Question

solve: -3/x-2 = 1 - 12/x^2-4

Answer 1

You want to find common terms here.

There's a very common, easy to spot little trick you can use here:

x^2 - 4 = (x-2)(x+2)

Or in general for any x and any y with an integer root:

x^2 - y = (x-y)(x+y)

So we can apply that here:

-3/(x-2) = (1-12)/(x-2)(x+2)

Solve for x now that it's easy.

Answer 2

how would I go about solving for x? @Rav

Answer 3

Continuing on:

-3/(x-2) = (1-12)/(x-2)(x+2)

See on both sides of the equation, you can see (x-2). We want that gone. Multiply both sides by (x-2):

-3 = (1-12)/(x+2)

Now multiply by (x+2) on both sides:
-3(x+2) = 1 - 12 = -11

Negate and simplify:

3x + 6 = 11

Move the 6:

3x = 11 - 6 = 5

Divide by 3:

x = 5/3

Answer 4

im kinda confuse because the only answers it could be is
A. -2
B. -5
C. -2,2
or D. 2.-5 @ rav

Answer 5

@Rav

Answer 6

Is it 1 - (12)/(x^2 - 4) or is it (1-12)/(x^2-4)?

Answer 7

its 1- (12(/(x^2-4)

Answer 8

Fixed:

-3/(x-2) = 1 - (12)/(x-2)(x+2)

-3 = (x-2) - (12)/(x+2)

Multiply by (x+2):
-3x - 6 = x^2 - 4 - 12

x^2 - 16 + 3x + 6 = 0

x^2 + 3x - 10 = 0

Now you can use the quadratic formula or you can go further with this method:

Now you want (x +/- a)(x +/- b) such that ab = 10 and a + b = 3

Can we have this? Yes: if |a| = 5 and |b| = 3, with one of them negative and one positive.

So that's:
x^2 + 3x - 10 = (x - 2)(x + 5) = 0

So now you substitute x-values until you find which answer fits.

Answer 9

okay thank you!

Answer 10

I got the answer B @Rav

Answer 11

-5 IS a solution to this problem. So is 2. Try out all answers! The correct answer is D.

Answer 12

ill try it again

Answer 13

thank you