# How do you solve 1/(x+3)+1/(x+5)=1?

Apr 26, 2018

There are two values of $x$ that make this equation true. They are $x = - 3 - \sqrt{2}$ and $x = - 3 + \sqrt{2}$.

#### Explanation:

$\frac{1}{x + 3} + \frac{1}{x + 5} = 1$

Lots of ways to solve this. I'm going to try to do it the easiest way I know how.

First subtract $\frac{1}{x + 5}$ from both sides.

$\frac{1}{x + 3} = 1 - \frac{1}{x + 5}$

Now combine the two terms on the right-hand side of the equation.

$\frac{1}{x + 3} = \frac{x + 5 - 1}{x + 5}$

Simplify the numerator on the right-hand side.

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

Multiply both sides by $x - 5$.

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

Multiply both sides by $x + 3$.

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

Now expand the right hand side.

$x + 5 = {x}^{2} + 7 x + 12$

Put this quadratic equation in standard form.

${x}^{2} + 6 x + 7 = 0$

Use the quadratic formula to determine $x$.

$x = \frac{- 6 \pm \sqrt{{6}^{2} - 4 \left(1\right) \left(7\right)}}{2 \left(1\right)} = \frac{- 6 \pm \sqrt{8}}{2} = - 3 \pm \sqrt{2}$

So there are two values of $x$ that make this equation true. They are $x = - 3 - \sqrt{2}$ and $x = - 3 + \sqrt{2}$.

Apr 26, 2018

$x = - 3 \pm \sqrt{2}$

#### Explanation:

Given: $\frac{1}{x + 3} + \frac{1}{x + 5} = 1$

One way to solve is to eliminate the denominators by multiplying both sides of the equation by the common denominator.

The common denominator = $\left(x + 3\right) \left(x + 5\right)$

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

Distribute and cancel:
$\frac{\cancel{x + 3} \left(x + 5\right)}{\cancel{x + 3}} + \frac{\left(x + 3\right) \cancel{x + 5}}{\cancel{x + 5}} = \left(x + 3\right) \left(x + 5\right)$
:
Add like terms and use FOIL
$x + 5 + x + 3 = \left(x + 3\right) \left(x + 5\right)$

$2 x + 8 = {x}^{2} + 5 x + 3 x + 15$

$2 x + 8 = {x}^{2} + 8 x + 15$

Subtract $2 x$ and $8$ from both sides:
${x}^{2} + 6 x + 7 = 0$

Use the quadratic formula to solve for $x$:

$x = \frac{- B \pm \sqrt{{B}^{2} - 4 A C}}{2 A}$,

where the equation is in the form $A {x}^{2} + B x + C = 0$

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot 7}}{2}$

$x = - 3 \pm \frac{\sqrt{8}}{2} = - 3 \pm \frac{\sqrt{4} \sqrt{2}}{2}$

$x = = - 3 \pm \frac{\cancel{2} \sqrt{2}}{\cancel{2}}$

$x = - 3 \pm \sqrt{2}$