how do you solve (x^2+ x)/(x^2- x) = -1?

Question

I'm doing logarithmic equations and i got stuck on this part. I guess i forgot how to do rational equations. any help is appreciated. Thanks so much.

Answer 1

OK - let's try and mske this a little easier

(x^2 + x) / (x^2 - x) = -1
x(x+1) / x (x-1) = -1
x+1 / x-1 = -1
x+1 = 1-x ; well as you can see the only solution in 0

But - if we place 0 in the original equation, the equation is undefined. So - there is no solution.

Hope that helps.

Answer 2

because x^2-x is denominator, it can not be 0
so x(x-1) can not be 0, so xcan not be 1 or 0
the eq could be written x^2+x= - x^2+x which is the same as 2x^2=0, or x=0 (which is not solution)

Answer 3

multiply both sides by x^2-x
you get x^2 + x = -x^2+1
add x^2 and subtract 1 from both sides
you get 2x^2+x-1=0
factor
(2x-1)(x+1)=0
Set each equal to 0
2x-1=0 x+1=0
2x=1 x=-1
x=1/2 or x=-1

Answer 4

x^2 + x = -x^2 +x
2x^2 = 0
x = 0

But then the eq is undefined, so there is no sol.

Answer 5

(x² + x) / (x² - x) = -1
x² + x = -1 * (x² - x)
x² + x = -x² + x
2x² + x ≠ x

I don't think there's a solution to this problem.