Negative Numbers and Inequalities
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.
System of Inequalities
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.
Graphing a system of inequalities
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.
Lesson Summary
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.
Lesson Objectives
When you complete this lesson you'll be able to split and unsplit absolute value inequalities and give systems of inequalities for absolute values.