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

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

Solve for x

$$x=-20$$

$$x=25$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,5 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.

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

Use the distributive property to multiply x-5 by 200+2x and combine like terms.

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

Subtract x\times 200 from both sides.

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

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

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

Divide both sides by 2.

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

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.

$$a+b=-5 ab=1\left(-500\right)=-500$$

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.

$$a=-25 b=20$$

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

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

Factor out x in the first and 20 in the second group.

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

Factor out common term x-25 by using distributive property.

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

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

$$x=25 x=-20$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,5 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.

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

Use the distributive property to multiply x-5 by 200+2x and combine like terms.

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

Subtract x\times 200 from both sides.

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

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

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

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}.

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

Square -10.

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

Multiply -4 times 2.

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

Multiply -8 times -1000.

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

Add 100 to 8000.

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

Take the square root of 8100.

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

The opposite of -10 is 10.

$$x=\frac{10±90}{2\times 2}$$

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 0,5 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.

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

Use the distributive property to multiply x-5 by 200+2x and combine like terms.

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

Subtract x\times 200 from both sides.

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

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

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

Add 1000 to both sides. Anything plus zero gives itself.

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

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.

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

Add \frac{5}{2} to both sides of the equation.

$$x=25 x=-20$$