Solve 6(1/x+1/x+5)=1

$$6( \frac{ 1 }{ x } + \frac{ 1 }{ x+5 } )=1$$

Solve for x

$$x=-3$$

$$x=10$$

Steps Using Factoring By Grouping

Divide both sides by 6.

$$\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}$$

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 6x\left(x+5\right), the least common multiple of x,x+5,6.

$$6x+30+6x=x\left(x+5\right)$$

Combine 6x and 6x to get 12x.

$$12x+30=x\left(x+5\right)$$

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

$$12x+30=x^{2}+5x$$

Subtract x^{2} from both sides.

$$12x+30-x^{2}=5x$$

Subtract 5x from both sides.

$$12x+30-x^{2}-5x=0$$

Combine 12x and -5x to get 7x.

$$7x+30-x^{2}=0$$

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

$$-x^{2}+7x+30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+30. To find a and b, set up a system to be solved.

$$a+b=7 ab=-30=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 7.

$$a=10 b=-3$$

Rewrite -x^{2}+7x+30 as \left(-x^{2}+10x\right)+\left(-3x+30\right).

$$\left(-x^{2}+10x\right)+\left(-3x+30\right)$$

Factor out -x in the first and -3 in the second group.

$$-x\left(x-10\right)-3\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(-x-3\right)$$

To find equation solutions, solve x-10=0 and -x-3=0.

$$x=10 x=-3$$

Steps Using the Quadratic Formula

Divide both sides by 6.

$$\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}$$

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 6x\left(x+5\right), the least common multiple of x,x+5,6.

$$6x+30+6x=x\left(x+5\right)$$

Combine 6x and 6x to get 12x.

$$12x+30=x\left(x+5\right)$$

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

$$12x+30=x^{2}+5x$$

Subtract x^{2} from both sides.

$$12x+30-x^{2}=5x$$

Subtract 5x from both sides.

$$12x+30-x^{2}-5x=0$$

Combine 12x and -5x to get 7x.

$$7x+30-x^{2}=0$$

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.

$$-x^{2}+7x+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 7 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-1\right)\times 30}}{2\left(-1\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-1\right)\times 30}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-7±\sqrt{49+4\times 30}}{2\left(-1\right)}$$

Multiply 4 times 30.

$$x=\frac{-7±\sqrt{49+120}}{2\left(-1\right)}$$

Add 49 to 120.

$$x=\frac{-7±\sqrt{169}}{2\left(-1\right)}$$

Take the square root of 169.

$$x=\frac{-7±13}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-7±13}{-2}$$

Now solve the equation x=\frac{-7±13}{-2} when ± is plus. Add -7 to 13.

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

Divide 6 by -2.

$$x=-3$$

Now solve the equation x=\frac{-7±13}{-2} when ± is minus. Subtract 13 from -7.

$$x=-\frac{20}{-2}$$

Divide -20 by -2.

$$x=10$$

The equation is now solved.

$$x=-3 x=10$$

Steps for Completing the Square

Divide both sides by 6.

$$\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}$$

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 6x\left(x+5\right), the least common multiple of x,x+5,6.

$$6x+30+6x=x\left(x+5\right)$$

Combine 6x and 6x to get 12x.

$$12x+30=x\left(x+5\right)$$

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

$$12x+30=x^{2}+5x$$

Subtract x^{2} from both sides.

$$12x+30-x^{2}=5x$$

Subtract 5x from both sides.

$$12x+30-x^{2}-5x=0$$

Combine 12x and -5x to get 7x.

$$7x+30-x^{2}=0$$

Subtract 30 from both sides. Anything subtracted from zero gives its negation.

$$7x-x^{2}=-30$$

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.

$$-x^{2}+7x=-30$$

Divide both sides by -1.

$$\frac{-x^{2}+7x}{-1}=-\frac{30}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{7}{-1}x=-\frac{30}{-1}$$

Divide 7 by -1.

$$x^{2}-7x=-\frac{30}{-1}$$

Divide -30 by -1.

$$x^{2}-7x=30$$

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

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

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=30+\frac{49}{4}$$

Add 30 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{169}{4}$$

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

$$\left(x-\frac{7}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{13}{2} x-\frac{7}{2}=-\frac{13}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=10 x=-3$$