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

Mar 21, 2016

Resolving this gives: "x^2+6x+7=0

I will let you finish that off (you need to use the formula).

#### Explanation:

The bottom number/expression (denominator) need to be the same to enable direct addition.

Multiply by 1 and you do not change the value. Multiply by 1 but in the form of $1 = \frac{x + 5}{x + 5}$ you do not change the value but you do change the way it looks.

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Multiply $\frac{1}{x + 3}$ by 1 but in the form of $1 = \frac{x + 5}{x + 5}$

Giving $\textcolor{b r o w n}{\frac{1}{x + 3} \times \frac{x + 5}{x + 5} = \frac{x + 5}{\left(x + 3\right) \left(x + 5\right)}}$

Multiply $\frac{1}{x + 5}$ by 1 but in the form of $1 = \frac{x + 3}{x + 3}$

Giving $\textcolor{b r o w n}{\frac{1}{x + 5} \times \frac{x + 3}{x + 3} = \frac{x + 3}{\left(x + 3\right) \left(x + 5\right)}}$

$\textcolor{b l u e}{\text{Putting it all together}}$

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

$\text{ } \frac{2 x + 8}{\left(x + 3\right) \left(x + 5\right)} = 1$

Multiply both sides by $\left(x + 3\right) \left(x + 5\right)$ giving

$\text{ } \left(2 x + 8\right) \times \frac{\cancel{\left(x + 3\right) \left(x + 5\right)}}{\cancel{\left(x + 3\right) \left(x + 5\right)}} = 1 \times \left(x + 3\right) \left(x + 5\right)$

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

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

Use the formula to solve for $x$

I will let you do that

$y = a {x}^{2} + b x + c \text{ }$ where $\text{ } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
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