Back To CourseMath 101: College Algebra
13 chapters | 102 lessons | 11 flashcard sets
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Luke has taught high school algebra and geometry, college calculus, and has a master's degree in education.
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.
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.
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.
If at any point you are solving a problem like one of these two and you end up with a statement where an absolute value of anything is either greater than or less than a negative number, you know something is up. For example, take |7x - 1| > -5. It doesn't matter what is happening on the inside of the absolute value, because we know that it will eventually be turned into a positive number. Therefore, this absolute value will always be bigger than -5; the inequality will always be true no matter what we substitute in for x, and therefore there are an infinite number of solutions to this problem. But on the other hand, if instead we had |7x - 1| < -5, the same logic tells us that it is impossible for us to make an absolute value smaller than -5, and therefore there are no solutions to this one.
We'll end this practice problem video with a system of two-variable absolute value inequalities. Graph y > (1/2)|x| - 5 and y < 2.
This is a system of inequalities because there is more than one inequality, and at least one of them has two variables. We can solve this problem simply by putting one inequality on the graph at a time and then figuring out where they overlap. First, let's start with graphing y > (1/2)|x| - 5. To do this, we'll have to remember how to use translations and also how to graph an absolute value. All absolute value graphs look more or less like 'V's, but this one is going to have a few differences. First off, the -5 on the end will pull the vertex of the graph down five places, so instead of it being at the origin (0,0), it will be pulled down to the point (0,-5). Secondly, the 1/2 in front of the absolute value will make the slope of the 'V' 1/2 instead of 1. That means that, starting from our vertex, we'll go up one and then over two in each direction to determine how steep the 'V' is, and we can sketch it in like this. We can leave the line of the 'V' solid because it is 'or equal to' from the inequality, but we still need to determine which part of the graph to shade. Substituting in (0,0) give us the inequality 0 > (1/2)(0) - 5, which can simplify down to 0 > -5, which is true. That means we can shade the area of the graph with (0,0) in it, and we fill in everything on the inside of the 'V' to get that two-variable inequality.
But this was a system of inequalities, so we've still got another piece to add to this graph, y < 2. Just like we began the first half of this problem by graphing the line where y was equal to the absolute value to get our 'V' and then determining which side to shade, we'll start graphing y < 2 by graphing where y=2 and then deciding which side of that line to shade. Graphing y= lines on a coordinate plane leaves us with horizontal lines, so y=2 looks like this. We need to make it a dotted line to indicate that it is strictly 'less than,' not 'equal to,' and then shade below it to indicate that y=0 works when I substitute it into the inequality (0 < 2). Now we've got a graph with a ton of shading everywhere. Because the solution to a system of inequalities is only where the shaded regions overlap, when we lay the y < 2 graph on top of our absolute value 'V', we find that our solution is only the triangular region in the middle of the graph, inside the 'V' but below the horizontal line.
To review, when splitting an absolute value inequality into two new ones to undo an absolute value, you must flip the inequality symbol on the one with the negative. Operations outside an absolute value must be undone before undoing the absolute value itself. Finally, systems of inequalities can be done with absolute values just like other lines, one graph at a time, where the solution is only the area where the shading overlaps.
When you complete this lesson you'll be able to split and unsplit absolute value inequalities and give systems of inequalities for absolute values.
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Back To CourseMath 101: College Algebra
13 chapters | 102 lessons | 11 flashcard sets