How do you solve and graph the compound inequality 4 < n + 6 < 9 ?

Mar 22, 2018

See below.

Explanation:

$4 < n + 6 < 9$

Subtract 6 from each part of the inequality:

$4 - 6 < n + 6 - 6 < 9 - 6$

$\textcolor{b l u e}{- 2 < n < 3}$

To graph:

Make two equations:

We have:

$n > - 2$ , $n + 2 > 0$

Equation $y = n + 2$

$n < 3$ , $n - 3 < 0$

Graph these two lines. This we give you the boundary between included and excluded regions. Remember to use a dotted line as these are of the for < , > and not of the form $\le , \ge$, so the line will not be an included region.

With these plotted, we have three regions A , B and C, we now test coordinates in each region to see which is an included or excluded region.

Region A:

coordinates:

$\left(- 3 , 2\right)$

$n + 2 > y$

$\left(- 3\right) + 2 > 2 \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ False

$n - 3 < y$

$\left(- 3\right) - 3 < 2 \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ True

Region B:

coordinates:

$\left(2 , 2\right)$

$n + 2 > y$

$\left(2\right) + 2 > 2 \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ True

$n - 3 < y$

$\left(2\right) - 3 < 2 \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ True

Region C

coordinates:

$\left(- 2 , 4\right)$

$n + 2 > y$

$\left(- 2\right) + 2 > 2 \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ False

$n - 3 < y$

$\left(4\right) - 3 < 2 \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ True

The only region where both conditions are met is region B. Shade region B