Solve x-1/x+1-x-2/x-1=-3

$$\frac { x - 1 } { x + 1 } - \frac { x - 2 } { x - 1 } = - 3$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=0$$

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.

$$\left(x-1\right)\left(x-1\right)-\left(x+1\right)\left(x-2\right)=-3\left(x-1\right)\left(x+1\right)$$

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

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

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

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

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

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

To find the opposite of x^{2}-x-2, find the opposite of each term.

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

Combine x^{2} and -x^{2} to get 0.

$$-2x+1+x+2=-3\left(x-1\right)\left(x+1\right)$$

Combine -2x and x to get -x.

$$-x+1+2=-3\left(x-1\right)\left(x+1\right)$$

Add 1 and 2 to get 3.

$$-x+3=-3\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply -3 by x-1.

$$-x+3=\left(-3x+3\right)\left(x+1\right)$$

Use the distributive property to multiply -3x+3 by x+1 and combine like terms.

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

Add 3x^{2} to both sides.

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

Subtract 3 from both sides.

$$-x+3+3x^{2}-3=0$$

Subtract 3 from 3 to get 0.

$$-x+3x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$3x^{2}-x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1}}{2\times 3}$$

Take the square root of 1.

$$x=\frac{-\left(-1\right)±1}{2\times 3}$$

The opposite of -1 is 1.

$$x=\frac{1±1}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{1±1}{6}$$

Now solve the equation x=\frac{1±1}{6} when ± is plus. Add 1 to 1.

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

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{1±1}{6} when ± is minus. Subtract 1 from 1.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

The equation is now solved.

$$x=\frac{1}{3} x=0$$

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.

$$\left(x-1\right)\left(x-1\right)-\left(x+1\right)\left(x-2\right)=-3\left(x-1\right)\left(x+1\right)$$

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

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

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

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

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

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

To find the opposite of x^{2}-x-2, find the opposite of each term.

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

Combine x^{2} and -x^{2} to get 0.

$$-2x+1+x+2=-3\left(x-1\right)\left(x+1\right)$$

Combine -2x and x to get -x.

$$-x+1+2=-3\left(x-1\right)\left(x+1\right)$$

Add 1 and 2 to get 3.

$$-x+3=-3\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply -3 by x-1.

$$-x+3=\left(-3x+3\right)\left(x+1\right)$$

Use the distributive property to multiply -3x+3 by x+1 and combine like terms.

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

Add 3x^{2} to both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

$$-x+3x^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$3x^{2}-x=0$$

Divide both sides by 3.

$$\frac{3x^{2}-x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{1}{3}x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}-\frac{1}{3}x=0$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{1}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{1}{6} x-\frac{1}{6}=-\frac{1}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{1}{3} x=0$$