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

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

Solve for x

$$x=8$$

$$x=0$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{2}x-1 by \frac{1}{3}x-2 and combine like terms.

$$\frac{1}{6}x^{2}-\frac{4}{3}x+2=2$$

Subtract 2 from both sides.

$$\frac{1}{6}x^{2}-\frac{4}{3}x+2-2=0$$

Subtract 2 from 2 to get 0.

$$\frac{1}{6}x^{2}-\frac{4}{3}x=0$$

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

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

Take the square root of \left(-\frac{4}{3}\right)^{2}.

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

The opposite of -\frac{4}{3} is \frac{4}{3}.

$$x=\frac{\frac{4}{3}±\frac{4}{3}}{2\times \frac{1}{6}}$$

Multiply 2 times \frac{1}{6}.

$$x=\frac{\frac{4}{3}±\frac{4}{3}}{\frac{1}{3}}$$

Now solve the equation x=\frac{\frac{4}{3}±\frac{4}{3}}{\frac{1}{3}} when ± is plus. Add \frac{4}{3} to \frac{4}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{8}{3}}{\frac{1}{3}}$$

Divide \frac{8}{3} by \frac{1}{3} by multiplying \frac{8}{3} by the reciprocal of \frac{1}{3}.

$$x=8$$

Now solve the equation x=\frac{\frac{4}{3}±\frac{4}{3}}{\frac{1}{3}} when ± is minus. Subtract \frac{4}{3} from \frac{4}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{\frac{1}{3}}$$

Divide 0 by \frac{1}{3} by multiplying 0 by the reciprocal of \frac{1}{3}.

$$x=0$$

The equation is now solved.

$$x=8 x=0$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{2}x-1 by \frac{1}{3}x-2 and combine like terms.

$$\frac{1}{6}x^{2}-\frac{4}{3}x+2=2$$

Subtract 2 from both sides.

$$\frac{1}{6}x^{2}-\frac{4}{3}x=2-2$$

Subtract 2 from 2 to get 0.

$$\frac{1}{6}x^{2}-\frac{4}{3}x=0$$

Multiply both sides by 6.

$$\frac{\frac{1}{6}x^{2}-\frac{4}{3}x}{\frac{1}{6}}=\frac{0}{\frac{1}{6}}$$

Dividing by \frac{1}{6} undoes the multiplication by \frac{1}{6}.

$$x^{2}+\left(-\frac{\frac{4}{3}}{\frac{1}{6}}\right)x=\frac{0}{\frac{1}{6}}$$

Divide -\frac{4}{3} by \frac{1}{6} by multiplying -\frac{4}{3} by the reciprocal of \frac{1}{6}.

$$x^{2}-8x=\frac{0}{\frac{1}{6}}$$

Divide 0 by \frac{1}{6} by multiplying 0 by the reciprocal of \frac{1}{6}.

$$x^{2}-8x=0$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=16$$

Factor x^{2}-8x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=4 x-4=-4$$

Add 4 to both sides of the equation.

$$x=8 x=0$$