Can you guys explain how to solve this (x/x-2)+(x-1/x+1)=-1

Question

jadonbb · Answer

$$(x \div x-2) + (x-1\div x+1)=-1$$

p0sitr0n · Answer

so um first put them all on a common denominator: $\frac{2x^2-2x+2}{(x-2)(x+1)}=-1 $
or $ 2x^2-2x+2 = (2-x)(x+1)$. Now solve it.

mathstudent55 · Answer

Although you wrote 

(x/x-2)+(x-1/x+1)=-1 

which means

$\dfrac{x}{x}-2+x-\dfrac{1}{x}+1=-1 $

you really meant this

$\dfrac{x}{x-2}+\dfrac{x-1}{x+1}=-1 $

right?

jadonbb · Answer

yes @mathstudent55

mathstudent55 · Answer

If the problem is really the last one above, then there are several ways of solving.
Here is one of them.

Add the fractions on the left side by using the LCD.
The LCD of x - 2 and x + 1 is (x - 2)(x + 1).

jadonbb · Answer

@mathstudent55 I actually solved it earlier today x=0,1 is i remember correctly

mathstudent55 · Answer

We need to multiply the left fraction by $\dfrac{x + 1}{x + 1}$ and the right fraction by $\dfrac{x - 2}{x - 2} $ to have a common denominator.

$\dfrac{x}{x-2} \times \dfrac{x + 1}{x + 1} +\dfrac{x-1}{x+1} \times \dfrac{x - 2}{x - 2} =-1$

$\dfrac{x(x + 1)}{(x+1)(x - 2)} +\dfrac{(x-1)(x - 2)}{(x+1)(x - 2)} =-1$

$\dfrac{x^2 + x}{(x+1)(x - 2)} +\dfrac{x^2 -2x - x + 2}{(x + 1)(x - 2)} =-1$

$\dfrac{2x^2 -2x + 2}{(x+1)(x - 2)} =-1$

$2x^2 -2x + 2 =-1(x+1)(x - 2)$

$2x^2 -2x + 2 = -(x^2 -2x + x - 2)$

$2x^2 -2x + 2 = -x^2 + x + 2$

$3x^2 - 3x = 0$

$3x(x - 1) = 0$

$x = 0$  or  $x - 1 = 0$

$x = 0$  or  $x = 1$

mathstudent55 · Answer

Now we check the original equation.
It has denominators x - 2 and x + 1.
That means the solution cannot be x = 2 or x = -1.
Since our solutions are x = 0 and x = 1, these solutions are acceptable, and they are the correct solution to the equation.