Solve 3/x-2+2/2-x=2-x

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

Solve for x (complex solution)

$$x=2+i$$

$$x=2-i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2, the least common multiple of x-2,2-x.

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

Subtract 2 from 3 to get 1.

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

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

$$1=2x-4-x\left(x-2\right)$$

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

$$1=2x-4-x^{2}+2x$$

Combine 2x and 2x to get 4x.

$$1=4x-4-x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$4x-4-x^{2}=1$$

Subtract 1 from both sides.

$$4x-4-x^{2}-1=0$$

Subtract 1 from -4 to get -5.

$$4x-5-x^{2}=0$$

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.

$$-x^{2}+4x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 4 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-4±\sqrt{16+4\left(-5\right)}}{2\left(-1\right)}$$

Multiply 4 times -5.

$$x=\frac{-4±\sqrt{16-20}}{2\left(-1\right)}$$

Add 16 to -20.

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

Take the square root of -4.

$$x=\frac{-4±2i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-4±2i}{-2}$$

Now solve the equation x=\frac{-4±2i}{-2} when ± is plus. Add -4 to 2i.

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

Divide -4+2i by -2.

$$x=2-i$$

Now solve the equation x=\frac{-4±2i}{-2} when ± is minus. Subtract 2i from -4.

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

Divide -4-2i by -2.

$$x=2+i$$

The equation is now solved.

$$x=2-i x=2+i$$

Steps for Completing the Square

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2, the least common multiple of x-2,2-x.

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

Subtract 2 from 3 to get 1.

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

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

$$1=2x-4-x\left(x-2\right)$$

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

$$1=2x-4-x^{2}+2x$$

Combine 2x and 2x to get 4x.

$$1=4x-4-x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$4x-4-x^{2}=1$$

Add 4 to both sides.

$$4x-x^{2}=1+4$$

Add 1 and 4 to get 5.

$$4x-x^{2}=5$$

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.

$$-x^{2}+4x=5$$

Divide both sides by -1.

$$\frac{-x^{2}+4x}{-1}=\frac{5}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{4}{-1}x=\frac{5}{-1}$$

Divide 4 by -1.

$$x^{2}-4x=\frac{5}{-1}$$

Divide 5 by -1.

$$x^{2}-4x=-5$$

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

$$x^{2}-4x+\left(-2\right)^{2}=-5+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-5+4$$

Add -5 to 4.

$$x^{2}-4x+4=-1$$

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=i x-2=-i$$

Add 2 to both sides of the equation.

$$x=2+i x=2-i$$