Solve x=2x+1/x-3

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

Solve for x

$$x = \frac{\sqrt{29} + 5}{2} \approx 5.192582404$$

$$x=\frac{5-\sqrt{29}}{2}\approx -0.192582404$$

Steps Using the Quadratic Formula

Subtract \frac{2x+1}{x-3} from both sides.

$$x-\frac{2x+1}{x-3}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x-3}{x-3}.

$$\frac{x\left(x-3\right)}{x-3}-\frac{2x+1}{x-3}=0$$

Since \frac{x\left(x-3\right)}{x-3} and \frac{2x+1}{x-3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x-3\right)-\left(2x+1\right)}{x-3}=0$$

Do the multiplications in x\left(x-3\right)-\left(2x+1\right).

$$\frac{x^{2}-3x-2x-1}{x-3}=0$$

Combine like terms in x^{2}-3x-2x-1.

$$\frac{x^{2}-5x-1}{x-3}=0$$

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3.

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

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

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)}}{2}$$

Multiply -4 times -1.

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

Add 25 to 4.

$$x=\frac{-\left(-5\right)±\sqrt{29}}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{29}}{2}$$

Now solve the equation x=\frac{5±\sqrt{29}}{2} when ± is plus. Add 5 to \sqrt{29}.

$$x=\frac{\sqrt{29}+5}{2}$$

Now solve the equation x=\frac{5±\sqrt{29}}{2} when ± is minus. Subtract \sqrt{29} from 5.

$$x=\frac{5-\sqrt{29}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{29}+5}{2} x=\frac{5-\sqrt{29}}{2}$$

Steps for Completing the Square

Subtract \frac{2x+1}{x-3} from both sides.

$$x-\frac{2x+1}{x-3}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x-3}{x-3}.

$$\frac{x\left(x-3\right)}{x-3}-\frac{2x+1}{x-3}=0$$

Since \frac{x\left(x-3\right)}{x-3} and \frac{2x+1}{x-3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x-3\right)-\left(2x+1\right)}{x-3}=0$$

Do the multiplications in x\left(x-3\right)-\left(2x+1\right).

$$\frac{x^{2}-3x-2x-1}{x-3}=0$$

Combine like terms in x^{2}-3x-2x-1.

$$\frac{x^{2}-5x-1}{x-3}=0$$

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3.

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

Add 1 to both sides. Anything plus zero gives itself.

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

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=1+\frac{25}{4}$$

Add 1 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{29}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{29}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{29}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{29}}{2} x-\frac{5}{2}=-\frac{\sqrt{29}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{29}+5}{2} x=\frac{5-\sqrt{29}}{2}$$