Solve 2x-5/x-1=x-1/x+1

$$\frac{ 2x-5 }{ x-1 } = \frac{ x-1 }{ x+1 }$$

Solve for x

$$x=-2$$

$$x=3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right), the least common multiple of x-1,x+1.

$$\left(x+1\right)\left(2x-5\right)=\left(x-1\right)\left(x-1\right)$$

Multiply x-1 and x-1 to get \left(x-1\right)^{2}.

$$\left(x+1\right)\left(2x-5\right)=\left(x-1\right)^{2}$$

Use the distributive property to multiply x+1 by 2x-5 and combine like terms.

$$2x^{2}-3x-5=\left(x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$2x^{2}-3x-5=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$2x^{2}-3x-5-x^{2}=-2x+1$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$x^{2}-3x-5=-2x+1$$

Add 2x to both sides.

$$x^{2}-3x-5+2x=1$$

Combine -3x and 2x to get -x.

$$x^{2}-x-5=1$$

Subtract 1 from both sides.

$$x^{2}-x-5-1=0$$

Subtract 1 from -5 to get -6.

$$x^{2}-x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2}$$

Add 1 to 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2}$$

Take the square root of 25.

$$x=\frac{-\left(-1\right)±5}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±5}{2}$$

Now solve the equation x=\frac{1±5}{2} when ± is plus. Add 1 to 5.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{1±5}{2} when ± is minus. Subtract 5 from 1.

$$x=-\frac{4}{2}$$

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=3 x=-2$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right), the least common multiple of x-1,x+1.

$$\left(x+1\right)\left(2x-5\right)=\left(x-1\right)\left(x-1\right)$$

Multiply x-1 and x-1 to get \left(x-1\right)^{2}.

$$\left(x+1\right)\left(2x-5\right)=\left(x-1\right)^{2}$$

Use the distributive property to multiply x+1 by 2x-5 and combine like terms.

$$2x^{2}-3x-5=\left(x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$2x^{2}-3x-5=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$2x^{2}-3x-5-x^{2}=-2x+1$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$x^{2}-3x-5=-2x+1$$

Add 2x to both sides.

$$x^{2}-3x-5+2x=1$$

Combine -3x and 2x to get -x.

$$x^{2}-x-5=1$$

Add 5 to both sides.

$$x^{2}-x=1+5$$

Add 1 and 5 to get 6.

$$x^{2}-x=6$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=6+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{25}{4}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{5}{2} x-\frac{1}{2}=-\frac{5}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=3 x=-2$$