Solve 1/x+1/frac{1{2}+x/x+1}=1/5

$$\frac { \frac { 1 } { x + 1 } } { \frac { 1 } { 2 } + \frac { x } { x + 1 } } = \frac { 1 } { 5 }$$

Solve for x

$$x=3$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and x+1 is 2\left(x+1\right). Multiply \frac{1}{2} times \frac{x+1}{x+1}. Multiply \frac{x}{x+1} times \frac{2}{2}.

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

Since \frac{x+1}{2\left(x+1\right)} and \frac{2x}{2\left(x+1\right)} have the same denominator, add them by adding their numerators.

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

Combine like terms in x+1+2x.

$$\frac{\frac{1}{x+1}}{\frac{3x+1}{2\left(x+1\right)}}=\frac{1}{5}$$

Variable x cannot be equal to -1 since division by zero is not defined. Divide \frac{1}{x+1} by \frac{3x+1}{2\left(x+1\right)} by multiplying \frac{1}{x+1} by the reciprocal of \frac{3x+1}{2\left(x+1\right)}.

$$\frac{2\left(x+1\right)}{\left(x+1\right)\left(3x+1\right)}=\frac{1}{5}$$

Use the distributive property to multiply 2 by x+1.

$$\frac{2x+2}{\left(x+1\right)\left(3x+1\right)}=\frac{1}{5}$$

Use the distributive property to multiply x+1 by 3x+1 and combine like terms.

$$\frac{2x+2}{3x^{2}+4x+1}=\frac{1}{5}$$

Subtract \frac{1}{5} from both sides.

$$\frac{2x+2}{3x^{2}+4x+1}-\frac{1}{5}=0$$

Factor 3x^{2}+4x+1.

$$\frac{2x+2}{\left(x+1\right)\left(3x+1\right)}-\frac{1}{5}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+1\right)\left(3x+1\right) and 5 is 5\left(x+1\right)\left(3x+1\right). Multiply \frac{2x+2}{\left(x+1\right)\left(3x+1\right)} times \frac{5}{5}. Multiply \frac{1}{5} times \frac{\left(x+1\right)\left(3x+1\right)}{\left(x+1\right)\left(3x+1\right)}.

$$\frac{5\left(2x+2\right)}{5\left(x+1\right)\left(3x+1\right)}-\frac{\left(x+1\right)\left(3x+1\right)}{5\left(x+1\right)\left(3x+1\right)}=0$$

Since \frac{5\left(2x+2\right)}{5\left(x+1\right)\left(3x+1\right)} and \frac{\left(x+1\right)\left(3x+1\right)}{5\left(x+1\right)\left(3x+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{5\left(2x+2\right)-\left(x+1\right)\left(3x+1\right)}{5\left(x+1\right)\left(3x+1\right)}=0$$

Do the multiplications in 5\left(2x+2\right)-\left(x+1\right)\left(3x+1\right).

$$\frac{10x+10-3x^{2}-x-3x-1}{5\left(x+1\right)\left(3x+1\right)}=0$$

Combine like terms in 10x+10-3x^{2}-x-3x-1.

$$\frac{6x+9-3x^{2}}{5\left(x+1\right)\left(3x+1\right)}=0$$

Variable x cannot be equal to any of the values -1,-\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by 5\left(x+1\right)\left(3x+1\right).

$$6x+9-3x^{2}=0$$

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.

$$-3x^{2}+6x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 6 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-3\right)\times 9}}{2\left(-3\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-3\right)\times 9}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-6±\sqrt{36+12\times 9}}{2\left(-3\right)}$$

Multiply 12 times 9.

$$x=\frac{-6±\sqrt{36+108}}{2\left(-3\right)}$$

Add 36 to 108.

$$x=\frac{-6±\sqrt{144}}{2\left(-3\right)}$$

Take the square root of 144.

$$x=\frac{-6±12}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-6±12}{-6}$$

Now solve the equation x=\frac{-6±12}{-6} when ± is plus. Add -6 to 12.

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

Divide 6 by -6.

$$x=-1$$

Now solve the equation x=\frac{-6±12}{-6} when ± is minus. Subtract 12 from -6.

$$x=-\frac{18}{-6}$$

Divide -18 by -6.

$$x=3$$

The equation is now solved.

$$x=-1 x=3$$

Variable x cannot be equal to -1.

$$x=3$$

Steps for Completing the Square

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and x+1 is 2\left(x+1\right). Multiply \frac{1}{2} times \frac{x+1}{x+1}. Multiply \frac{x}{x+1} times \frac{2}{2}.

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

Since \frac{x+1}{2\left(x+1\right)} and \frac{2x}{2\left(x+1\right)} have the same denominator, add them by adding their numerators.

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

Combine like terms in x+1+2x.

$$\frac{\frac{1}{x+1}}{\frac{3x+1}{2\left(x+1\right)}}=\frac{1}{5}$$

Variable x cannot be equal to -1 since division by zero is not defined. Divide \frac{1}{x+1} by \frac{3x+1}{2\left(x+1\right)} by multiplying \frac{1}{x+1} by the reciprocal of \frac{3x+1}{2\left(x+1\right)}.

$$\frac{2\left(x+1\right)}{\left(x+1\right)\left(3x+1\right)}=\frac{1}{5}$$

Use the distributive property to multiply 2 by x+1.

$$\frac{2x+2}{\left(x+1\right)\left(3x+1\right)}=\frac{1}{5}$$

Use the distributive property to multiply x+1 by 3x+1 and combine like terms.

$$\frac{2x+2}{3x^{2}+4x+1}=\frac{1}{5}$$

Variable x cannot be equal to any of the values -1,-\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by 5\left(x+1\right)\left(3x+1\right), the least common multiple of 3x^{2}+4x+1,5.

$$5\left(2x+2\right)=\left(x+1\right)\left(3x+1\right)$$

Use the distributive property to multiply 5 by 2x+2.

$$10x+10=\left(x+1\right)\left(3x+1\right)$$

Use the distributive property to multiply x+1 by 3x+1 and combine like terms.

$$10x+10=3x^{2}+4x+1$$

Subtract 3x^{2} from both sides.

$$10x+10-3x^{2}=4x+1$$

Subtract 4x from both sides.

$$10x+10-3x^{2}-4x=1$$

Combine 10x and -4x to get 6x.

$$6x+10-3x^{2}=1$$

Subtract 10 from both sides.

$$6x-3x^{2}=1-10$$

Subtract 10 from 1 to get -9.

$$6x-3x^{2}=-9$$

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.

$$-3x^{2}+6x=-9$$

Divide both sides by -3.

$$\frac{-3x^{2}+6x}{-3}=-\frac{9}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{6}{-3}x=-\frac{9}{-3}$$

Divide 6 by -3.

$$x^{2}-2x=-\frac{9}{-3}$$

Divide -9 by -3.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=3+1$$

Add 3 to 1.

$$x^{2}-2x+1=4$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-1=2 x-1=-2$$

Add 1 to both sides of the equation.

$$x=3 x=-1$$

Variable x cannot be equal to -1.

$$x=3$$