Big Ideas Math Integrated I, 2016
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Big Ideas Math Integrated I, 2016 View details
7. Systems of Linear Inequalities
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Exercise 39 Page 262

Split the absolute value inequality into a system of linear inequalities using the fact that is equivalent to for

System:
Graph:

Practice makes perfect

Absolute value inequalities can be solved using a compound inequality. Let's start by considering an example.

Example

Consider an absolute value inequality.
The above means that the distance to from is less than
We see that, for this example, must be greater than and less than

Writing the System

Let's apply the same reasoning to the given inequality, where
We can combine the two inequalities that form the above compound inequality to form a system.

Graphing the System

Let's consider both of these inequalities individually before we combine them to find the solution set for the entire system.

First, we need to graph the boundary line, which will be dashed because the inequality is strict. Remember that we are also told that and must restrict our domain accordingly.

Now we must choose which side of the line to shade. We can choose any arbitrary point within the domain. Let's test the point If it produces a true statement, we shade the region which contains the point. Otherwise, we shade the opposite region.
We should shade the side of the line that does not contain this point. Recall, again, that

Once again, we need to graph our boundary line first. This line is also dashed because we have a strict inequality. Remember that we must restrict our domain according to the fact that

Now we must choose which side of the line to shade. We can choose any arbitrary point within the domain to test. Let's test the point
We shade the side of the line that contains this point. Recall that

Combined solution set

Now we can combine the graphs to see where they overlap.

Finally, we can cut away all the unnecessary parts and leave only the solution set for the system of inequalities.