Solve 2/x-3+3/x-2=3

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

Solve for x

$$x = \frac{\sqrt{7} + 10}{3} \approx 4.215250437$$

$$x = \frac{10 - \sqrt{7}}{3} \approx 2.45141623$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 2,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x-2\right), the least common multiple of x-3,x-2.

$$\left(x-2\right)\times 2+\left(x-3\right)\times 3=3\left(x-3\right)\left(x-2\right)$$

Use the distributive property to multiply x-2 by 2.

$$2x-4+\left(x-3\right)\times 3=3\left(x-3\right)\left(x-2\right)$$

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

$$2x-4+3x-9=3\left(x-3\right)\left(x-2\right)$$

Combine 2x and 3x to get 5x.

$$5x-4-9=3\left(x-3\right)\left(x-2\right)$$

Subtract 9 from -4 to get -13.

$$5x-13=3\left(x-3\right)\left(x-2\right)$$

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

$$5x-13=\left(3x-9\right)\left(x-2\right)$$

Use the distributive property to multiply 3x-9 by x-2 and combine like terms.

$$5x-13=3x^{2}-15x+18$$

Subtract 3x^{2} from both sides.

$$5x-13-3x^{2}=-15x+18$$

Add 15x to both sides.

$$5x-13-3x^{2}+15x=18$$

Combine 5x and 15x to get 20x.

$$20x-13-3x^{2}=18$$

Subtract 18 from both sides.

$$20x-13-3x^{2}-18=0$$

Subtract 18 from -13 to get -31.

$$20x-31-3x^{2}=0$$

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.

$$-3x^{2}+20x-31=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 20 for b, and -31 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-3\right)\left(-31\right)}}{2\left(-3\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-3\right)\left(-31\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-20±\sqrt{400+12\left(-31\right)}}{2\left(-3\right)}$$

Multiply 12 times -31.

$$x=\frac{-20±\sqrt{400-372}}{2\left(-3\right)}$$

Add 400 to -372.

$$x=\frac{-20±\sqrt{28}}{2\left(-3\right)}$$

Take the square root of 28.

$$x=\frac{-20±2\sqrt{7}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-20±2\sqrt{7}}{-6}$$

Now solve the equation x=\frac{-20±2\sqrt{7}}{-6} when ± is plus. Add -20 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-20}{-6}$$

Divide -20+2\sqrt{7} by -6.

$$x=\frac{10-\sqrt{7}}{3}$$

Now solve the equation x=\frac{-20±2\sqrt{7}}{-6} when ± is minus. Subtract 2\sqrt{7} from -20.

$$x=\frac{-2\sqrt{7}-20}{-6}$$

Divide -20-2\sqrt{7} by -6.

$$x=\frac{\sqrt{7}+10}{3}$$

The equation is now solved.

$$x=\frac{10-\sqrt{7}}{3} x=\frac{\sqrt{7}+10}{3}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 2,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x-2\right), the least common multiple of x-3,x-2.

$$\left(x-2\right)\times 2+\left(x-3\right)\times 3=3\left(x-3\right)\left(x-2\right)$$

Use the distributive property to multiply x-2 by 2.

$$2x-4+\left(x-3\right)\times 3=3\left(x-3\right)\left(x-2\right)$$

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

$$2x-4+3x-9=3\left(x-3\right)\left(x-2\right)$$

Combine 2x and 3x to get 5x.

$$5x-4-9=3\left(x-3\right)\left(x-2\right)$$

Subtract 9 from -4 to get -13.

$$5x-13=3\left(x-3\right)\left(x-2\right)$$

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

$$5x-13=\left(3x-9\right)\left(x-2\right)$$

Use the distributive property to multiply 3x-9 by x-2 and combine like terms.

$$5x-13=3x^{2}-15x+18$$

Subtract 3x^{2} from both sides.

$$5x-13-3x^{2}=-15x+18$$

Add 15x to both sides.

$$5x-13-3x^{2}+15x=18$$

Combine 5x and 15x to get 20x.

$$20x-13-3x^{2}=18$$

Add 13 to both sides.

$$20x-3x^{2}=18+13$$

Add 18 and 13 to get 31.

$$20x-3x^{2}=31$$

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.

$$-3x^{2}+20x=31$$

Divide both sides by -3.

$$\frac{-3x^{2}+20x}{-3}=\frac{31}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{20}{-3}x=\frac{31}{-3}$$

Divide 20 by -3.

$$x^{2}-\frac{20}{3}x=\frac{31}{-3}$$

Divide 31 by -3.

$$x^{2}-\frac{20}{3}x=-\frac{31}{3}$$

Divide -\frac{20}{3}, the coefficient of the x term, by 2 to get -\frac{10}{3}. Then add the square of -\frac{10}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{20}{3}x+\left(-\frac{10}{3}\right)^{2}=-\frac{31}{3}+\left(-\frac{10}{3}\right)^{2}$$

Square -\frac{10}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=-\frac{31}{3}+\frac{100}{9}$$

Add -\frac{31}{3} to \frac{100}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=\frac{7}{9}$$

Factor x^{2}-\frac{20}{3}x+\frac{100}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{10}{3}\right)^{2}=\frac{7}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{10}{3}\right)^{2}}=\sqrt{\frac{7}{9}}$$

Simplify.

$$x-\frac{10}{3}=\frac{\sqrt{7}}{3} x-\frac{10}{3}=-\frac{\sqrt{7}}{3}$$

Add \frac{10}{3} to both sides of the equation.

$$x=\frac{\sqrt{7}+10}{3} x=\frac{10-\sqrt{7}}{3}$$