Solve 1/x^2-7x+12+1/x^2-6x+8=x/x^2-5x+6

$$\frac { 1 } { x ^ { 2 } - 7 x + 12 } + \frac { 1 } { x ^ { 2 } - 6 x + 8 } = \frac { x } { x ^ { 2 } - 5 x + 6 }$$

Solve for x

$$x=1$$

$$x=5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 2,3,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x-3\right)\left(x-2\right), the least common multiple of x^{2}-7x+12,x^{2}-6x+8,x^{2}-5x+6.

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

Combine x and x to get 2x.

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

Subtract 3 from -2 to get -5.

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

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

$$2x-5=x^{2}-4x$$

Subtract x^{2} from both sides.

$$2x-5-x^{2}=-4x$$

Add 4x to both sides.

$$2x-5-x^{2}+4x=0$$

Combine 2x and 4x to get 6x.

$$6x-5-x^{2}=0$$

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

$$-x^{2}+6x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-5\right)=5$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=5 b=1$$

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

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

Factor out -x in -x^{2}+5x.

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

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(-x+1\right)$$

To find equation solutions, solve x-5=0 and -x+1=0.

$$x=5 x=1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 2,3,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x-3\right)\left(x-2\right), the least common multiple of x^{2}-7x+12,x^{2}-6x+8,x^{2}-5x+6.

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

Combine x and x to get 2x.

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

Subtract 3 from -2 to get -5.

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

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

$$2x-5=x^{2}-4x$$

Subtract x^{2} from both sides.

$$2x-5-x^{2}=-4x$$

Add 4x to both sides.

$$2x-5-x^{2}+4x=0$$

Combine 2x and 4x to get 6x.

$$6x-5-x^{2}=0$$

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.

$$-x^{2}+6x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-6±\sqrt{36+4\left(-5\right)}}{2\left(-1\right)}$$

Multiply 4 times -5.

$$x=\frac{-6±\sqrt{36-20}}{2\left(-1\right)}$$

Add 36 to -20.

$$x=\frac{-6±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

$$x=\frac{-6±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-6±4}{-2}$$

Now solve the equation x=\frac{-6±4}{-2} when ± is plus. Add -6 to 4.

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

Divide -2 by -2.

$$x=1$$

Now solve the equation x=\frac{-6±4}{-2} when ± is minus. Subtract 4 from -6.

$$x=-\frac{10}{-2}$$

Divide -10 by -2.

$$x=5$$

The equation is now solved.

$$x=1 x=5$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 2,3,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x-3\right)\left(x-2\right), the least common multiple of x^{2}-7x+12,x^{2}-6x+8,x^{2}-5x+6.

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

Combine x and x to get 2x.

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

Subtract 3 from -2 to get -5.

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

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

$$2x-5=x^{2}-4x$$

Subtract x^{2} from both sides.

$$2x-5-x^{2}=-4x$$

Add 4x to both sides.

$$2x-5-x^{2}+4x=0$$

Combine 2x and 4x to get 6x.

$$6x-5-x^{2}=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$6x-x^{2}=5$$

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.

$$-x^{2}+6x=5$$

Divide both sides by -1.

$$\frac{-x^{2}+6x}{-1}=\frac{5}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{6}{-1}x=\frac{5}{-1}$$

Divide 6 by -1.

$$x^{2}-6x=\frac{5}{-1}$$

Divide 5 by -1.

$$x^{2}-6x=-5$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-5+9$$

Add -5 to 9.

$$x^{2}-6x+9=4$$

Factor x^{2}-6x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-3=2 x-3=-2$$

Add 3 to both sides of the equation.

$$x=5 x=1$$