Solve 300/x-300/x+5=2

$$\frac { 300 } { x } - \frac { 300 } { x + 5 } = 2$$

Solve for x

$$x=-30$$

$$x=25$$

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.

$$\left(x+5\right)\times 300-x\times 300=2x\left(x+5\right)$$

Use the distributive property to multiply x+5 by 300.

$$300x+1500-x\times 300=2x\left(x+5\right)$$

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

$$300x+1500-x\times 300=2x^{2}+10x$$

Subtract 2x^{2} from both sides.

$$300x+1500-x\times 300-2x^{2}=10x$$

Subtract 10x from both sides.

$$300x+1500-x\times 300-2x^{2}-10x=0$$

Combine 300x and -10x to get 290x.

$$290x+1500-x\times 300-2x^{2}=0$$

Multiply -1 and 300 to get -300.

$$290x+1500-300x-2x^{2}=0$$

Combine 290x and -300x to get -10x.

$$-10x+1500-2x^{2}=0$$

Divide both sides by 2.

$$-5x+750-x^{2}=0$$

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

$$-x^{2}-5x+750=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+750. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-750=-750$$

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 -750.

$$1,-750 2,-375 3,-250 5,-150 6,-125 10,-75 15,-50 25,-30$$

Calculate the sum for each pair.

$$1-750=-749 2-375=-373 3-250=-247 5-150=-145 6-125=-119 10-75=-65 15-50=-35 25-30=-5$$

The solution is the pair that gives sum -5.

$$a=25 b=-30$$

Rewrite -x^{2}-5x+750 as \left(-x^{2}+25x\right)+\left(-30x+750\right).

$$\left(-x^{2}+25x\right)+\left(-30x+750\right)$$

Factor out x in the first and 30 in the second group.

$$x\left(-x+25\right)+30\left(-x+25\right)$$

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

$$\left(-x+25\right)\left(x+30\right)$$

To find equation solutions, solve -x+25=0 and x+30=0.

$$x=25 x=-30$$

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.

$$\left(x+5\right)\times 300-x\times 300=2x\left(x+5\right)$$

Use the distributive property to multiply x+5 by 300.

$$300x+1500-x\times 300=2x\left(x+5\right)$$

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

$$300x+1500-x\times 300=2x^{2}+10x$$

Subtract 2x^{2} from both sides.

$$300x+1500-x\times 300-2x^{2}=10x$$

Subtract 10x from both sides.

$$300x+1500-x\times 300-2x^{2}-10x=0$$

Combine 300x and -10x to get 290x.

$$290x+1500-x\times 300-2x^{2}=0$$

Multiply -1 and 300 to get -300.

$$290x+1500-300x-2x^{2}=0$$

Combine 290x and -300x to get -10x.

$$-10x+1500-2x^{2}=0$$

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.

$$-2x^{2}-10x+1500=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -10 for b, and 1500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-2\right)\times 1500}}{2\left(-2\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-2\right)\times 1500}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-10\right)±\sqrt{100+8\times 1500}}{2\left(-2\right)}$$

Multiply 8 times 1500.

$$x=\frac{-\left(-10\right)±\sqrt{100+12000}}{2\left(-2\right)}$$

Add 100 to 12000.

$$x=\frac{-\left(-10\right)±\sqrt{12100}}{2\left(-2\right)}$$

Take the square root of 12100.

$$x=\frac{-\left(-10\right)±110}{2\left(-2\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±110}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{10±110}{-4}$$

Now solve the equation x=\frac{10±110}{-4} when ± is plus. Add 10 to 110.

$$x=\frac{120}{-4}$$

Divide 120 by -4.

$$x=-30$$

Now solve the equation x=\frac{10±110}{-4} when ± is minus. Subtract 110 from 10.

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

Divide -100 by -4.

$$x=25$$

The equation is now solved.

$$x=-30 x=25$$

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.

$$\left(x+5\right)\times 300-x\times 300=2x\left(x+5\right)$$

Use the distributive property to multiply x+5 by 300.

$$300x+1500-x\times 300=2x\left(x+5\right)$$

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

$$300x+1500-x\times 300=2x^{2}+10x$$

Subtract 2x^{2} from both sides.

$$300x+1500-x\times 300-2x^{2}=10x$$

Subtract 10x from both sides.

$$300x+1500-x\times 300-2x^{2}-10x=0$$

Combine 300x and -10x to get 290x.

$$290x+1500-x\times 300-2x^{2}=0$$

Subtract 1500 from both sides. Anything subtracted from zero gives its negation.

$$290x-x\times 300-2x^{2}=-1500$$

Multiply -1 and 300 to get -300.

$$290x-300x-2x^{2}=-1500$$

Combine 290x and -300x to get -10x.

$$-10x-2x^{2}=-1500$$

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.

$$-2x^{2}-10x=-1500$$

Divide both sides by -2.

$$\frac{-2x^{2}-10x}{-2}=-\frac{1500}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{10}{-2}\right)x=-\frac{1500}{-2}$$

Divide -10 by -2.

$$x^{2}+5x=-\frac{1500}{-2}$$

Divide -1500 by -2.

$$x^{2}+5x=750$$

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.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=750+\left(\frac{5}{2}\right)^{2}$$

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

$$x^{2}+5x+\frac{25}{4}=750+\frac{25}{4}$$

Add 750 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{3025}{4}$$

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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{3025}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{3025}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{55}{2} x+\frac{5}{2}=-\frac{55}{2}$$

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

$$x=25 x=-30$$