Solve g(x)=x/3x-2=x

$$g ( x ) = \frac { x } { 3 x - 2 } = x$$

Solve for x

$$x=1$$

$$x=0$$

Steps Using Factoring

Variable x cannot be equal to \frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by 3x-2.

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

Use the distributive property to multiply x by 3x-2.

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

Subtract 3x^{2} from both sides.

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

Add 2x to both sides.

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

Combine x and 2x to get 3x.

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

Factor out x.

$$x\left(3-3x\right)=0$$

To find equation solutions, solve x=0 and 3-3x=0.

$$x=0 x=1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to \frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by 3x-2.

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

Use the distributive property to multiply x by 3x-2.

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

Subtract 3x^{2} from both sides.

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

Add 2x to both sides.

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

Combine x and 2x to get 3x.

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

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.

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

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

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

Take the square root of 3^{2}.

$$x=\frac{-3±3}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-3±3}{-6}$$

Now solve the equation x=\frac{-3±3}{-6} when ± is plus. Add -3 to 3.

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

Divide 0 by -6.

$$x=0$$

Now solve the equation x=\frac{-3±3}{-6} when ± is minus. Subtract 3 from -3.

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

Divide -6 by -6.

$$x=1$$

The equation is now solved.

$$x=0 x=1$$

Steps for Completing the Square

Variable x cannot be equal to \frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by 3x-2.

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

Use the distributive property to multiply x by 3x-2.

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

Subtract 3x^{2} from both sides.

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

Add 2x to both sides.

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

Combine x and 2x to get 3x.

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

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.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 3 by -3.

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

Divide 0 by -3.

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

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.

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

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

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$x=1 x=0$$