# How do you graph the compound inequality 3p+6<8-p and 5p+8>=p+6?

Apr 27, 2015

The first step is to bring both inequalities to a simplest possible form using invariant transformations (that is, those that produce equivalent inequalities).

$3 p + 6 < 8 - p$
Add $p$ to both sides and subtract $6$ from both sides.
$4 p < 2$
Divide both sides by $4$.
$p < \frac{1}{2}$ (simplified inequality 1)

$5 p + 8 \ge p + 6$
Subtract $p$ and subtract $8$ from both sides.
$4 p \ge - 2$
Divide both sides by $4$.
$p \ge - \frac{1}{2}$ (simplified inequality 2)

Now it's easy to combine both simplified inequalities (1) and (2).
The first one restricts $p$ from above to be less than $\frac{1}{2}$.
The second one restricts $p$ from below to be greater or equal to $- \frac{1}{2}$.
Combining these restrictions, we come to an interval $p$ is supposed to be in:
$- \frac{1}{2} \le p < \frac{1}{2}$

Graphically, it is represented by an interval on the X-axis $\left[- \frac{1}{2} , \frac{1}{2}\right)$ where a square bracket on the left indicates that the left border point $p = - \frac{1}{2}$ is included into an interval, while the parenthesis on the right indicates that the right border point $p = \frac{1}{2}$ is not included into an interval.
Usually, an arrow on the side of strong inequality ($p = \frac{1}{2}$) might indicate that an end point is not included (not on this graph).
graph{sqrt(x+1/2)0+sqrt(1/2-x) 0 [-1, 1, -0.5, 0.5]}