# How do you solve and write the following in interval notation:  (x + 5)/(x + 9) &lt; 3?

Feb 10, 2018

$x \in \left(- \infty , - 11\right) \cup \left(- 9 , \infty\right)$

#### Explanation:

Given:

$\frac{x + 5}{x + 9} < 3$

I would like first to note what you can't do, which is to simply multiply both sides by $\left(x + 9\right)$. The problem is that if $x < - 9$ then $\left(x + 9\right) < 0$, so you would have to reverse the inequality too, while if $x > - 9$ then you would leave the inequality as $<$. So you would need to split into cases - messy.

So what can we do?

Noting first that $x \ne - 9$, we can multiply both sides of the inequality by ${\left(x + 9\right)}^{2}$ which is positive, to get:

$\left(x + 5\right) \left(x + 9\right) < 3 {\left(x + 9\right)}^{2}$

which multiplies out to:

${x}^{2} + 14 x + 45 < 3 {x}^{2} + 54 x + 243$

Then subtracting ${x}^{2} + 14 x + 45$ from both sides we get:

$0 < 2 {x}^{2} + 40 x + 198 = 2 \left({x}^{2} + 20 x + 99\right) = 2 \left(x + 9\right) \left(x + 11\right)$

The right hand side is a quadratic with positive leading coefficient, which intersects the $x$ axis at $x = - 11$ and $x = - 9$.

Therefore it is positive when:

$x < - 11 \text{ }$ or $\text{ } x > - 9$

That is:

$x \in \left(- \infty , - 11\right) \cup \left(- 9 , \infty\right)$