Solve x-3/x^2-5x+6=2

$$\frac{ x-3 }{ { x }^{ 2 } -5x+6 } = 2$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Factoring By Grouping

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).

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

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

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

Use the distributive property to multiply 2x-6 by x-2 and combine like terms.

$$x-3=2x^{2}-10x+12$$

Subtract 2x^{2} from both sides.

$$x-3-2x^{2}=-10x+12$$

Add 10x to both sides.

$$x-3-2x^{2}+10x=12$$

Combine x and 10x to get 11x.

$$11x-3-2x^{2}=12$$

Subtract 12 from both sides.

$$11x-3-2x^{2}-12=0$$

Subtract 12 from -3 to get -15.

$$11x-15-2x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-2x^{2}+11x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=11 ab=-2\left(-15\right)=30$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 11.

$$a=6 b=5$$

Rewrite -2x^{2}+11x-15 as \left(-2x^{2}+6x\right)+\left(5x-15\right).

$$\left(-2x^{2}+6x\right)+\left(5x-15\right)$$

Factor out 2x in the first and -5 in the second group.

$$2x\left(-x+3\right)-5\left(-x+3\right)$$

Factor out common term -x+3 by using distributive property.

$$\left(-x+3\right)\left(2x-5\right)$$

To find equation solutions, solve -x+3=0 and 2x-5=0.

$$x=3 x=\frac{5}{2}$$

Variable x cannot be equal to 3.

$$x=\frac{5}{2}$$

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).

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

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

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

Use the distributive property to multiply 2x-6 by x-2 and combine like terms.

$$x-3=2x^{2}-10x+12$$

Subtract 2x^{2} from both sides.

$$x-3-2x^{2}=-10x+12$$

Add 10x to both sides.

$$x-3-2x^{2}+10x=12$$

Combine x and 10x to get 11x.

$$11x-3-2x^{2}=12$$

Subtract 12 from both sides.

$$11x-3-2x^{2}-12=0$$

Subtract 12 from -3 to get -15.

$$11x-15-2x^{2}=0$$

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.

$$-2x^{2}+11x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 11 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-2\right)\left(-15\right)}}{2\left(-2\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-2\right)\left(-15\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-11±\sqrt{121+8\left(-15\right)}}{2\left(-2\right)}$$

Multiply 8 times -15.

$$x=\frac{-11±\sqrt{121-120}}{2\left(-2\right)}$$

Add 121 to -120.

$$x=\frac{-11±\sqrt{1}}{2\left(-2\right)}$$

Take the square root of 1.

$$x=\frac{-11±1}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-11±1}{-4}$$

Now solve the equation x=\frac{-11±1}{-4} when ± is plus. Add -11 to 1.

$$x=-\frac{10}{-4}$$

Reduce the fraction \frac{-10}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{-11±1}{-4} when ± is minus. Subtract 1 from -11.

$$x=-\frac{12}{-4}$$

Divide -12 by -4.

$$x=3$$

The equation is now solved.

$$x=\frac{5}{2} x=3$$

Variable x cannot be equal to 3.

$$x=\frac{5}{2}$$

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).

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

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

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

Use the distributive property to multiply 2x-6 by x-2 and combine like terms.

$$x-3=2x^{2}-10x+12$$

Subtract 2x^{2} from both sides.

$$x-3-2x^{2}=-10x+12$$

Add 10x to both sides.

$$x-3-2x^{2}+10x=12$$

Combine x and 10x to get 11x.

$$11x-3-2x^{2}=12$$

Add 3 to both sides.

$$11x-2x^{2}=12+3$$

Add 12 and 3 to get 15.

$$11x-2x^{2}=15$$

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.

$$-2x^{2}+11x=15$$

Divide both sides by -2.

$$\frac{-2x^{2}+11x}{-2}=\frac{15}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{11}{-2}x=\frac{15}{-2}$$

Divide 11 by -2.

$$x^{2}-\frac{11}{2}x=\frac{15}{-2}$$

Divide 15 by -2.

$$x^{2}-\frac{11}{2}x=-\frac{15}{2}$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=-\frac{15}{2}+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=-\frac{15}{2}+\frac{121}{16}$$

Add -\frac{15}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{1}{16}$$

Factor x^{2}-\frac{11}{2}x+\frac{121}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{1}{4} x-\frac{11}{4}=-\frac{1}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$x=3 x=\frac{5}{2}$$

Variable x cannot be equal to 3.

$$x=\frac{5}{2}$$