Rational Expressions

[BMWrider]

Problem:
x/x-2 - x+1/x = 8/x^2 - 2x

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?

 

[stapel]

Problem:
x/x-2 - x+1/x = 8/x^2 - 2x

Does the above mean any of the following?

. . . . .x/(x - 2) - x + 1/x = 8/x^2 - 2x

. . . . .x/(x - 2) - (x + 1)/x = 8/x^2 - 2x

. . . . .x/(x - 2) - x + 1/x = 8/(x^2 - 2x)

. . . . .x/(x - 2) - (x + 1)/x = 8/x^2 - 2x)

Or something else?

x/x-2 - x+1/x = 8/x(x-2)

LCD x(x-2) = x^2-(x-2)(x+1) = 8

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?

I think you mean the following:

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:

. . .\(\displaystyle \L (x)(x)\, -\, (x\, +\, 1)(x\, -\, 2)\, =\, 8\)

. . .\(\displaystyle \L x^2 \, -\, (x^2\,+\,1x\,-\,2x\,- 2)\,=\,8\)

. . .\(\displaystyle \L x^2\, -\, (x^2\, -\, x\, -\, 2)\, =\, 8\)

. . .\(\displaystyle \L x^2\, -\,x^2\, +\, x\, +\, 2\, =\, 8\)

. . .\(\displaystyle \L x\, +\, 2\, =\, 8\)

...etc, etc. But how did you get "x = -10"?

Please forgive my confusion and be complete in your reply. Thank you!

Eliz.

 

[BMWrider]

Rational Expression Response

Correct, I multiplied by the LCD to cancel x(x-2) leaving 8

I factored the denominator of x^2 - 2x to x(x-2),

how did you come up with (x+1)(x-2)? I think the x got omitted by mistake.

 

[stapel]

how did you come up with (x + 1)(x - 2)? I think the x got omitted by mistake.

You came up with the same result for the second fraction, after multiplying through:

x^2 - (x - 2)(x + 1) = 8

At which point in my steps do you feel I made the error?

Also, what x are you saying was omitted, and from which point?

Please be specific. Thank you!

Eliz.

 

[Deleted member 4993]

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?

 

[BMWrider]

Thank you, I had a feeling it had something to do with how I applied the signs when removing the parenthesis from the trinomial.

This makes sense!