Solve 2/3-x+1/2=6/x(3-x)

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

Solve for x

$$x=4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by 2x\left(x-3\right), the least common multiple of 3-x,2,x\left(3-x\right).

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

Multiply -2 and 2 to get -4.

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

Multiply 2 and \frac{1}{2} to get 1.

$$-4x+x\left(x-3\right)=-2\times 6$$

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

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

Combine -4x and -3x to get -7x.

$$-7x+x^{2}=-2\times 6$$

Multiply -2 and 6 to get -12.

$$-7x+x^{2}=-12$$

Add 12 to both sides.

$$-7x+x^{2}+12=0$$

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.

$$x^{2}-7x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 12}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 12}}{2}$$

Multiply -4 times 12.

$$x=\frac{-\left(-7\right)±\sqrt{49-48}}{2}$$

Add 49 to -48.

$$x=\frac{-\left(-7\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-\left(-7\right)±1}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±1}{2}$$

Now solve the equation x=\frac{7±1}{2} when ± is plus. Add 7 to 1.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{7±1}{2} when ± is minus. Subtract 1 from 7.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

The equation is now solved.

$$x=4 x=3$$

Variable x cannot be equal to 3.

$$x=4$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by 2x\left(x-3\right), the least common multiple of 3-x,2,x\left(3-x\right).

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

Multiply -2 and 2 to get -4.

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

Multiply 2 and \frac{1}{2} to get 1.

$$-4x+x\left(x-3\right)=-2\times 6$$

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

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

Combine -4x and -3x to get -7x.

$$-7x+x^{2}=-2\times 6$$

Multiply -2 and 6 to get -12.

$$-7x+x^{2}=-12$$

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.

$$x^{2}-7x=-12$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-12+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-12+\frac{49}{4}$$

Add -12 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{1}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{1}{2} x-\frac{7}{2}=-\frac{1}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=4 x=3$$

Variable x cannot be equal to 3.

$$x=4$$