Solve x/2+x=x/2/2-x

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

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=0$$

Steps Using Factoring

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

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

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

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

Express \left(2+x\right)\times \frac{x}{2} as a single fraction.

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

Use the distributive property to multiply 2+x by x.

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

Divide each term of 2x+x^{2} by 2 to get x+\frac{1}{2}x^{2}.

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

To find the opposite of x+\frac{1}{2}x^{2}, find the opposite of each term.

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

Add x to both sides.

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

Combine -2x and x to get -x.

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

Add \frac{1}{2}x^{2} to both sides.

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

Combine x^{2} and \frac{1}{2}x^{2} to get \frac{3}{2}x^{2}.

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

Factor out x.

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

To find equation solutions, solve x=0 and \frac{3x}{2}-1=0.

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

Steps Using the Quadratic Formula

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

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

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

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

Express \left(2+x\right)\times \frac{x}{2} as a single fraction.

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

Use the distributive property to multiply 2+x by x.

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

Divide each term of 2x+x^{2} by 2 to get x+\frac{1}{2}x^{2}.

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

To find the opposite of x+\frac{1}{2}x^{2}, find the opposite of each term.

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

Add x to both sides.

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

Combine -2x and x to get -x.

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

Add \frac{1}{2}x^{2} to both sides.

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

Combine x^{2} and \frac{1}{2}x^{2} to get \frac{3}{2}x^{2}.

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

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

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

Take the square root of 1.

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

The opposite of -1 is 1.

$$x=\frac{1±1}{2\times \frac{3}{2}}$$

Multiply 2 times \frac{3}{2}.

$$x=\frac{1±1}{3}$$

Now solve the equation x=\frac{1±1}{3} when ± is plus. Add 1 to 1.

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

Now solve the equation x=\frac{1±1}{3} when ± is minus. Subtract 1 from 1.

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

Divide 0 by 3.

$$x=0$$

The equation is now solved.

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

Steps for Completing the Square

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

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

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

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

Express \left(2+x\right)\times \frac{x}{2} as a single fraction.

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

Use the distributive property to multiply 2+x by x.

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

Divide each term of 2x+x^{2} by 2 to get x+\frac{1}{2}x^{2}.

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

To find the opposite of x+\frac{1}{2}x^{2}, find the opposite of each term.

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

Add x to both sides.

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

Combine -2x and x to get -x.

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

Add \frac{1}{2}x^{2} to both sides.

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

Combine x^{2} and \frac{1}{2}x^{2} to get \frac{3}{2}x^{2}.

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

Divide both sides of the equation by \frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

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

Dividing by \frac{3}{2} undoes the multiplication by \frac{3}{2}.

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

Divide -1 by \frac{3}{2} by multiplying -1 by the reciprocal of \frac{3}{2}.

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

Divide 0 by \frac{3}{2} by multiplying 0 by the reciprocal of \frac{3}{2}.

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

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{1}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

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

Add \frac{1}{3} to both sides of the equation.

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