# How do you solve (x+10)/(x+3)<2?

Jul 31, 2017

The solution is $x \in \left(- 3 , 4\right)$

#### Explanation:

We cannot do crossing over

$\frac{x + 10}{x + 3} < 2$

Let's rewrite the inequality

$\frac{x + 10}{x + 3} - 2 < 0$

$\frac{\left(x + 10\right) - 2 \left(x + 3\right)}{x + 3} < 0$

$\frac{\left(x + 10 - 2 x - 6\right)}{x + 3} < 0$

$\frac{\left(4 - x\right)}{x + 3} < 0$

Let $f \left(x\right) = \frac{4 - x}{x + 3}$

We build the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a}$$- 3$$\textcolor{w h i t e}{a a a a a a a a}$$4$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 3$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$4 - x$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$-$

Therefore,

$f \left(x\right) < 0$ when $x \in \left(- 3 , 4\right)$

graph{(x+10)/(x+3)-2 [-16.02, 16.02, -8.01, 8.02]}