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

Dec 20, 2016

There are two cases to investigate

#### Explanation:

(1)
$x > - 3 \to x + 3 > 0$
We may multiply both sides by $x + 3$ and the $<$ sign stays:
$x + 1 < 2 \left(x + 3\right) \to x + 1 < 2 x + 6 \to$ subtract $x$ and $6$
$\cancel{x} - \cancel{x} + 1 - 6 < 2 x - x + \cancel{6} - \cancel{6} \to$
$- 5 < x \to x > - 5$
This, with the first assumption, leads to $x > - 3$
(2)
$x < - 3 \to x + 3 < 0$
Multiply by $x + 3$ but the $<$sign becomes $>$
$x + 1 > 2 x + 6 \to$ subtract $x$ and $6$
$\cancel{x} - \cancel{x} + 1 - 6 > 2 x - x + \cancel{6} - \cancel{6} \to$
$- 5 > x \to x < - 5$

Combined: $x < - 5 \mathmr{and} x > - 3$

graph{(x+1)/(x+3) [-10, 10, -5, 5]}