How do you solve 1/x+1/(x+3)=2?

Dec 20, 2016

$x = - 1 \pm \sqrt{\frac{5}{2}}$.

Explanation:

To add fractions, we need to find a common denominator. For example, to do $\frac{1}{3} + \frac{1}{5}$, we would multiply $\frac{1}{3}$ by $\frac{5}{5}$ to get $\frac{5}{15}$ and $\frac{1}{5}$ by $\frac{3}{3}$ to get $\frac{3}{15}$, then add them to get $\frac{8}{15}$.

Similarly, $\frac{1}{x} = \frac{1}{x} \cdot \frac{x + 3}{x + 3} = \frac{x + 3}{x \left(x + 3\right)}$.

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

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

So now we have $\frac{2 x + 3}{x \left(x + 3\right)} = 2$.

Multiply both sides by $x \left(x + 3\right)$ to get $2 x + 3 = 2 x \left(x + 3\right)$.

Expand the right side to get $2 x + 3 = 2 {x}^{2} + 6 x$.

Rearrange terms to get $2 {x}^{2} + 4 x - 3 = 0$.

This is a quadratic equation with solutions $x = - 1 \pm \sqrt{\frac{5}{2}}$.