 2 1
______ + ______ = _____
a^2 -36 a - 6 a + 6

Answer 1

for starters \[(a^2-36) = (a-6)(a+6)\] then you can find a common denominator and solve for a

Answer 2

-6

Answer 3

Could you guys tell me how you got your answer please??

Answer 4

a should be -9 i think

Answer 5

you have a/[(a-6)(a+6)] + 2(a+6)/[(a-6)(a+6)] = 1(a-6)/[(a-6)(a+6)]
withe [(a-6)(a+6)] as your common denominator

Answer 6

this becomes (a + 2a + 12)/[(a-6)(a+6)] = (a-6)/[(a-6)(a+6)]

Answer 7

then (3a+12)/[(a-6)(a+6)] = (a-6)/[(a-6)(a+6)] becomes just (3a+12) = (a-6) by multiplying both sides by the denominator and then you can solve for a

Answer 8

sure

a 2 1
______ + ______ = _____
a^2 -36 a - 6 a + 6

a
------- + 2/a-6==1/a+6
(a−6)(a+6)


taking 1/a-6 common on lhs

it becomes

(1/a-6)(a/a+6 +2)=1/a+6

(1/a-6)(3a+6/a+6)=1/a+6

cancel a+6 on both sides

solve x

Answer 9

(3a+6)/(a-6)=1

3a+6=a-6

2a=-12

a=-6

Answer 10

doesnt that make the denominator zero for the rhs of the original equation?

Answer 11

I think he's right cause it can't be -9. When i put -9 my teacher marked it wrong so it can't be that. But anyways thanks guys

Answer 12

~.~ im not convinced that this is -6. If you plug in -9 into the original equation you get the right answer
http://www.wolframalpha.com/input/?i=%28-9%29%2F%28%28-9%29^2-36%29%2B2%2F%28%28-9%29-6%29+-+1%2F%28%28-9%29%2B6%29
should equal zero when i subtract both sides by 1/(a+6) where a=-9

if a=-6 then you have a denominator being zero which dividing by zero is impossible.
if you plug in -9 into a calculator you get that both sides of the equation are the same and -6 is uncomputable since its dividing by zero

Answer 13

i agree with you

Answer 14

yea.. i see what your saying