Solve 2(x-3)+3(x-1)/(x-3)(x-1)=1

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

Solve for x

$$x = \frac{\sqrt{33} + 9}{2} \approx 7.372281323$$

$$x = \frac{9 - \sqrt{33}}{2} \approx 1.627718677$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 1,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x-1\right).

$$2\left(x-3\right)+3\left(x-1\right)=\left(x-3\right)\left(x-1\right)$$

Use the distributive property to multiply 2 by x-3.

$$2x-6+3\left(x-1\right)=\left(x-3\right)\left(x-1\right)$$

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

$$2x-6+3x-3=\left(x-3\right)\left(x-1\right)$$

Combine 2x and 3x to get 5x.

$$5x-6-3=\left(x-3\right)\left(x-1\right)$$

Subtract 3 from -6 to get -9.

$$5x-9=\left(x-3\right)\left(x-1\right)$$

Use the distributive property to multiply x-3 by x-1 and combine like terms.

$$5x-9=x^{2}-4x+3$$

Subtract x^{2} from both sides.

$$5x-9-x^{2}=-4x+3$$

Add 4x to both sides.

$$5x-9-x^{2}+4x=3$$

Combine 5x and 4x to get 9x.

$$9x-9-x^{2}=3$$

Subtract 3 from both sides.

$$9x-9-x^{2}-3=0$$

Subtract 3 from -9 to get -12.

$$9x-12-x^{2}=0$$

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.

$$-x^{2}+9x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 9 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-9±\sqrt{81+4\left(-12\right)}}{2\left(-1\right)}$$

Multiply 4 times -12.

$$x=\frac{-9±\sqrt{81-48}}{2\left(-1\right)}$$

Add 81 to -48.

$$x=\frac{-9±\sqrt{33}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-9±\sqrt{33}}{-2}$$

Now solve the equation x=\frac{-9±\sqrt{33}}{-2} when ± is plus. Add -9 to \sqrt{33}.

$$x=\frac{\sqrt{33}-9}{-2}$$

Divide -9+\sqrt{33} by -2.

$$x=\frac{9-\sqrt{33}}{2}$$

Now solve the equation x=\frac{-9±\sqrt{33}}{-2} when ± is minus. Subtract \sqrt{33} from -9.

$$x=\frac{-\sqrt{33}-9}{-2}$$

Divide -9-\sqrt{33} by -2.

$$x=\frac{\sqrt{33}+9}{2}$$

The equation is now solved.

$$x=\frac{9-\sqrt{33}}{2} x=\frac{\sqrt{33}+9}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 1,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x-1\right).

$$2\left(x-3\right)+3\left(x-1\right)=\left(x-3\right)\left(x-1\right)$$

Use the distributive property to multiply 2 by x-3.

$$2x-6+3\left(x-1\right)=\left(x-3\right)\left(x-1\right)$$

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

$$2x-6+3x-3=\left(x-3\right)\left(x-1\right)$$

Combine 2x and 3x to get 5x.

$$5x-6-3=\left(x-3\right)\left(x-1\right)$$

Subtract 3 from -6 to get -9.

$$5x-9=\left(x-3\right)\left(x-1\right)$$

Use the distributive property to multiply x-3 by x-1 and combine like terms.

$$5x-9=x^{2}-4x+3$$

Subtract x^{2} from both sides.

$$5x-9-x^{2}=-4x+3$$

Add 4x to both sides.

$$5x-9-x^{2}+4x=3$$

Combine 5x and 4x to get 9x.

$$9x-9-x^{2}=3$$

Add 9 to both sides.

$$9x-x^{2}=3+9$$

Add 3 and 9 to get 12.

$$9x-x^{2}=12$$

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.

$$-x^{2}+9x=12$$

Divide both sides by -1.

$$\frac{-x^{2}+9x}{-1}=\frac{12}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{9}{-1}x=\frac{12}{-1}$$

Divide 9 by -1.

$$x^{2}-9x=\frac{12}{-1}$$

Divide 12 by -1.

$$x^{2}-9x=-12$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-12+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-12+\frac{81}{4}$$

Add -12 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{33}{4}$$

Factor x^{2}-9x+\frac{81}{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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{33}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{33}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{\sqrt{33}}{2} x-\frac{9}{2}=-\frac{\sqrt{33}}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=\frac{\sqrt{33}+9}{2} x=\frac{9-\sqrt{33}}{2}$$