Solve 1000/x-1000/x+5=10

$$\frac { 1000 } { x } - \frac { 1000 } { x + 5 } = 10$$

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.

$$\left(x+5\right)\times 1000-x\times 1000=10x\left(x+5\right)$$

Use the distributive property to multiply x+5 by 1000.

$$1000x+5000-x\times 1000=10x\left(x+5\right)$$

Use the distributive property to multiply 10x by x+5.

$$1000x+5000-x\times 1000=10x^{2}+50x$$

Subtract 10x^{2} from both sides.

$$1000x+5000-x\times 1000-10x^{2}=50x$$

Subtract 50x from both sides.

$$1000x+5000-x\times 1000-10x^{2}-50x=0$$

Combine 1000x and -50x to get 950x.

$$950x+5000-x\times 1000-10x^{2}=0$$

Multiply -1 and 1000 to get -1000.

$$950x+5000-1000x-10x^{2}=0$$

Combine 950x and -1000x to get -50x.

$$-50x+5000-10x^{2}=0$$

Divide both sides by 10.

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

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.

$$\left(x+5\right)\times 1000-x\times 1000=10x\left(x+5\right)$$

Use the distributive property to multiply x+5 by 1000.

$$1000x+5000-x\times 1000=10x\left(x+5\right)$$

Use the distributive property to multiply 10x by x+5.

$$1000x+5000-x\times 1000=10x^{2}+50x$$

Subtract 10x^{2} from both sides.

$$1000x+5000-x\times 1000-10x^{2}=50x$$

Subtract 50x from both sides.

$$1000x+5000-x\times 1000-10x^{2}-50x=0$$

Combine 1000x and -50x to get 950x.

$$950x+5000-x\times 1000-10x^{2}=0$$

Multiply -1 and 1000 to get -1000.

$$950x+5000-1000x-10x^{2}=0$$

Combine 950x and -1000x to get -50x.

$$-50x+5000-10x^{2}=0$$

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.

$$-10x^{2}-50x+5000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, -50 for b, and 5000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\left(-10\right)\times 5000}}{2\left(-10\right)}$$

Square -50.

$$x=\frac{-\left(-50\right)±\sqrt{2500-4\left(-10\right)\times 5000}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-\left(-50\right)±\sqrt{2500+40\times 5000}}{2\left(-10\right)}$$

Multiply 40 times 5000.

$$x=\frac{-\left(-50\right)±\sqrt{2500+200000}}{2\left(-10\right)}$$

Add 2500 to 200000.

$$x=\frac{-\left(-50\right)±\sqrt{202500}}{2\left(-10\right)}$$

Take the square root of 202500.

$$x=\frac{-\left(-50\right)±450}{2\left(-10\right)}$$

The opposite of -50 is 50.

$$x=\frac{50±450}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{50±450}{-20}$$

Now solve the equation x=\frac{50±450}{-20} when ± is plus. Add 50 to 450.

$$x=\frac{500}{-20}$$

Divide 500 by -20.

$$x=-25$$

Now solve the equation x=\frac{50±450}{-20} when ± is minus. Subtract 450 from 50.

$$x=-\frac{400}{-20}$$

Divide -400 by -20.

$$x=20$$

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.

$$\left(x+5\right)\times 1000-x\times 1000=10x\left(x+5\right)$$

Use the distributive property to multiply x+5 by 1000.

$$1000x+5000-x\times 1000=10x\left(x+5\right)$$

Use the distributive property to multiply 10x by x+5.

$$1000x+5000-x\times 1000=10x^{2}+50x$$

Subtract 10x^{2} from both sides.

$$1000x+5000-x\times 1000-10x^{2}=50x$$

Subtract 50x from both sides.

$$1000x+5000-x\times 1000-10x^{2}-50x=0$$

Combine 1000x and -50x to get 950x.

$$950x+5000-x\times 1000-10x^{2}=0$$

Subtract 5000 from both sides. Anything subtracted from zero gives its negation.

$$950x-x\times 1000-10x^{2}=-5000$$

Multiply -1 and 1000 to get -1000.

$$950x-1000x-10x^{2}=-5000$$

Combine 950x and -1000x to get -50x.

$$-50x-10x^{2}=-5000$$

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.

$$-10x^{2}-50x=-5000$$

Divide both sides by -10.

$$\frac{-10x^{2}-50x}{-10}=-\frac{5000}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\left(-\frac{50}{-10}\right)x=-\frac{5000}{-10}$$

Divide -50 by -10.

$$x^{2}+5x=-\frac{5000}{-10}$$

Divide -5000 by -10.

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

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.

$$x=20 x=-25$$