Solve 200/x-200/x+10=1

$$\frac { 200 } { x } - \frac { 200 } { x + 10 } = 1$$

Solve for x

$$x=-50$$

$$x=40$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -10,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+10\right), the least common multiple of x,x+10.

$$\left(x+10\right)\times 200-x\times 200=x\left(x+10\right)$$

Use the distributive property to multiply x+10 by 200.

$$200x+2000-x\times 200=x\left(x+10\right)$$

Use the distributive property to multiply x by x+10.

$$200x+2000-x\times 200=x^{2}+10x$$

Subtract x^{2} from both sides.

$$200x+2000-x\times 200-x^{2}=10x$$

Subtract 10x from both sides.

$$200x+2000-x\times 200-x^{2}-10x=0$$

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

$$190x+2000-x\times 200-x^{2}=0$$

Multiply -1 and 200 to get -200.

$$190x+2000-200x-x^{2}=0$$

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

$$-10x+2000-x^{2}=0$$

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

$$-x^{2}-10x+2000=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+2000. To find a and b, set up a system to be solved.

$$a+b=-10 ab=-2000=-2000$$

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 -2000.

$$1,-2000 2,-1000 4,-500 5,-400 8,-250 10,-200 16,-125 20,-100 25,-80 40,-50$$

Calculate the sum for each pair.

$$1-2000=-1999 2-1000=-998 4-500=-496 5-400=-395 8-250=-242 10-200=-190 16-125=-109 20-100=-80 25-80=-55 40-50=-10$$

The solution is the pair that gives sum -10.

$$a=40 b=-50$$

Rewrite -x^{2}-10x+2000 as \left(-x^{2}+40x\right)+\left(-50x+2000\right).

$$\left(-x^{2}+40x\right)+\left(-50x+2000\right)$$

Factor out x in the first and 50 in the second group.

$$x\left(-x+40\right)+50\left(-x+40\right)$$

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

$$\left(-x+40\right)\left(x+50\right)$$

To find equation solutions, solve -x+40=0 and x+50=0.

$$x=40 x=-50$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -10,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+10\right), the least common multiple of x,x+10.

$$\left(x+10\right)\times 200-x\times 200=x\left(x+10\right)$$

Use the distributive property to multiply x+10 by 200.

$$200x+2000-x\times 200=x\left(x+10\right)$$

Use the distributive property to multiply x by x+10.

$$200x+2000-x\times 200=x^{2}+10x$$

Subtract x^{2} from both sides.

$$200x+2000-x\times 200-x^{2}=10x$$

Subtract 10x from both sides.

$$200x+2000-x\times 200-x^{2}-10x=0$$

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

$$190x+2000-x\times 200-x^{2}=0$$

Multiply -1 and 200 to get -200.

$$190x+2000-200x-x^{2}=0$$

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

$$-10x+2000-x^{2}=0$$

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.

$$-x^{2}-10x+2000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -10 for b, and 2000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-1\right)\times 2000}}{2\left(-1\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-1\right)\times 2000}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-10\right)±\sqrt{100+4\times 2000}}{2\left(-1\right)}$$

Multiply 4 times 2000.

$$x=\frac{-\left(-10\right)±\sqrt{100+8000}}{2\left(-1\right)}$$

Add 100 to 8000.

$$x=\frac{-\left(-10\right)±\sqrt{8100}}{2\left(-1\right)}$$

Take the square root of 8100.

$$x=\frac{-\left(-10\right)±90}{2\left(-1\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±90}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{10±90}{-2}$$

Now solve the equation x=\frac{10±90}{-2} when ± is plus. Add 10 to 90.

$$x=\frac{100}{-2}$$

Divide 100 by -2.

$$x=-50$$

Now solve the equation x=\frac{10±90}{-2} when ± is minus. Subtract 90 from 10.

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

Divide -80 by -2.

$$x=40$$

The equation is now solved.

$$x=-50 x=40$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -10,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+10\right), the least common multiple of x,x+10.

$$\left(x+10\right)\times 200-x\times 200=x\left(x+10\right)$$

Use the distributive property to multiply x+10 by 200.

$$200x+2000-x\times 200=x\left(x+10\right)$$

Use the distributive property to multiply x by x+10.

$$200x+2000-x\times 200=x^{2}+10x$$

Subtract x^{2} from both sides.

$$200x+2000-x\times 200-x^{2}=10x$$

Subtract 10x from both sides.

$$200x+2000-x\times 200-x^{2}-10x=0$$

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

$$190x+2000-x\times 200-x^{2}=0$$

Subtract 2000 from both sides. Anything subtracted from zero gives its negation.

$$190x-x\times 200-x^{2}=-2000$$

Multiply -1 and 200 to get -200.

$$190x-200x-x^{2}=-2000$$

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

$$-10x-x^{2}=-2000$$

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.

$$-x^{2}-10x=-2000$$

Divide both sides by -1.

$$\frac{-x^{2}-10x}{-1}=-\frac{2000}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{10}{-1}\right)x=-\frac{2000}{-1}$$

Divide -10 by -1.

$$x^{2}+10x=-\frac{2000}{-1}$$

Divide -2000 by -1.

$$x^{2}+10x=2000$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=2000+5^{2}$$

Square 5.

$$x^{2}+10x+25=2000+25$$

Add 2000 to 25.

$$x^{2}+10x+25=2025$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+5=45 x+5=-45$$

Subtract 5 from both sides of the equation.

$$x=40 x=-50$$