Solve 200/x-200/x+5=2

$$\frac{ 200 }{ x } - \frac{ 200 }{ x+5 } = 2$$

Solve for x

$$x=-25$$

$$x=20$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -5,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+5\right), the least common multiple of x,x+5.

$$\left(x+5\right)\times 200-x\times 200=2x\left(x+5\right)$$

Use the distributive property to multiply x+5 by 200.

$$200x+1000-x\times 200=2x\left(x+5\right)$$

Use the distributive property to multiply 2x by x+5.

$$200x+1000-x\times 200=2x^{2}+10x$$

Subtract 2x^{2} from both sides.

$$200x+1000-x\times 200-2x^{2}=10x$$

Subtract 10x from both sides.

$$200x+1000-x\times 200-2x^{2}-10x=0$$

Combine 200x and -10x to get 190x.

$$190x+1000-x\times 200-2x^{2}=0$$

Multiply -1 and 200 to get -200.

$$190x+1000-200x-2x^{2}=0$$

Combine 190x and -200x to get -10x.

$$-10x+1000-2x^{2}=0$$

Divide both sides by 2.

$$-5x+500-x^{2}=0$$

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

$$-x^{2}-5x+500=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+500. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-500=-500$$

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 -500.

$$1,-500 2,-250 4,-125 5,-100 10,-50 20,-25$$

Calculate the sum for each pair.

$$1-500=-499 2-250=-248 4-125=-121 5-100=-95 10-50=-40 20-25=-5$$

The solution is the pair that gives sum -5.

$$a=20 b=-25$$

Rewrite -x^{2}-5x+500 as \left(-x^{2}+20x\right)+\left(-25x+500\right).

$$\left(-x^{2}+20x\right)+\left(-25x+500\right)$$

Factor out x in the first and 25 in the second group.

$$x\left(-x+20\right)+25\left(-x+20\right)$$

Factor out common term -x+20 by using distributive property.

$$\left(-x+20\right)\left(x+25\right)$$

To find equation solutions, solve -x+20=0 and x+25=0.

$$x=20 x=-25$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -5,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+5\right), the least common multiple of x,x+5.

$$\left(x+5\right)\times 200-x\times 200=2x\left(x+5\right)$$

Use the distributive property to multiply x+5 by 200.

$$200x+1000-x\times 200=2x\left(x+5\right)$$

Use the distributive property to multiply 2x by x+5.

$$200x+1000-x\times 200=2x^{2}+10x$$

Subtract 2x^{2} from both sides.

$$200x+1000-x\times 200-2x^{2}=10x$$

Subtract 10x from both sides.

$$200x+1000-x\times 200-2x^{2}-10x=0$$

Combine 200x and -10x to get 190x.

$$190x+1000-x\times 200-2x^{2}=0$$

Multiply -1 and 200 to get -200.

$$190x+1000-200x-2x^{2}=0$$

Combine 190x and -200x to get -10x.

$$-10x+1000-2x^{2}=0$$

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.

$$-2x^{2}-10x+1000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -10 for b, and 1000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-2\right)\times 1000}}{2\left(-2\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-2\right)\times 1000}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-10\right)±\sqrt{100+8\times 1000}}{2\left(-2\right)}$$

Multiply 8 times 1000.

$$x=\frac{-\left(-10\right)±\sqrt{100+8000}}{2\left(-2\right)}$$

Add 100 to 8000.

$$x=\frac{-\left(-10\right)±\sqrt{8100}}{2\left(-2\right)}$$

Take the square root of 8100.

$$x=\frac{-\left(-10\right)±90}{2\left(-2\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±90}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{10±90}{-4}$$

Now solve the equation x=\frac{10±90}{-4} when ± is plus. Add 10 to 90.

$$x=\frac{100}{-4}$$

Divide 100 by -4.

$$x=-25$$

Now solve the equation x=\frac{10±90}{-4} when ± is minus. Subtract 90 from 10.

$$x=-\frac{80}{-4}$$

Divide -80 by -4.

$$x=20$$

The equation is now solved.

$$x=-25 x=20$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -5,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+5\right), the least common multiple of x,x+5.

$$\left(x+5\right)\times 200-x\times 200=2x\left(x+5\right)$$

Use the distributive property to multiply x+5 by 200.

$$200x+1000-x\times 200=2x\left(x+5\right)$$

Use the distributive property to multiply 2x by x+5.

$$200x+1000-x\times 200=2x^{2}+10x$$

Subtract 2x^{2} from both sides.

$$200x+1000-x\times 200-2x^{2}=10x$$

Subtract 10x from both sides.

$$200x+1000-x\times 200-2x^{2}-10x=0$$

Combine 200x and -10x to get 190x.

$$190x+1000-x\times 200-2x^{2}=0$$

Subtract 1000 from both sides. Anything subtracted from zero gives its negation.

$$190x-x\times 200-2x^{2}=-1000$$

Multiply -1 and 200 to get -200.

$$190x-200x-2x^{2}=-1000$$

Combine 190x and -200x to get -10x.

$$-10x-2x^{2}=-1000$$

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.

$$-2x^{2}-10x=-1000$$

Divide both sides by -2.

$$\frac{-2x^{2}-10x}{-2}=-\frac{1000}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{10}{-2}\right)x=-\frac{1000}{-2}$$

Divide -10 by -2.

$$x^{2}+5x=-\frac{1000}{-2}$$

Divide -1000 by -2.

$$x^{2}+5x=500$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=500+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=500+\frac{25}{4}$$

Add 500 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{2025}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{2025}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{2025}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{45}{2} x+\frac{5}{2}=-\frac{45}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=20 x=-25$$