Inequality Review
There is one big difference between absolute value equations and absolute value inequalities that we should quickly review before jumping into some practice problems, and that is: when we split an inequality into two different ones in order to undo an absolute value, we need to remember to flip the sign on the one we match with the negative. Other than that, all the rules are the same and we should be good to go.
I also want to mention that because this is a practice problem video and we'll be doing about four practice problems, I encourage you to pause the video when you see the problem. Try it on your own and see how far you get. If you get stuck, or if you finish the whole thing and want to check your answer, watch the video to see if I did it the same way you did it and if you got the right answer. That way you'll be able to focus in on what you did wrong, or you can skip through it quickly if you already know you got it right.
OR Compound Inequality
Graph of an OR compound inequality
We'll practice that one difference in this first question here: Solve and graph |x + 4| > 5.
Because there is nothing going on the outside of the absolute value, we can begin by splitting the inequality up to undo the absolute value. That leaves us with two inequalities: one, x + 4 > 5, and another, x + 4 < -5. So not only did we set it to -5, we also flipped it around to a 'less than' sign. We still need to solve each inequality for x, which means undoing the + 4 on both of them. You can undo addition with subtraction, and doing that for both inequalities gives us our solved inequality, x > 1 or x < -9. This is now a compound inequality, because there are two inequalities in one problem.
Graphing this compound inequality is as easy as putting both graphs on the same number line one at a time. We can begin by putting an open circle at 1 and drawing an arrow to the right. It's an open circle because it's only 'greater than,' not 'equal to', and the arrow goes to the right because we want all numbers that are bigger than 1. Putting the other one on there means an open circle at -9 and an arrow going to the left. We see that our graph is complete, and it looks like it's an OR compound inequality because the graphs are going in opposite directions. That means that either being bigger than 1 or being less than -9 is enough to satisfy this inequality. It does not have to be both; it can be one or the other. We actually could have known that it was an OR inequality from the beginning just by realizing first that all one-variable absolute value inequalities give us compound inequalities and that problems where the absolute value is greater than something lead us to OR examples.
AND Compound Inequality
Our second problem might at first seem like it will also be an OR compound inequality, but there is a difference that will make it end up not being exactly as it might initially seem. The problem, solve and graph -2|x| -1 > -9, also has an absolute value and a > symbol, but this time there are mathematical operations on the outside of the absolute value. This means before we split the inequality into two, we need to undo the -1 and the times -2 on the outside.
The outermost step is the -1, so we undo that with addition. We next undo multiplication of -2 with division of -2, and now we must remember the rule of inequalities that tells us to flip the symbol whenever multiplying or dividing by a negative number. This leaves us with the resulting inequality as |x| < 4, and now we realize that this will end up being an AND compound inequality because the absolute value is now less than 4 instead of greater than like in the beginning. Splitting the inequality and flipping one of the symbols leaves us with our solved inequality: x < 4 and x> -4.
To graph this, I can again do one at a time and put a closed circle (because it's 'or equal to') at 4 and draw an arrow to the left, then put another closed circle at -4 and draw an arrow to the right. Because the arrows are pointing toward each other, we can just connect the two dots and we end up with our AND compound inequality graph looking like so.
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