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

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

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

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 x+1.

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

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

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

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

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

Combine 3x and -x to get 2x.

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

Subtract 3x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

Add 1 to both sides.

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

Add 1 and 1 to get 2.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -1 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-1\right)±\sqrt{1+12\times 2}}{2\left(-3\right)}$$

Multiply 12 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\left(-3\right)}$$

Add 1 to 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\left(-3\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-1\right)±5}{2\left(-3\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±5}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{1±5}{-6}$$

Now solve the equation x=\frac{1±5}{-6} when ± is plus. Add 1 to 5.

$$x=\frac{6}{-6}$$

Divide 6 by -6.

$$x=-1$$

Now solve the equation x=\frac{1±5}{-6} when ± is minus. Subtract 5 from 1.

$$x=-\frac{4}{-6}$$

Reduce the fraction \frac{-4}{-6} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Variable x cannot be equal to -1.

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

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 x+1.

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

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

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

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

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

Combine 3x and -x to get 2x.

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

Subtract 3x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

Subtract 1 from both sides.

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

Subtract 1 from -1 to get -2.

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

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.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide -1 by -3.

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

Divide -2 by -3.

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

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.

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

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

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

Add \frac{2}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{25}{36}$$

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}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x+\frac{1}{6}=\frac{5}{6} x+\frac{1}{6}=-\frac{5}{6}$$

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

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

Variable x cannot be equal to -1.

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