fractional equation problem?

Question

(x-1)/2x-2x/(x-1)
the answer is 5x^2-2x+1/2x^2-2x

but my answer is -3x^2-2x+1/2x^2-2x

please give me the whole process hint, as I am really dummy to algebra.

Answer 1

your answer is correct. the difference is in the sign of the (4x^2).
this is just the numerator (top):
(x-1)(x-1) - (2x)(2x)
x^2 -2x +1 - 4x^2
-3x^2 -2x +1
we arrive at your answer.

if you misread the problem as minus instead of plus, then:
(x-1)(x-1) + (2x)(2x)
x^2 - 2x + 1 + 4x^2
5x^2 -2x +1
the book answer.

Answer 2

you mix the fraction using for denominator 2x (x-1)which is the product of denominators

so the first fraction is (x-1)(x-1)/2x(x-1) = (x^2-2x +1)/2x (x-1)

and the second is 4x^2 / 2x (x-1)

So ,as you have the same denominator , you can do the operation

AND I FOUND as YOU -3x^2-2x+1/ 2x^2 -x


Where did you found the wrong answer???

Answer 3

(x-1)/2x-2x/(x-1) =
LCF(2x; x-1) = 2x(x-1)
[(x-1)² - (2x)² ] / 2x(x-1) =
[(x² - 2x + 1) +4x²] / 2x² - 2x =
x² - 2x + 1 + 4x² / 2x² - 2x
5x² - 2x + 1 / 2x² - 2x or
5x² - 2x + 1 / 2x(x-1)

For the result 5x², it was considered minus as signal of fraction (-2x/x-1). If so, for me, both results should be consider correct because the expression is not clear. ><

Answer 4

I agree with the above poster. I got the answer you did, provided that you typed the problem correctly above.