Solve 18/x+4/x+5=3

$$\frac { 18 } { x } + \frac { 4 } { x + 5 } = 3$$

Solve for x

$$x = \frac{\sqrt{1129} + 7}{6} \approx 6.766765872$$

$$x=\frac{7-\sqrt{1129}}{6}\approx -4.433432539$$

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.

$$\left(x+5\right)\times 18+x\times 4=3x\left(x+5\right)$$

Use the distributive property to multiply x+5 by 18.

$$18x+90+x\times 4=3x\left(x+5\right)$$

Combine 18x and x\times 4 to get 22x.

$$22x+90=3x\left(x+5\right)$$

Use the distributive property to multiply 3x by x+5.

$$22x+90=3x^{2}+15x$$

Subtract 3x^{2} from both sides.

$$22x+90-3x^{2}=15x$$

Subtract 15x from both sides.

$$22x+90-3x^{2}-15x=0$$

Combine 22x and -15x to get 7x.

$$7x+90-3x^{2}=0$$

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.

$$-3x^{2}+7x+90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 7 for b, and 90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-3\right)\times 90}}{2\left(-3\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-3\right)\times 90}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-7±\sqrt{49+12\times 90}}{2\left(-3\right)}$$

Multiply 12 times 90.

$$x=\frac{-7±\sqrt{49+1080}}{2\left(-3\right)}$$

Add 49 to 1080.

$$x=\frac{-7±\sqrt{1129}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-7±\sqrt{1129}}{-6}$$

Now solve the equation x=\frac{-7±\sqrt{1129}}{-6} when ± is plus. Add -7 to \sqrt{1129}.

$$x=\frac{\sqrt{1129}-7}{-6}$$

Divide -7+\sqrt{1129} by -6.

$$x=\frac{7-\sqrt{1129}}{6}$$

Now solve the equation x=\frac{-7±\sqrt{1129}}{-6} when ± is minus. Subtract \sqrt{1129} from -7.

$$x=\frac{-\sqrt{1129}-7}{-6}$$

Divide -7-\sqrt{1129} by -6.

$$x=\frac{\sqrt{1129}+7}{6}$$

The equation is now solved.

$$x=\frac{7-\sqrt{1129}}{6} x=\frac{\sqrt{1129}+7}{6}$$

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.

$$\left(x+5\right)\times 18+x\times 4=3x\left(x+5\right)$$

Use the distributive property to multiply x+5 by 18.

$$18x+90+x\times 4=3x\left(x+5\right)$$

Combine 18x and x\times 4 to get 22x.

$$22x+90=3x\left(x+5\right)$$

Use the distributive property to multiply 3x by x+5.

$$22x+90=3x^{2}+15x$$

Subtract 3x^{2} from both sides.

$$22x+90-3x^{2}=15x$$

Subtract 15x from both sides.

$$22x+90-3x^{2}-15x=0$$

Combine 22x and -15x to get 7x.

$$7x+90-3x^{2}=0$$

Subtract 90 from both sides. Anything subtracted from zero gives its negation.

$$7x-3x^{2}=-90$$

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.

$$-3x^{2}+7x=-90$$

Divide both sides by -3.

$$\frac{-3x^{2}+7x}{-3}=-\frac{90}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{7}{-3}x=-\frac{90}{-3}$$

Divide 7 by -3.

$$x^{2}-\frac{7}{3}x=-\frac{90}{-3}$$

Divide -90 by -3.

$$x^{2}-\frac{7}{3}x=30$$

Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{3}x+\left(-\frac{7}{6}\right)^{2}=30+\left(-\frac{7}{6}\right)^{2}$$

Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=30+\frac{49}{36}$$

Add 30 to \frac{49}{36}.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=\frac{1129}{36}$$

Factor x^{2}-\frac{7}{3}x+\frac{49}{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}.

$$\left(x-\frac{7}{6}\right)^{2}=\frac{1129}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{6}\right)^{2}}=\sqrt{\frac{1129}{36}}$$

Simplify.

$$x-\frac{7}{6}=\frac{\sqrt{1129}}{6} x-\frac{7}{6}=-\frac{\sqrt{1129}}{6}$$

Add \frac{7}{6} to both sides of the equation.

$$x=\frac{\sqrt{1129}+7}{6} x=\frac{7-\sqrt{1129}}{6}$$