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

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

Solve for x

$$x=\sqrt{5}+2\approx 4.236067977$$

$$x=2-\sqrt{5}\approx -0.236067977$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-1\right).

$$2\left(3x+1\right)=x\times 2\left(x-1\right)$$

Use the distributive property to multiply 2 by 3x+1.

$$6x+2=x\times 2\left(x-1\right)$$

Use the distributive property to multiply x\times 2 by x-1.

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

Multiply -1 and 2 to get -2.

$$6x+2=2x^{2}-2x$$

Subtract 2x^{2} from both sides.

$$6x+2-2x^{2}=-2x$$

Add 2x to both sides.

$$6x+2-2x^{2}+2x=0$$

Combine 6x and 2x to get 8x.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 8 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-2\right)\times 2}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-8±\sqrt{64+8\times 2}}{2\left(-2\right)}$$

Multiply 8 times 2.

$$x=\frac{-8±\sqrt{64+16}}{2\left(-2\right)}$$

Add 64 to 16.

$$x=\frac{-8±\sqrt{80}}{2\left(-2\right)}$$

Take the square root of 80.

$$x=\frac{-8±4\sqrt{5}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-8±4\sqrt{5}}{-4}$$

Now solve the equation x=\frac{-8±4\sqrt{5}}{-4} when ± is plus. Add -8 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}-8}{-4}$$

Divide -8+4\sqrt{5} by -4.

$$x=2-\sqrt{5}$$

Now solve the equation x=\frac{-8±4\sqrt{5}}{-4} when ± is minus. Subtract 4\sqrt{5} from -8.

$$x=\frac{-4\sqrt{5}-8}{-4}$$

Divide -8-4\sqrt{5} by -4.

$$x=\sqrt{5}+2$$

The equation is now solved.

$$x=2-\sqrt{5} x=\sqrt{5}+2$$

Steps for Completing the Square

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-1\right).

$$2\left(3x+1\right)=x\times 2\left(x-1\right)$$

Use the distributive property to multiply 2 by 3x+1.

$$6x+2=x\times 2\left(x-1\right)$$

Use the distributive property to multiply x\times 2 by x-1.

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

Multiply -1 and 2 to get -2.

$$6x+2=2x^{2}-2x$$

Subtract 2x^{2} from both sides.

$$6x+2-2x^{2}=-2x$$

Add 2x to both sides.

$$6x+2-2x^{2}+2x=0$$

Combine 6x and 2x to get 8x.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -2.

$$\frac{-2x^{2}+8x}{-2}=-\frac{2}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{8}{-2}x=-\frac{2}{-2}$$

Divide 8 by -2.

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

Divide -2 by -2.

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

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.

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

Square -2.

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

Add 1 to 4.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=\sqrt{5} x-2=-\sqrt{5}$$

Add 2 to both sides of the equation.

$$x=\sqrt{5}+2 x=2-\sqrt{5}$$