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

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

Solve for x

$$x=\frac{\sqrt{13}-1}{6}\approx 0.434258546$$

$$x=\frac{-\sqrt{13}-1}{6}\approx -0.767591879$$

Steps Using the Quadratic Formula

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

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

Multiply x-1 and x-1 to get \left(x-1\right)^{2}.

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

Multiply 2x+1 and 2x+1 to get \left(2x+1\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

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

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

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

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

$$x^{2}-2x+1=4x^{2}+4x+1+6x^{2}-3x-3$$

Combine 4x^{2} and 6x^{2} to get 10x^{2}.

$$x^{2}-2x+1=10x^{2}+4x+1-3x-3$$

Combine 4x and -3x to get x.

$$x^{2}-2x+1=10x^{2}+x+1-3$$

Subtract 3 from 1 to get -2.

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

Subtract 10x^{2} from both sides.

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

Combine x^{2} and -10x^{2} to get -9x^{2}.

$$-9x^{2}-2x+1=x-2$$

Subtract x from both sides.

$$-9x^{2}-2x+1-x=-2$$

Combine -2x and -x to get -3x.

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

Add 2 to both sides.

$$-9x^{2}-3x+1+2=0$$

Add 1 and 2 to get 3.

$$-9x^{2}-3x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, -3 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-9\right)\times 3}}{2\left(-9\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-9\right)\times 3}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-\left(-3\right)±\sqrt{9+36\times 3}}{2\left(-9\right)}$$

Multiply 36 times 3.

$$x=\frac{-\left(-3\right)±\sqrt{9+108}}{2\left(-9\right)}$$

Add 9 to 108.

$$x=\frac{-\left(-3\right)±\sqrt{117}}{2\left(-9\right)}$$

Take the square root of 117.

$$x=\frac{-\left(-3\right)±3\sqrt{13}}{2\left(-9\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±3\sqrt{13}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{3±3\sqrt{13}}{-18}$$

Now solve the equation x=\frac{3±3\sqrt{13}}{-18} when ± is plus. Add 3 to 3\sqrt{13}.

$$x=\frac{3\sqrt{13}+3}{-18}$$

Divide 3+3\sqrt{13} by -18.

$$x=\frac{-\sqrt{13}-1}{6}$$

Now solve the equation x=\frac{3±3\sqrt{13}}{-18} when ± is minus. Subtract 3\sqrt{13} from 3.

$$x=\frac{3-3\sqrt{13}}{-18}$$

Divide 3-3\sqrt{13} by -18.

$$x=\frac{\sqrt{13}-1}{6}$$

The equation is now solved.

$$x=\frac{-\sqrt{13}-1}{6} x=\frac{\sqrt{13}-1}{6}$$

Steps for Completing the Square

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

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

Multiply x-1 and x-1 to get \left(x-1\right)^{2}.

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

Multiply 2x+1 and 2x+1 to get \left(2x+1\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

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

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

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

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

$$x^{2}-2x+1=4x^{2}+4x+1+6x^{2}-3x-3$$

Combine 4x^{2} and 6x^{2} to get 10x^{2}.

$$x^{2}-2x+1=10x^{2}+4x+1-3x-3$$

Combine 4x and -3x to get x.

$$x^{2}-2x+1=10x^{2}+x+1-3$$

Subtract 3 from 1 to get -2.

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

Subtract 10x^{2} from both sides.

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

Combine x^{2} and -10x^{2} to get -9x^{2}.

$$-9x^{2}-2x+1=x-2$$

Subtract x from both sides.

$$-9x^{2}-2x+1-x=-2$$

Combine -2x and -x to get -3x.

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

Subtract 1 from both sides.

$$-9x^{2}-3x=-2-1$$

Subtract 1 from -2 to get -3.

$$-9x^{2}-3x=-3$$

Divide both sides by -9.

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

Dividing by -9 undoes the multiplication by -9.

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

Reduce the fraction \frac{-3}{-9} to lowest terms by extracting and canceling out 3.

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

Reduce the fraction \frac{-3}{-9} to lowest terms by extracting and canceling out 3.

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

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{1}{3}+\frac{1}{36}$$

Add \frac{1}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{13}{36}$$

Factor x^{2}+\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{13}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{13}{36}}$$

Simplify.

$$x+\frac{1}{6}=\frac{\sqrt{13}}{6} x+\frac{1}{6}=-\frac{\sqrt{13}}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$x=\frac{\sqrt{13}-1}{6} x=\frac{-\sqrt{13}-1}{6}$$