Solving and Graphing Absolute Value Inequalities- Practice Problems

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.

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 1 or x -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.

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.

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.

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.