All steps for: x/x-2 + x-1/x+1= -1

[tex]\dfrac{x}{x-2}+\dfrac{x-1}{x+1}=-1\qquad\text{subtract}\ \dfrac{x-1}{x+1}\ \text{from both sides}\\\\\dfrac{x}{x-2}=-1-\dfrac{x-1}{x+1}\\\\\dfrac{x}{x-2}=\dfrac{-(x+1)}{x+1}+\dfrac{-(x-1)}{x+1}\\\\\dfrac{x}{x-2}=\dfrac{-(x+1)-(x-1)}{x+1}\\\\\dfrac{x}{x-2}=\dfrac{-x-1-x+1}{x+1}\\\\\dfrac{x}{x-2}=\dfrac{-2x}{x+1}\qquad\text{cross multiply}[/tex]

[tex]x(x+1)=-2x(x-2)\qquad\text{use the distributive property}\\\\(x)(x)+(x)(1)=(-2x)(x)+(-2x)(-2)\\\\x^2+x=-2x^2+4x\qquad\text{add}\ 2x^2\ \text{to both sides}\\\\3x^2+x=4x\qquad\text{subtract 4x from both sides}\\\\3x^2-3x=0\qquad\text{distributive}\\\\3x(x-1)=0\iff 3x=0\ \vee\ x-1=0\\\\x=0\in D\ \vee\ x=1\in D[/tex]

jcherry99 jcherry99 · Answer 2 · 2019-06-13T23:04:51+02:00

Answer:

Step-by-step explanation:

I'm taking this to mean

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

Multiply through by (x - 2)*(x + 1) to get rid of the denominator on the left.

x(x + 1) + (x - 1)(x - 2) = -1 * (x - 2)(x + 1)

Remove the brackets on the left and right.

Be careful about the right side. Do it in two steps (or three)

x^2 + x + x^2 - 3x + 2 = - (x^2 - 2x + x - 2)

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

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

Bring the right side to the left

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

3x^2 - 3x = 0

Factor this

x*(3x - 3) =0

x = 0

3x - 3 = 0

Add 3 to both sides.

3x = 3

Divide by 3

x = 3/3

So either x = 0

or

x = 1

Just to confirm that that is correct, a graph is included which shows the x roots are 0 and 1