Solve 40/x-40/x+20=1/3

$$\frac { 40 } { x } - \frac { 40 } { x + 20 } = \frac { 1 } { 3 }$$

Solve for x

$$x=-60$$

$$x=40$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -20,0 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x+20\right), the least common multiple of x,x+20,3.

$$\left(3x+60\right)\times 40-3x\times 40=x\left(x+20\right)$$

Use the distributive property to multiply 3x+60 by 40.

$$120x+2400-3x\times 40=x\left(x+20\right)$$

Multiply 3 and 40 to get 120.

$$120x+2400-120x=x\left(x+20\right)$$

Use the distributive property to multiply x by x+20.

$$120x+2400-120x=x^{2}+20x$$

Subtract x^{2} from both sides.

$$120x+2400-120x-x^{2}=20x$$

Subtract 20x from both sides.

$$120x+2400-120x-x^{2}-20x=0$$

Combine 120x and -20x to get 100x.

$$100x+2400-120x-x^{2}=0$$

Combine 100x and -120x to get -20x.

$$-20x+2400-x^{2}=0$$

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

$$-x^{2}-20x+2400=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+2400. To find a and b, set up a system to be solved.

$$a+b=-20 ab=-2400=-2400$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -2400.

$$1,-2400 2,-1200 3,-800 4,-600 5,-480 6,-400 8,-300 10,-240 12,-200 15,-160 16,-150 20,-120 24,-100 25,-96 30,-80 32,-75 40,-60 48,-50$$

Calculate the sum for each pair.

$$1-2400=-2399 2-1200=-1198 3-800=-797 4-600=-596 5-480=-475 6-400=-394 8-300=-292 10-240=-230 12-200=-188 15-160=-145 16-150=-134 20-120=-100 24-100=-76 25-96=-71 30-80=-50 32-75=-43 40-60=-20 48-50=-2$$

The solution is the pair that gives sum -20.

$$a=40 b=-60$$

Rewrite -x^{2}-20x+2400 as \left(-x^{2}+40x\right)+\left(-60x+2400\right).

$$\left(-x^{2}+40x\right)+\left(-60x+2400\right)$$

Factor out x in the first and 60 in the second group.

$$x\left(-x+40\right)+60\left(-x+40\right)$$

Factor out common term -x+40 by using distributive property.

$$\left(-x+40\right)\left(x+60\right)$$

To find equation solutions, solve -x+40=0 and x+60=0.

$$x=40 x=-60$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -20,0 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x+20\right), the least common multiple of x,x+20,3.

$$\left(3x+60\right)\times 40-3x\times 40=x\left(x+20\right)$$

Use the distributive property to multiply 3x+60 by 40.

$$120x+2400-3x\times 40=x\left(x+20\right)$$

Multiply 3 and 40 to get 120.

$$120x+2400-120x=x\left(x+20\right)$$

Use the distributive property to multiply x by x+20.

$$120x+2400-120x=x^{2}+20x$$

Subtract x^{2} from both sides.

$$120x+2400-120x-x^{2}=20x$$

Subtract 20x from both sides.

$$120x+2400-120x-x^{2}-20x=0$$

Combine 120x and -20x to get 100x.

$$100x+2400-120x-x^{2}=0$$

Combine 100x and -120x to get -20x.

$$-20x+2400-x^{2}=0$$

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.

$$-x^{2}-20x+2400=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -20 for b, and 2400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-1\right)\times 2400}}{2\left(-1\right)}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\left(-1\right)\times 2400}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-20\right)±\sqrt{400+4\times 2400}}{2\left(-1\right)}$$

Multiply 4 times 2400.

$$x=\frac{-\left(-20\right)±\sqrt{400+9600}}{2\left(-1\right)}$$

Add 400 to 9600.

$$x=\frac{-\left(-20\right)±\sqrt{10000}}{2\left(-1\right)}$$

Take the square root of 10000.

$$x=\frac{-\left(-20\right)±100}{2\left(-1\right)}$$

The opposite of -20 is 20.

$$x=\frac{20±100}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{20±100}{-2}$$

Now solve the equation x=\frac{20±100}{-2} when ± is plus. Add 20 to 100.

$$x=\frac{120}{-2}$$

Divide 120 by -2.

$$x=-60$$

Now solve the equation x=\frac{20±100}{-2} when ± is minus. Subtract 100 from 20.

$$x=-\frac{80}{-2}$$

Divide -80 by -2.

$$x=40$$

The equation is now solved.

$$x=-60 x=40$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -20,0 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x+20\right), the least common multiple of x,x+20,3.

$$\left(3x+60\right)\times 40-3x\times 40=x\left(x+20\right)$$

Use the distributive property to multiply 3x+60 by 40.

$$120x+2400-3x\times 40=x\left(x+20\right)$$

Multiply 3 and 40 to get 120.

$$120x+2400-120x=x\left(x+20\right)$$

Use the distributive property to multiply x by x+20.

$$120x+2400-120x=x^{2}+20x$$

Subtract x^{2} from both sides.

$$120x+2400-120x-x^{2}=20x$$

Subtract 20x from both sides.

$$120x+2400-120x-x^{2}-20x=0$$

Combine 120x and -20x to get 100x.

$$100x+2400-120x-x^{2}=0$$

Subtract 2400 from both sides. Anything subtracted from zero gives its negation.

$$100x-120x-x^{2}=-2400$$

Combine 100x and -120x to get -20x.

$$-20x-x^{2}=-2400$$

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.

$$-x^{2}-20x=-2400$$

Divide both sides by -1.

$$\frac{-x^{2}-20x}{-1}=-\frac{2400}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{20}{-1}\right)x=-\frac{2400}{-1}$$

Divide -20 by -1.

$$x^{2}+20x=-\frac{2400}{-1}$$

Divide -2400 by -1.

$$x^{2}+20x=2400$$

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

$$x^{2}+20x+10^{2}=2400+10^{2}$$

Square 10.

$$x^{2}+20x+100=2400+100$$

Add 2400 to 100.

$$x^{2}+20x+100=2500$$

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

$$\left(x+10\right)^{2}=2500$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+10\right)^{2}}=\sqrt{2500}$$

Simplify.

$$x+10=50 x+10=-50$$

Subtract 10 from both sides of the equation.

$$x=40 x=-60$$