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

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

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=0$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by x-1.

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

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

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

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

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

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

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

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

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

Subtract 3x^{2} from both sides.

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

Combine x^{2} and -3x^{2} to get -2x^{2}.

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

Add 3x to both sides.

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

Combine -2x and 3x to get x.

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

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

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

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

Take the square root of 1^{2}.

$$x=\frac{-1±1}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-1±1}{-4}$$

Now solve the equation x=\frac{-1±1}{-4} when ± is plus. Add -1 to 1.

$$x=\frac{0}{-4}$$

Divide 0 by -4.

$$x=0$$

Now solve the equation x=\frac{-1±1}{-4} when ± is minus. Subtract 1 from -1.

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

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=0 x=\frac{1}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by x-1.

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

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

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

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

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

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

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

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

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

Subtract 3x^{2} from both sides.

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

Combine x^{2} and -3x^{2} to get -2x^{2}.

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

Add 3x to both sides.

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

Combine -2x and 3x to get x.

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 1 by -2.

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

Divide 0 by -2.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{1}{4} x-\frac{1}{4}=-\frac{1}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{1}{2} x=0$$