Solve 36/x(x-12)-3/x-12=3

$$\frac { 36 } { x ( x - 12 ) } - \frac { 3 } { x - 12 } = 3$$

Solve for x

$$x=-1$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,12 since division by zero is not defined. Multiply both sides of the equation by x\left(x-12\right), the least common multiple of x\left(x-12\right),x-12.

$$36-x\times 3=3x\left(x-12\right)$$

Use the distributive property to multiply 3x by x-12.

$$36-x\times 3=3x^{2}-36x$$

Subtract 3x^{2} from both sides.

$$36-x\times 3-3x^{2}=-36x$$

Add 36x to both sides.

$$36-x\times 3-3x^{2}+36x=0$$

Multiply -1 and 3 to get -3.

$$36-3x-3x^{2}+36x=0$$

Combine -3x and 36x to get 33x.

$$36+33x-3x^{2}=0$$

Divide both sides by 3.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=11 ab=-12=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 11.

$$a=12 b=-1$$

Rewrite -x^{2}+11x+12 as \left(-x^{2}+12x\right)+\left(-x+12\right).

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

Factor out -x in the first and -1 in the second group.

$$-x\left(x-12\right)-\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(-x-1\right)$$

To find equation solutions, solve x-12=0 and -x-1=0.

$$x=12 x=-1$$

Variable x cannot be equal to 12.

$$x=-1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,12 since division by zero is not defined. Multiply both sides of the equation by x\left(x-12\right), the least common multiple of x\left(x-12\right),x-12.

$$36-x\times 3=3x\left(x-12\right)$$

Use the distributive property to multiply 3x by x-12.

$$36-x\times 3=3x^{2}-36x$$

Subtract 3x^{2} from both sides.

$$36-x\times 3-3x^{2}=-36x$$

Add 36x to both sides.

$$36-x\times 3-3x^{2}+36x=0$$

Multiply -1 and 3 to get -3.

$$36-3x-3x^{2}+36x=0$$

Combine -3x and 36x to get 33x.

$$36+33x-3x^{2}=0$$

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.

$$-3x^{2}+33x+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 33 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-33±\sqrt{33^{2}-4\left(-3\right)\times 36}}{2\left(-3\right)}$$

Square 33.

$$x=\frac{-33±\sqrt{1089-4\left(-3\right)\times 36}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-33±\sqrt{1089+12\times 36}}{2\left(-3\right)}$$

Multiply 12 times 36.

$$x=\frac{-33±\sqrt{1089+432}}{2\left(-3\right)}$$

Add 1089 to 432.

$$x=\frac{-33±\sqrt{1521}}{2\left(-3\right)}$$

Take the square root of 1521.

$$x=\frac{-33±39}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-33±39}{-6}$$

Now solve the equation x=\frac{-33±39}{-6} when ± is plus. Add -33 to 39.

$$x=\frac{6}{-6}$$

Divide 6 by -6.

$$x=-1$$

Now solve the equation x=\frac{-33±39}{-6} when ± is minus. Subtract 39 from -33.

$$x=-\frac{72}{-6}$$

Divide -72 by -6.

$$x=12$$

The equation is now solved.

$$x=-1 x=12$$

Variable x cannot be equal to 12.

$$x=-1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,12 since division by zero is not defined. Multiply both sides of the equation by x\left(x-12\right), the least common multiple of x\left(x-12\right),x-12.

$$36-x\times 3=3x\left(x-12\right)$$

Use the distributive property to multiply 3x by x-12.

$$36-x\times 3=3x^{2}-36x$$

Subtract 3x^{2} from both sides.

$$36-x\times 3-3x^{2}=-36x$$

Add 36x to both sides.

$$36-x\times 3-3x^{2}+36x=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$-x\times 3-3x^{2}+36x=-36$$

Multiply -1 and 3 to get -3.

$$-3x-3x^{2}+36x=-36$$

Combine -3x and 36x to get 33x.

$$33x-3x^{2}=-36$$

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.

$$-3x^{2}+33x=-36$$

Divide both sides by -3.

$$\frac{-3x^{2}+33x}{-3}=-\frac{36}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{33}{-3}x=-\frac{36}{-3}$$

Divide 33 by -3.

$$x^{2}-11x=-\frac{36}{-3}$$

Divide -36 by -3.

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

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

$$x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=12+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-11x+\frac{121}{4}=12+\frac{121}{4}$$

Add 12 to \frac{121}{4}.

$$x^{2}-11x+\frac{121}{4}=\frac{169}{4}$$

Factor x^{2}-11x+\frac{121}{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}.

$$\left(x-\frac{11}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x-\frac{11}{2}=\frac{13}{2} x-\frac{11}{2}=-\frac{13}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$x=12 x=-1$$

Variable x cannot be equal to 12.

$$x=-1$$