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

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{15}i+3}{2}\approx 1.5-1.936491673i$$

$$x=\frac{3+\sqrt{15}i}{2}\approx 1.5+1.936491673i$$

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.

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

Subtract 2 from -1 to get -3.

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

Use the distributive property to multiply \frac{1}{2}x-\frac{1}{2} by x.

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

Subtract \frac{1}{2}x^{2} from both sides.

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

Add \frac{1}{2}x to both sides.

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

Combine x and \frac{1}{2}x to get \frac{3}{2}x.

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

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.

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

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

$$x=\frac{-\frac{3}{2}±\sqrt{\left(\frac{3}{2}\right)^{2}-4\left(-\frac{1}{2}\right)\left(-3\right)}}{2\left(-\frac{1}{2}\right)}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}-4\left(-\frac{1}{2}\right)\left(-3\right)}}{2\left(-\frac{1}{2}\right)}$$

Multiply -4 times -\frac{1}{2}.

$$x=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}+2\left(-3\right)}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -3.

$$x=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}-6}}{2\left(-\frac{1}{2}\right)}$$

Add \frac{9}{4} to -6.

$$x=\frac{-\frac{3}{2}±\sqrt{-\frac{15}{4}}}{2\left(-\frac{1}{2}\right)}$$

Take the square root of -\frac{15}{4}.

$$x=\frac{-\frac{3}{2}±\frac{\sqrt{15}i}{2}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -\frac{1}{2}.

$$x=\frac{-\frac{3}{2}±\frac{\sqrt{15}i}{2}}{-1}$$

Now solve the equation x=\frac{-\frac{3}{2}±\frac{\sqrt{15}i}{2}}{-1} when ± is plus. Add -\frac{3}{2} to \frac{i\sqrt{15}}{2}.

$$x=\frac{-3+\sqrt{15}i}{-2}$$

Divide \frac{-3+i\sqrt{15}}{2} by -1.

$$x=\frac{-\sqrt{15}i+3}{2}$$

Now solve the equation x=\frac{-\frac{3}{2}±\frac{\sqrt{15}i}{2}}{-1} when ± is minus. Subtract \frac{i\sqrt{15}}{2} from -\frac{3}{2}.

$$x=\frac{-\sqrt{15}i-3}{-2}$$

Divide \frac{-3-i\sqrt{15}}{2} by -1.

$$x=\frac{3+\sqrt{15}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{15}i+3}{2} x=\frac{3+\sqrt{15}i}{2}$$

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.

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

Subtract 2 from -1 to get -3.

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

Use the distributive property to multiply \frac{1}{2}x-\frac{1}{2} by x.

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

Subtract \frac{1}{2}x^{2} from both sides.

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

Add \frac{1}{2}x to both sides.

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

Combine x and \frac{1}{2}x to get \frac{3}{2}x.

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

Add 3 to both sides. Anything plus zero gives itself.

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

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.

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

Multiply both sides by -2.

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

Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.

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

Divide \frac{3}{2} by -\frac{1}{2} by multiplying \frac{3}{2} by the reciprocal of -\frac{1}{2}.

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

Divide 3 by -\frac{1}{2} by multiplying 3 by the reciprocal of -\frac{1}{2}.

$$x^{2}-3x=-6$$

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-6+\frac{9}{4}$$

Add -6 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=-\frac{15}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=-\frac{15}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{15}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{15}i}{2} x-\frac{3}{2}=-\frac{\sqrt{15}i}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3+\sqrt{15}i}{2} x=\frac{-\sqrt{15}i+3}{2}$$