Solve x=1/2-frac{1{2-x}}

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

Solve for x

$$x=1$$

Steps Using Factoring By Grouping

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{2-x}{2-x}.

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

Since \frac{2\left(2-x\right)}{2-x} and \frac{1}{2-x} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2\left(2-x\right)-1.

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

Combine like terms in 4-2x-1.

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

Variable x cannot be equal to 2 since division by zero is not defined. Divide 1 by \frac{3-2x}{2-x} by multiplying 1 by the reciprocal of \frac{3-2x}{2-x}.

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

Subtract \frac{2-x}{3-2x} from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{3-2x}{3-2x}.

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

Since \frac{x\left(3-2x\right)}{3-2x} and \frac{2-x}{3-2x} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in x\left(3-2x\right)-\left(2-x\right).

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

Combine like terms in 3x-2x^{2}-2+x.

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

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

$$4x-2x^{2}-2=0$$

Divide both sides by 2.

$$2x-x^{2}-1=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=2 ab=-\left(-1\right)=1$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=1$$

Rewrite -x^{2}+2x-1 as \left(-x^{2}+x\right)+\left(x-1\right).

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

Factor out -x in -x^{2}+x.

$$-x\left(x-1\right)+x-1$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(-x+1\right)$$

To find equation solutions, solve x-1=0 and -x+1=0.

$$x=1 x=1$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{2-x}{2-x}.

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

Since \frac{2\left(2-x\right)}{2-x} and \frac{1}{2-x} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2\left(2-x\right)-1.

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

Combine like terms in 4-2x-1.

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

Variable x cannot be equal to 2 since division by zero is not defined. Divide 1 by \frac{3-2x}{2-x} by multiplying 1 by the reciprocal of \frac{3-2x}{2-x}.

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

Subtract \frac{2-x}{3-2x} from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{3-2x}{3-2x}.

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

Since \frac{x\left(3-2x\right)}{3-2x} and \frac{2-x}{3-2x} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in x\left(3-2x\right)-\left(2-x\right).

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

Combine like terms in 3x-2x^{2}-2+x.

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

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

$$4x-2x^{2}-2=0$$

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.

$$-2x^{2}+4x-2=0$$

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

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

Square 4.

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

Multiply -4 times -2.

$$x=\frac{-4±\sqrt{16+8\left(-2\right)}}{2\left(-2\right)}$$

Multiply 8 times -2.

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

Add 16 to -16.

$$x=\frac{-4±\sqrt{0}}{2\left(-2\right)}$$

Take the square root of 0.

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

Multiply 2 times -2.

$$x=-\frac{4}{-4}$$

Divide -4 by -4.

$$x=1$$

Steps for Completing the Square

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{2-x}{2-x}.

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

Since \frac{2\left(2-x\right)}{2-x} and \frac{1}{2-x} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2\left(2-x\right)-1.

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

Combine like terms in 4-2x-1.

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

Variable x cannot be equal to 2 since division by zero is not defined. Divide 1 by \frac{3-2x}{2-x} by multiplying 1 by the reciprocal of \frac{3-2x}{2-x}.

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

Subtract \frac{2-x}{3-2x} from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{3-2x}{3-2x}.

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

Since \frac{x\left(3-2x\right)}{3-2x} and \frac{2-x}{3-2x} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in x\left(3-2x\right)-\left(2-x\right).

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

Combine like terms in 3x-2x^{2}-2+x.

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

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

$$4x-2x^{2}-2=0$$

Add 2 to both sides. Anything plus zero gives itself.

$$4x-2x^{2}=2$$

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.

$$-2x^{2}+4x=2$$

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 4 by -2.

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

Divide 2 by -2.

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

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

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

Add -1 to 1.

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

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

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-1=0 x-1=0$$

Add 1 to both sides of the equation.

$$x=1 x=1$$

The equation is now solved. Solutions are the same.

$$x=1$$