Solve 360/x-360/x+5=1

$$\frac { 360 } { x } - \frac { 360 } { x + 5 } = 1$$

Solve for x

$$x=-45$$

$$x=40$$

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.

$$\left(x+5\right)\times 360-x\times 360=x\left(x+5\right)$$

Use the distributive property to multiply x+5 by 360.

$$360x+1800-x\times 360=x\left(x+5\right)$$

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

$$360x+1800-x\times 360=x^{2}+5x$$

Subtract x^{2} from both sides.

$$360x+1800-x\times 360-x^{2}=5x$$

Subtract 5x from both sides.

$$360x+1800-x\times 360-x^{2}-5x=0$$

Combine 360x and -5x to get 355x.

$$355x+1800-x\times 360-x^{2}=0$$

Multiply -1 and 360 to get -360.

$$355x+1800-360x-x^{2}=0$$

Combine 355x and -360x to get -5x.

$$-5x+1800-x^{2}=0$$

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

$$-x^{2}-5x+1800=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+1800. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-1800=-1800$$

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 -1800.

$$1,-1800 2,-900 3,-600 4,-450 5,-360 6,-300 8,-225 9,-200 10,-180 12,-150 15,-120 18,-100 20,-90 24,-75 25,-72 30,-60 36,-50 40,-45$$

Calculate the sum for each pair.

$$1-1800=-1799 2-900=-898 3-600=-597 4-450=-446 5-360=-355 6-300=-294 8-225=-217 9-200=-191 10-180=-170 12-150=-138 15-120=-105 18-100=-82 20-90=-70 24-75=-51 25-72=-47 30-60=-30 36-50=-14 40-45=-5$$

The solution is the pair that gives sum -5.

$$a=40 b=-45$$

Rewrite -x^{2}-5x+1800 as \left(-x^{2}+40x\right)+\left(-45x+1800\right).

$$\left(-x^{2}+40x\right)+\left(-45x+1800\right)$$

Factor out x in the first and 45 in the second group.

$$x\left(-x+40\right)+45\left(-x+40\right)$$

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

$$\left(-x+40\right)\left(x+45\right)$$

To find equation solutions, solve -x+40=0 and x+45=0.

$$x=40 x=-45$$

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.

$$\left(x+5\right)\times 360-x\times 360=x\left(x+5\right)$$

Use the distributive property to multiply x+5 by 360.

$$360x+1800-x\times 360=x\left(x+5\right)$$

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

$$360x+1800-x\times 360=x^{2}+5x$$

Subtract x^{2} from both sides.

$$360x+1800-x\times 360-x^{2}=5x$$

Subtract 5x from both sides.

$$360x+1800-x\times 360-x^{2}-5x=0$$

Combine 360x and -5x to get 355x.

$$355x+1800-x\times 360-x^{2}=0$$

Multiply -1 and 360 to get -360.

$$355x+1800-360x-x^{2}=0$$

Combine 355x and -360x to get -5x.

$$-5x+1800-x^{2}=0$$

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.

$$-x^{2}-5x+1800=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -5 for b, and 1800 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)\times 1800}}{2\left(-1\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\times 1800}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-5\right)±\sqrt{25+4\times 1800}}{2\left(-1\right)}$$

Multiply 4 times 1800.

$$x=\frac{-\left(-5\right)±\sqrt{25+7200}}{2\left(-1\right)}$$

Add 25 to 7200.

$$x=\frac{-\left(-5\right)±\sqrt{7225}}{2\left(-1\right)}$$

Take the square root of 7225.

$$x=\frac{-\left(-5\right)±85}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±85}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{5±85}{-2}$$

Now solve the equation x=\frac{5±85}{-2} when ± is plus. Add 5 to 85.

$$x=\frac{90}{-2}$$

Divide 90 by -2.

$$x=-45$$

Now solve the equation x=\frac{5±85}{-2} when ± is minus. Subtract 85 from 5.

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

Divide -80 by -2.

$$x=40$$

The equation is now solved.

$$x=-45 x=40$$

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.

$$\left(x+5\right)\times 360-x\times 360=x\left(x+5\right)$$

Use the distributive property to multiply x+5 by 360.

$$360x+1800-x\times 360=x\left(x+5\right)$$

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

$$360x+1800-x\times 360=x^{2}+5x$$

Subtract x^{2} from both sides.

$$360x+1800-x\times 360-x^{2}=5x$$

Subtract 5x from both sides.

$$360x+1800-x\times 360-x^{2}-5x=0$$

Combine 360x and -5x to get 355x.

$$355x+1800-x\times 360-x^{2}=0$$

Subtract 1800 from both sides. Anything subtracted from zero gives its negation.

$$355x-x\times 360-x^{2}=-1800$$

Multiply -1 and 360 to get -360.

$$355x-360x-x^{2}=-1800$$

Combine 355x and -360x to get -5x.

$$-5x-x^{2}=-1800$$

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.

$$-x^{2}-5x=-1800$$

Divide both sides by -1.

$$\frac{-x^{2}-5x}{-1}=-\frac{1800}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{5}{-1}\right)x=-\frac{1800}{-1}$$

Divide -5 by -1.

$$x^{2}+5x=-\frac{1800}{-1}$$

Divide -1800 by -1.

$$x^{2}+5x=1800$$

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.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=1800+\left(\frac{5}{2}\right)^{2}$$

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

$$x^{2}+5x+\frac{25}{4}=1800+\frac{25}{4}$$

Add 1800 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{7225}{4}$$

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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{7225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{7225}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{85}{2} x+\frac{5}{2}=-\frac{85}{2}$$

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

$$x=40 x=-45$$