# How do you solve: [ (x-2) / (x+3) ] < [ (x+1) / (x) ]?

Oct 20, 2015

$x \in \left(- \frac{1}{2} , + \infty\right) \text{\} \left\{0\right\}$

#### Explanation:

The inequality given to you looks like this

$\frac{x - 2}{x + 3} < \frac{x + 1}{x}$

Right from the start, you know that any solution interval must not contain the values of $x$ that will make the two denominators equal to zero.

More specifically, you need to have

$x + 3 \ne 0 \implies x \ne - 3 \text{ }$ and $\text{ } x \ne 0$

With that in mind, use the common denominator of the two fractions, which is equal to $\left(x + 3\right) \cdot x$, to get rid of the denominators.

More specifically, multiply the first fraction by $1 = \frac{x}{x}$ and the second fraction by $1 = \frac{x + 3}{x + 3}$.

This will get you

$\frac{x - 2}{x + 3} \cdot \frac{x}{x} < \frac{x + 1}{x} \cdot \frac{x + 3}{x + 3}$

$\frac{x \left(x - 2\right)}{x \left(x + 3\right)} < \frac{\left(x + 1\right) \left(x + 3\right)}{x \left(x + 3\right)}$

This is equivalent to

$x \left(x - 2\right) < \left(x + 1\right) \left(x + 3\right)$

Expand the parantheses to get

$\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} - 2 x < \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} + x + 3 x + 3$

Rearrange the inequality to isolate $x$ on one side

$- 6 x < 3 \implies x > \frac{3}{\left(- 6\right)} \iff x > - \frac{1}{2}$

This means that any value of $x$ that is greater than $- \frac{1}{2}$, except $x = 0$, will be a solution to the original inequality.

Therefore, the solution interval will be $x \in \left(- \frac{1}{2} , + \infty\right) \text{\} \left\{0\right\}$