SOLUTION: Can someone please tell me what I did wrong, IF I did indeed do anything wrong? This is the equation: (4x/x-2) = x+ (8/x-2) This is what I did to get my answer: (4x/x-2)

Algebra ->  Equations -> SOLUTION: Can someone please tell me what I did wrong, IF I did indeed do anything wrong? This is the equation: (4x/x-2) = x+ (8/x-2) This is what I did to get my answer: (4x/x-2)       Log On


   



Question 241381: Can someone please tell me what I did wrong, IF I did indeed do anything wrong?
This is the equation:
(4x/x-2) = x+ (8/x-2)
This is what I did to get my answer:
(4x/x-2) = x+ (8/x-2)
4x = x(x-2) + 8
4x = +x%5E2+ - 2x + 8
-4x -4x
+x%5E2+ - 6x + 8
(x-4)(x-2)


Your help would be greatly appreciate, as I desperately need it!

Found 2 solutions by Alan3354, josmiceli:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
(4x/x-2) = x+ (8/x-2)
This is what I did to get my answer:
(4x/x-2) = x+ (8/x-2)
4x = x(x-2) + 8
----------------
4x = x^2 - 2x + 8
x^2 - 6x + 8 = 0
(x-4)*(x-2) = 0
x = 4
x = 2 (2 causes the dreaded "Divide by 0" error, so reject it)
-----
I don't see how you did, but you got to the same place.
4x = - 2x + 8
-4x -4x
- 6x + 8
(x-4)(x-2)

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
It's not absolutly clear what you're doing,
although you have the right idea
4x%2F%28x-2%29+=+x+%2B+8%2F%28x-2%29
Multiply both sides by x-2
4x+=+x%2A%28x-2%29+%2B+8
4x+=+x%5E2+-+2x+%2B+8
x%5E2+-+6x+%2B+8+=+0
%28x+-+4%29%28x+-+2%29+=+0
The solutions are
x+=+4
x+=+2
These are the values that make the equation true
But the problem is that division by 0 is not
allowed, so x+=+2 must be rejected
I'll test the x+=+4 solution
4x%2F%28x-2%29+=+x+%2B+8%2F%28x-2%29
%284%2A4%29%2F%284-2%29+=+4+%2B+8%2F%284-2%29
16%2F2+=+4+%2B+8%2F2
8+=+4+%2B+4
OK