Solve 45/x(x+15)+3/x+15=1

$$\frac { 45 } { x ( x + 15 ) } + \frac { 3 } { x + 15 } = 1$$

Solve for x

$$x=3$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -15,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+15\right), the least common multiple of x\left(x+15\right),x+15.

$$45+x\times 3=x\left(x+15\right)$$

Use the distributive property to multiply x by x+15.

$$45+x\times 3=x^{2}+15x$$

Subtract x^{2} from both sides.

$$45+x\times 3-x^{2}=15x$$

Subtract 15x from both sides.

$$45+x\times 3-x^{2}-15x=0$$

Combine x\times 3 and -15x to get -12x.

$$45-12x-x^{2}=0$$

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

$$-x^{2}-12x+45=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+45. To find a and b, set up a system to be solved.

$$a+b=-12 ab=-45=-45$$

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 -45.

$$1,-45 3,-15 5,-9$$

Calculate the sum for each pair.

$$1-45=-44 3-15=-12 5-9=-4$$

The solution is the pair that gives sum -12.

$$a=3 b=-15$$

Rewrite -x^{2}-12x+45 as \left(-x^{2}+3x\right)+\left(-15x+45\right).

$$\left(-x^{2}+3x\right)+\left(-15x+45\right)$$

Factor out x in the first and 15 in the second group.

$$x\left(-x+3\right)+15\left(-x+3\right)$$

Factor out common term -x+3 by using distributive property.

$$\left(-x+3\right)\left(x+15\right)$$

To find equation solutions, solve -x+3=0 and x+15=0.

$$x=3 x=-15$$

Variable x cannot be equal to -15.

$$x=3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -15,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+15\right), the least common multiple of x\left(x+15\right),x+15.

$$45+x\times 3=x\left(x+15\right)$$

Use the distributive property to multiply x by x+15.

$$45+x\times 3=x^{2}+15x$$

Subtract x^{2} from both sides.

$$45+x\times 3-x^{2}=15x$$

Subtract 15x from both sides.

$$45+x\times 3-x^{2}-15x=0$$

Combine x\times 3 and -15x to get -12x.

$$45-12x-x^{2}=0$$

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.

$$-x^{2}-12x+45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -12 for b, and 45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-1\right)\times 45}}{2\left(-1\right)}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\left(-1\right)\times 45}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-12\right)±\sqrt{144+4\times 45}}{2\left(-1\right)}$$

Multiply 4 times 45.

$$x=\frac{-\left(-12\right)±\sqrt{144+180}}{2\left(-1\right)}$$

Add 144 to 180.

$$x=\frac{-\left(-12\right)±\sqrt{324}}{2\left(-1\right)}$$

Take the square root of 324.

$$x=\frac{-\left(-12\right)±18}{2\left(-1\right)}$$

The opposite of -12 is 12.

$$x=\frac{12±18}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{12±18}{-2}$$

Now solve the equation x=\frac{12±18}{-2} when ± is plus. Add 12 to 18.

$$x=\frac{30}{-2}$$

Divide 30 by -2.

$$x=-15$$

Now solve the equation x=\frac{12±18}{-2} when ± is minus. Subtract 18 from 12.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

The equation is now solved.

$$x=-15 x=3$$

Variable x cannot be equal to -15.

$$x=3$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -15,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+15\right), the least common multiple of x\left(x+15\right),x+15.

$$45+x\times 3=x\left(x+15\right)$$

Use the distributive property to multiply x by x+15.

$$45+x\times 3=x^{2}+15x$$

Subtract x^{2} from both sides.

$$45+x\times 3-x^{2}=15x$$

Subtract 15x from both sides.

$$45+x\times 3-x^{2}-15x=0$$

Combine x\times 3 and -15x to get -12x.

$$45-12x-x^{2}=0$$

Subtract 45 from both sides. Anything subtracted from zero gives its negation.

$$-12x-x^{2}=-45$$

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.

$$-x^{2}-12x=-45$$

Divide both sides by -1.

$$\frac{-x^{2}-12x}{-1}=-\frac{45}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{12}{-1}\right)x=-\frac{45}{-1}$$

Divide -12 by -1.

$$x^{2}+12x=-\frac{45}{-1}$$

Divide -45 by -1.

$$x^{2}+12x=45$$

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+12x+6^{2}=45+6^{2}$$

Square 6.

$$x^{2}+12x+36=45+36$$

Add 45 to 36.

$$x^{2}+12x+36=81$$

Factor x^{2}+12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+6\right)^{2}=81$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{81}$$

Simplify.

$$x+6=9 x+6=-9$$

Subtract 6 from both sides of the equation.

$$x=3 x=-15$$

Variable x cannot be equal to -15.

$$x=3$$