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

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

Solve for x

$$x=\frac{\sqrt{65}}{20}-\frac{1}{4}\approx 0.153112887$$

$$x=-\frac{\sqrt{65}}{20}-\frac{1}{4}\approx -0.653112887$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{1}{2},\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-1\right)\left(2x+1\right), the least common multiple of 2x+1,1-2x.

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

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

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

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

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

Combine -x and -4x to get -5x.

$$2x^{2}-5x-2=3\left(2x-1\right)\left(2x+1\right)$$

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

$$2x^{2}-5x-2=\left(6x-3\right)\left(2x+1\right)$$

Use the distributive property to multiply 6x-3 by 2x+1 and combine like terms.

$$2x^{2}-5x-2=12x^{2}-3$$

Subtract 12x^{2} from both sides.

$$2x^{2}-5x-2-12x^{2}=-3$$

Combine 2x^{2} and -12x^{2} to get -10x^{2}.

$$-10x^{2}-5x-2=-3$$

Add 3 to both sides.

$$-10x^{2}-5x-2+3=0$$

Add -2 and 3 to get 1.

$$-10x^{2}-5x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, -5 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-10\right)}}{2\left(-10\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-10\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-\left(-5\right)±\sqrt{25+40}}{2\left(-10\right)}$$

Add 25 to 40.

$$x=\frac{-\left(-5\right)±\sqrt{65}}{2\left(-10\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{65}}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{5±\sqrt{65}}{-20}$$

Now solve the equation x=\frac{5±\sqrt{65}}{-20} when ± is plus. Add 5 to \sqrt{65}.

$$x=\frac{\sqrt{65}+5}{-20}$$

Divide 5+\sqrt{65} by -20.

$$x=-\frac{\sqrt{65}}{20}-\frac{1}{4}$$

Now solve the equation x=\frac{5±\sqrt{65}}{-20} when ± is minus. Subtract \sqrt{65} from 5.

$$x=\frac{5-\sqrt{65}}{-20}$$

Divide 5-\sqrt{65} by -20.

$$x=\frac{\sqrt{65}}{20}-\frac{1}{4}$$

The equation is now solved.

$$x=-\frac{\sqrt{65}}{20}-\frac{1}{4} x=\frac{\sqrt{65}}{20}-\frac{1}{4}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{1}{2},\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-1\right)\left(2x+1\right), the least common multiple of 2x+1,1-2x.

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

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

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

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

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

Combine -x and -4x to get -5x.

$$2x^{2}-5x-2=3\left(2x-1\right)\left(2x+1\right)$$

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

$$2x^{2}-5x-2=\left(6x-3\right)\left(2x+1\right)$$

Use the distributive property to multiply 6x-3 by 2x+1 and combine like terms.

$$2x^{2}-5x-2=12x^{2}-3$$

Subtract 12x^{2} from both sides.

$$2x^{2}-5x-2-12x^{2}=-3$$

Combine 2x^{2} and -12x^{2} to get -10x^{2}.

$$-10x^{2}-5x-2=-3$$

Add 2 to both sides.

$$-10x^{2}-5x=-3+2$$

Add -3 and 2 to get -1.

$$-10x^{2}-5x=-1$$

Divide both sides by -10.

$$\frac{-10x^{2}-5x}{-10}=-\frac{1}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\left(-\frac{5}{-10}\right)x=-\frac{1}{-10}$$

Reduce the fraction \frac{-5}{-10} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{1}{2}x=-\frac{1}{-10}$$

Divide -1 by -10.

$$x^{2}+\frac{1}{2}x=\frac{1}{10}$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{1}{10}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{10}+\frac{1}{16}$$

Add \frac{1}{10} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{13}{80}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{13}{80}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{13}{80}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{65}}{20} x+\frac{1}{4}=-\frac{\sqrt{65}}{20}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{\sqrt{65}}{20}-\frac{1}{4} x=-\frac{\sqrt{65}}{20}-\frac{1}{4}$$