Does the above mean any of the following?BMWrider said:Problem:
x/x-2 - x+1/x = 8/x^2 - 2x
I'm sorry, but I don't understand what you're doing here...? How did you get that x(x - 2) equalled 8? Where did the fractions go? Did you multiply through at some point by the LCM?BMWrider said:x/x-2 - x+1/x = 8/x(x-2)
LCD x(x-2) = x^2-(x-2)(x+1) = 8
...etc, etc. But how did you get "x = -10"?Q) Solve \(\displaystyle \L \frac{x}{x\, -\, 2}\, -\, \frac{x\,+\,1}{x}\, =\, \frac{8}{x^2\,-\,2}\) and check your solution.
A) First, I factored the denominator on the right-hand side to get x(x - 2), and found the lowest common denominator to be x(x - 2). Multiplying through by this, I got:
. . .\(\displaystyle \L (x)(x)\, -\, (x\, +\, 1)(x\, -\, 2)\, =\, 8\)
. . .\(\displaystyle \L x^2 \, -\, (x^2\,+\,1x\,-\,2x\,- 2)\,=\,8\)
. . .\(\displaystyle \L x^2\, -\, (x^2\, -\, x\, -\, 2)\, =\, 8\)
. . .\(\displaystyle \L x^2\, -\,x^2\, +\, x\, +\, 2\, =\, 8\)
. . .\(\displaystyle \L x\, +\, 2\, =\, 8\)
You came up with the same result for the second fraction, after multiplying through:BMWrider said:how did you come up with (x + 1)(x - 2)? I think the x got omitted by mistake.
At which point in my steps do you feel I made the error?BMWrider said:x^2 - (x - 2)(x + 1) = 8
BMWrider said:Problem:
x/(x-2) - (x+1)/x = 8/(x^2 - 2x)
[x/(x-2) - (x+1)/x] * x * (x-2) = 8/(x^2 - 2x) * x * (x-2)
x*x - (x+1)(x-2) = 8
x^2 - (x^2 - 2x + x - 2) = 8
x^2 - (x^2 - x - 2) = 8
x^2 - x^2 + x + 2 = 8
x + 2 = 8
and so on....
Work:
x/x-2 - x+1/x = 8/x(x-2) LCD x(x-2) =
x^2-(x-2)(x+1) = 8 =
x^2 - x^2 + x - 2x - 2 = 8 =
-x = 10
x = -10
Trying to check the problem for -10 does not seem to work... any suggestions?