How do you graph the compound inequality 3p+6<8-p3p+6<8p and 5p+8>=p+65p+8p+6?

1 Answer
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).

3p+6<8-p3p+6<8p
Add pp to both sides and subtract 66 from both sides.
4p<24p<2
Divide both sides by 44.
p<1/2p<12 (simplified inequality 1)

5p+8 >= p+65p+8p+6
Subtract pp and subtract 88 from both sides.
4p>=-24p2
Divide both sides by 44.
p >= -1/2p12 (simplified inequality 2)

Now it's easy to combine both simplified inequalities (1) and (2).
The first one restricts pp from above to be less than 1/212.
The second one restricts pp from below to be greater or equal to -1/212.
Combining these restrictions, we come to an interval pp is supposed to be in:
-1/2 <= p < 1/212p<12

Graphically, it is represented by an interval on the X-axis [-1/2,1/2)[12,12) where a square bracket on the left indicates that the left border point p=-1/2p=12 is included into an interval, while the parenthesis on the right indicates that the right border point p=1/2p=12 is not included into an interval.
Usually, an arrow on the side of strong inequality (p=1/2p=12) 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]}