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2(x^2 - 9) / (x^2 + x - 6) + 2/(x-2) =?

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by lee3381 » Thu Jan 14, 2016 4:44 am
2(x^2 - 9) / (x^2 + x - 6) + 2/(x-2) = ?

The answer is 2.

Can anyone explain this answer?
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by chetan.sharma » Thu Jan 14, 2016 8:19 am
lee3381 wrote:2(x^2 - 9) / (x^2 + x - 6) + 2/(x-2) = ?

The answer is 2.

Can anyone explain this answer?
hi,
first let me factorize different equations ..
(x^2 - 9)=(x-3)(x+3)...
(x^2 + x - 6)= x^2+3x-2x-6=x(x+3)-2(x+3)=(x-2)(x+3)..

now lets substitute these values in main equation..
2(x^2 - 9) / (x^2 + x - 6) + 2/(x-2)..
2(x-3)(x+3) / (x-2)(x+3) + 2/(x-2)=2(x-3)/(x-2) + 2/(x-2)...
{2(x-3)+2}/(x-2) = (2x-6+2)/(x-2)..
(2x-4)/(x-2)= 2(x-2)/(x-2)=2..

Hope it helps .
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by [email protected] » Thu Jan 14, 2016 12:01 pm
Hi All,

This prompt might be confusing to certain Test Takers, so here are a couple of things to note about it:

1) It's not 'formatted' particularly well. Having brackets around the first 'fraction' would probably help...

[2(X^2 - 9) / (X^2 + X - 6)] + 2/(X-2)

In this way, you can see that you're adding together 2 distinct fractions.

2) You're NOT trying to solve for the value of X. You're attempting to simplify the given equation. The math itself involves Classic Quadratics and 'canceling out' terms.

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