Solve x-4=3/x-3

$$x - 4 = \frac { 3 } { x - 3 }$$

Solve for x

$$x = \frac{\sqrt{13} + 7}{2} \approx 5.302775638$$

$$x = \frac{7 - \sqrt{13}}{2} \approx 1.697224362$$

Steps Using the Quadratic Formula

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

$$\left(x-3\right)x+\left(x-3\right)\left(-4\right)=3$$

Use the distributive property to multiply x-3 by x.

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

Use the distributive property to multiply x-3 by -4.

$$x^{2}-3x-4x+12=3$$

Combine -3x and -4x to get -7x.

$$x^{2}-7x+12=3$$

Subtract 3 from both sides.

$$x^{2}-7x+12-3=0$$

Subtract 3 from 12 to get 9.

$$x^{2}-7x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 9}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 9}}{2}$$

Multiply -4 times 9.

$$x=\frac{-\left(-7\right)±\sqrt{49-36}}{2}$$

Add 49 to -36.

$$x=\frac{-\left(-7\right)±\sqrt{13}}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{13}}{2}$$

Now solve the equation x=\frac{7±\sqrt{13}}{2} when ± is plus. Add 7 to \sqrt{13}.

$$x=\frac{\sqrt{13}+7}{2}$$

Now solve the equation x=\frac{7±\sqrt{13}}{2} when ± is minus. Subtract \sqrt{13} from 7.

$$x=\frac{7-\sqrt{13}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{13}+7}{2} x=\frac{7-\sqrt{13}}{2}$$

Steps for Completing the Square

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

$$\left(x-3\right)x+\left(x-3\right)\left(-4\right)=3$$

Use the distributive property to multiply x-3 by x.

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

Use the distributive property to multiply x-3 by -4.

$$x^{2}-3x-4x+12=3$$

Combine -3x and -4x to get -7x.

$$x^{2}-7x+12=3$$

Subtract 12 from both sides.

$$x^{2}-7x=3-12$$

Subtract 12 from 3 to get -9.

$$x^{2}-7x=-9$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-9+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-9+\frac{49}{4}$$

Add -9 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{13}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{13}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{\sqrt{13}}{2} x-\frac{7}{2}=-\frac{\sqrt{13}}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=\frac{\sqrt{13}+7}{2} x=\frac{7-\sqrt{13}}{2}$$