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

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

Solve for x

$$x=0$$

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.

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

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

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

Combine 3x and x to get 4x.

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

Subtract 3x^{2} from both sides.

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

Subtract 4x from both sides.

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

Combine x and -4x to get -3x.

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

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

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.

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

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

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

Take the square root of \left(-3\right)^{2}.

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

The opposite of -3 is 3.

$$x=\frac{3±3}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{3±3}{-6}$$

Now solve the equation x=\frac{3±3}{-6} when ± is plus. Add 3 to 3.

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

Divide 6 by -6.

$$x=-1$$

Now solve the equation x=\frac{3±3}{-6} when ± is minus. Subtract 3 from 3.

$$x=\frac{0}{-6}$$

Divide 0 by -6.

$$x=0$$

The equation is now solved.

$$x=-1 x=0$$

Variable x cannot be equal to -1.

$$x=0$$

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.

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

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

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

Combine 3x and x to get 4x.

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

Subtract 3x^{2} from both sides.

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

Subtract 4x from both sides.

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

Combine x and -4x to get -3x.

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

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

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.

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

Divide both sides by -3.

$$\frac{-3x^{2}-3x}{-3}=\frac{0}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{3}{-3}\right)x=\frac{0}{-3}$$

Divide -3 by -3.

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

Divide 0 by -3.

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

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Factor x^{2}+x+\frac{1}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=0 x=-1$$

Variable x cannot be equal to -1.

$$x=0$$