Can someone please help me with this absolute value inequality, can I draw a graph for it? |(x+4)/(x+2)|<=1?

Mar 27, 2018

$\left(- \infty , - 3\right]$

Explanation:

$| \frac{x + 4}{x + 2} | \le 1$

We know by the absolute value property that we have to solve both:

$\frac{x + 4}{x + 2} \le 1$ and $- \left(\frac{x + 4}{x + 2}\right) \le 1$

For:

$\frac{x + 4}{x + 2} \le 1$

Subtract $1$

$\frac{x + 4}{x + 2} - 1 \le 0$

$\frac{\left(x + 4\right) - \left(x + 2\right)}{x + 2} \le 0$

Simplify:

$\frac{2}{x + 2} \le 0$

Divide by 2:

$\frac{1}{x + 2} \le 0$

There is no solution for zero, (undefined division by zero)

only $x < - 2$

For:

$- \left(\frac{x + 4}{x + 2}\right) \le 1$

Subtract 1:

$- \frac{x + 4}{x + 2} - 1 < + 0$

Multiply by $- 1$:

$\frac{x + 4}{x + 2} + 1 \le 0$

$\frac{2 x + 6}{x + 2} \le 0$

Solving for zero:

$\frac{2 x + 6}{x + 2} = 0$

$x = - 3$

We now need to look at :

$| \frac{x + 4}{x + 2} | - 1 \le 0$

For:

$x < - 3$

$0 \le | \frac{x + 4}{x + 2} | < 1$

So:

$| \frac{x + 4}{x + 2} | - 1 \le 0$

For $x > - 3$, $x \ne - 2$

$| \frac{x + 4}{x + 2} | > 1$

So:

$| \frac{x + 4}{x + 2} | - 1 > 0$

So only: $x \le - 3$

Solution set:

$\left(- \infty , - 3\right]$