Solve 12x=6x/x-1

$$12 x = \frac { 6 x } { x - 1 }$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x=0$$

Steps Using Factoring

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

$$12x\left(x-1\right)=6x$$

Use the distributive property to multiply 12x by x-1.

$$12x^{2}-12x=6x$$

Subtract 6x from both sides.

$$12x^{2}-12x-6x=0$$

Combine -12x and -6x to get -18x.

$$12x^{2}-18x=0$$

Factor out x.

$$x\left(12x-18\right)=0$$

To find equation solutions, solve x=0 and 12x-18=0.

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

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.

$$12x\left(x-1\right)=6x$$

Use the distributive property to multiply 12x by x-1.

$$12x^{2}-12x=6x$$

Subtract 6x from both sides.

$$12x^{2}-12x-6x=0$$

Combine -12x and -6x to get -18x.

$$12x^{2}-18x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -18 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}}}{2\times 12}$$

Take the square root of \left(-18\right)^{2}.

$$x=\frac{-\left(-18\right)±18}{2\times 12}$$

The opposite of -18 is 18.

$$x=\frac{18±18}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{18±18}{24}$$

Now solve the equation x=\frac{18±18}{24} when ± is plus. Add 18 to 18.

$$x=\frac{36}{24}$$

Reduce the fraction \frac{36}{24} to lowest terms by extracting and canceling out 12.

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

Now solve the equation x=\frac{18±18}{24} when ± is minus. Subtract 18 from 18.

$$x=\frac{0}{24}$$

Divide 0 by 24.

$$x=0$$

The equation is now solved.

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

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.

$$12x\left(x-1\right)=6x$$

Use the distributive property to multiply 12x by x-1.

$$12x^{2}-12x=6x$$

Subtract 6x from both sides.

$$12x^{2}-12x-6x=0$$

Combine -12x and -6x to get -18x.

$$12x^{2}-18x=0$$

Divide both sides by 12.

$$\frac{12x^{2}-18x}{12}=\frac{0}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\left(-\frac{18}{12}\right)x=\frac{0}{12}$$

Reduce the fraction \frac{-18}{12} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{3}{2}x=\frac{0}{12}$$

Divide 0 by 12.

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

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{9}{16}$$

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

$$\left(x-\frac{3}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{3}{4} x-\frac{3}{4}=-\frac{3}{4}$$

Add \frac{3}{4} to both sides of the equation.

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