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

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

Solve for x

$$x=\frac{\sqrt{33}}{12}+\frac{1}{4}\approx 0.728713554$$

$$x=-\frac{\sqrt{33}}{12}+\frac{1}{4}\approx -0.228713554$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3x.

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

Multiply 3 and -2 to get -6.

$$1-6x=2x\times 3x+3x\left(-3\right)$$

Multiply x and x to get x^{2}.

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

Multiply 2 and 3 to get 6.

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

Multiply 3 and -3 to get -9.

$$1-6x=6x^{2}-9x$$

Subtract 6x^{2} from both sides.

$$1-6x-6x^{2}=-9x$$

Add 9x to both sides.

$$1-6x-6x^{2}+9x=0$$

Combine -6x and 9x to get 3x.

$$1+3x-6x^{2}=0$$

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.

$$-6x^{2}+3x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 3 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 3.

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

Multiply -4 times -6.

$$x=\frac{-3±\sqrt{9+24}}{2\left(-6\right)}$$

Add 9 to 24.

$$x=\frac{-3±\sqrt{33}}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-3±\sqrt{33}}{-12}$$

Now solve the equation x=\frac{-3±\sqrt{33}}{-12} when ± is plus. Add -3 to \sqrt{33}.

$$x=\frac{\sqrt{33}-3}{-12}$$

Divide -3+\sqrt{33} by -12.

$$x=-\frac{\sqrt{33}}{12}+\frac{1}{4}$$

Now solve the equation x=\frac{-3±\sqrt{33}}{-12} when ± is minus. Subtract \sqrt{33} from -3.

$$x=\frac{-\sqrt{33}-3}{-12}$$

Divide -3-\sqrt{33} by -12.

$$x=\frac{\sqrt{33}}{12}+\frac{1}{4}$$

The equation is now solved.

$$x=-\frac{\sqrt{33}}{12}+\frac{1}{4} x=\frac{\sqrt{33}}{12}+\frac{1}{4}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3x.

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

Multiply 3 and -2 to get -6.

$$1-6x=2x\times 3x+3x\left(-3\right)$$

Multiply x and x to get x^{2}.

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

Multiply 2 and 3 to get 6.

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

Multiply 3 and -3 to get -9.

$$1-6x=6x^{2}-9x$$

Subtract 6x^{2} from both sides.

$$1-6x-6x^{2}=-9x$$

Add 9x to both sides.

$$1-6x-6x^{2}+9x=0$$

Combine -6x and 9x to get 3x.

$$1+3x-6x^{2}=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$3x-6x^{2}=-1$$

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.

$$-6x^{2}+3x=-1$$

Divide both sides by -6.

$$\frac{-6x^{2}+3x}{-6}=-\frac{1}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{3}{-6}x=-\frac{1}{-6}$$

Reduce the fraction \frac{3}{-6} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{1}{2}x=-\frac{1}{-6}$$

Divide -1 by -6.

$$x^{2}-\frac{1}{2}x=\frac{1}{6}$$

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.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{1}{6}+\left(-\frac{1}{4}\right)^{2}$$

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

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{6}+\frac{1}{16}$$

Add \frac{1}{6} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{11}{48}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{11}{48}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{11}{48}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{33}}{12} x-\frac{1}{4}=-\frac{\sqrt{33}}{12}$$

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

$$x=\frac{\sqrt{33}}{12}+\frac{1}{4} x=-\frac{\sqrt{33}}{12}+\frac{1}{4}$$