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

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

Solve for x

$$x=3$$

$$x=0$$

Steps Using Factoring

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by x+1.

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

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

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

Subtract 3x from both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

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

Factor out x.

$$x\left(x-3\right)=0$$

To find equation solutions, solve x=0 and x-3=0.

$$x=0 x=3$$

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.

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

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

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

Subtract 3x from both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

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

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

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

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

$$x=\frac{-\left(-3\right)±3}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±3}{2}$$

Now solve the equation x=\frac{3±3}{2} when ± is plus. Add 3 to 3.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{3±3}{2} when ± is minus. Subtract 3 from 3.

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=3 x=0$$

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.

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

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

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

Subtract 3x from both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=\frac{9}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=3 x=0$$