Solve 300/x-3=300/x+5

$$\frac { 300 } { x } - 3 = \frac { 300 } { x + 5 }$$

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.

$$\left(x+5\right)\times 300+x\left(x+5\right)\left(-3\right)=x\times 300$$

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

$$300x+1500+x\left(x+5\right)\left(-3\right)=x\times 300$$

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

$$300x+1500+\left(x^{2}+5x\right)\left(-3\right)=x\times 300$$

Use the distributive property to multiply x^{2}+5x by -3.

$$300x+1500-3x^{2}-15x=x\times 300$$

Combine 300x and -15x to get 285x.

$$285x+1500-3x^{2}=x\times 300$$

Subtract x\times 300 from both sides.

$$285x+1500-3x^{2}-x\times 300=0$$

Combine 285x and -x\times 300 to get -15x.

$$-15x+1500-3x^{2}=0$$

Divide both sides by 3.

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

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.

$$\left(x+5\right)\times 300+x\left(x+5\right)\left(-3\right)=x\times 300$$

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

$$300x+1500+x\left(x+5\right)\left(-3\right)=x\times 300$$

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

$$300x+1500+\left(x^{2}+5x\right)\left(-3\right)=x\times 300$$

Use the distributive property to multiply x^{2}+5x by -3.

$$300x+1500-3x^{2}-15x=x\times 300$$

Combine 300x and -15x to get 285x.

$$285x+1500-3x^{2}=x\times 300$$

Subtract x\times 300 from both sides.

$$285x+1500-3x^{2}-x\times 300=0$$

Combine 285x and -x\times 300 to get -15x.

$$-15x+1500-3x^{2}=0$$

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.

$$-3x^{2}-15x+1500=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -15 for b, and 1500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\left(-3\right)\times 1500}}{2\left(-3\right)}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\left(-3\right)\times 1500}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-15\right)±\sqrt{225+12\times 1500}}{2\left(-3\right)}$$

Multiply 12 times 1500.

$$x=\frac{-\left(-15\right)±\sqrt{225+18000}}{2\left(-3\right)}$$

Add 225 to 18000.

$$x=\frac{-\left(-15\right)±\sqrt{18225}}{2\left(-3\right)}$$

Take the square root of 18225.

$$x=\frac{-\left(-15\right)±135}{2\left(-3\right)}$$

The opposite of -15 is 15.

$$x=\frac{15±135}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{15±135}{-6}$$

Now solve the equation x=\frac{15±135}{-6} when ± is plus. Add 15 to 135.

$$x=\frac{150}{-6}$$

Divide 150 by -6.

$$x=-25$$

Now solve the equation x=\frac{15±135}{-6} when ± is minus. Subtract 135 from 15.

$$x=-\frac{120}{-6}$$

Divide -120 by -6.

$$x=20$$

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.

$$\left(x+5\right)\times 300+x\left(x+5\right)\left(-3\right)=x\times 300$$

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

$$300x+1500+x\left(x+5\right)\left(-3\right)=x\times 300$$

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

$$300x+1500+\left(x^{2}+5x\right)\left(-3\right)=x\times 300$$

Use the distributive property to multiply x^{2}+5x by -3.

$$300x+1500-3x^{2}-15x=x\times 300$$

Combine 300x and -15x to get 285x.

$$285x+1500-3x^{2}=x\times 300$$

Subtract x\times 300 from both sides.

$$285x+1500-3x^{2}-x\times 300=0$$

Combine 285x and -x\times 300 to get -15x.

$$-15x+1500-3x^{2}=0$$

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

$$-15x-3x^{2}=-1500$$

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.

$$-3x^{2}-15x=-1500$$

Divide both sides by -3.

$$\frac{-3x^{2}-15x}{-3}=-\frac{1500}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{15}{-3}\right)x=-\frac{1500}{-3}$$

Divide -15 by -3.

$$x^{2}+5x=-\frac{1500}{-3}$$

Divide -1500 by -3.

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

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.

$$x=20 x=-25$$