Students have learned how to solve inequalities in seventh grade (see posts 16 and 17). The lesson on solving inequalities in algebra is therefore a review, though it may contain more complex problems that have variables on both sides of the inequality, as well as containing distributions.
What is new to students is solving absolute value inequalities such as |x+2| < 5. Compound inequalities are new. These involve combining inequalities such as x<2 and x> -5 into one statement: -5 < x < 2. Students seem to have difficulty doing this at first. I’ve found that graphing helps with this concept, and both graphing and compound inequalities are part and parcel to absolute value inequalities which, like absolute value equations, have two solutions.
This lesson includes a discussion of the distinction between “and” and “or” inequalities. The former include inequalities such as x<2 and x > -5 in which there are numbers that satisfy both inequalities as seen by the resulting compound inequality -5 < x < 2. On the other hand, “or” inequalities are those in which each inequality does not have numbers in common; e.g., x > 2 and x < -5. In this case, x is either greater than 2 or less than -5, but not both.
I also address more complex inequalities such as -2 < x + 4 ≤ 5. These inequalities are simplified by subtracting 4 from each side resulting in -6 < x ≤ 1. Finally, students are to go the opposite direction and express -6 < x ≤ 1 as two inequalities; i.e., x > -6 and x ≤ 1. This amounts to writing one of the inequalities backwards which results in a reversal of the inequality sign. Knowing how to do this allows students to later write inequalities so that the variable is on the left hand side—a convention that results in ease of interpretation.
Warm-Ups.
1. Solve. |x – 15| = 15 Answer: x – 15 = 15;, x = 30, and x – 15 = -15, x = 0.
2. Solve. David is 17 pounds lighter than Paul. Their total weight is 259 pounds. Find David’s weight. Answer: Let x = Paul’s weight, then x – 17 = David’s weight. x + x – 17 = 259; 2x = 276; x = 138, x – 17 = 121. OR, Let x = David’s weight; then x + 17 = Paul’s weight. x + x + 17 = 259; 2x = 242; x = 121 and x + 17 = 138.
3. Solve. x -2 < 2x -3 Answer: x > 1
4. Write an equation and solve. Peter has $180 and Ryan has $150. How much money must Peter tive Ryan so that they each will have an equal amount of money? Answer: Let x = amount of money Peter pays and Ryan receives; then 180 – x = 150 + x; 2x = 30; x = $15
5. Simplify -4 <x + 5 < 16 Answer: -9 < x < 11.
Problem 3 is an extension of solving equations with variables on both sides. They might solve it by adding 2 to each side and subtracting 2x from each side resulting in –x < -1. Each side is then divided by -1 to obtain x > 1. Alternatively, students may add 3 to each side and subtract x from each side resulting in 1 < x. Reversing the inequality, it is x > 1.
Getting the Wrong Answer. I write on the board:
“Yesterday I said that if we make the right hand side negative for inequalities, it won’t work. Let’s look at this. Go ahead and solve making the right hand side negative.” They obtain x < 5 and x < -5.
“Let’s see if that’s right,” I’ll say. The first inequality seems to work with a few test numbers that are less than 5 plugged into the left hand side. But for the second inequality, I’ll ask for some numbers less than -5. Trying -6 results in |-6| on the left hand side, which equals 6. “Is 6 less than 5?”
“In this case, we have to use the formal definition of absolute value which, if you recall, if x < 0 then |x| = -x. So let’s see what happens if we make the left hand side negative.” I write:
“What’s next?” I’ll ask and will hear that I should divide both sides by -1. If I do not hear that—and that may be the case at this early stage in an algebra course—I remind them that this is the same as:
This usually serves to remind them what to do and the end result is x > -5.
“This makes more sense. So we have x <5 and x > -5 Let’s graph it.” (They know how to graph inequalities from a previous lesson.)
I’ll then nudge the class to tell me how we can write this as one inequality statement rather than two, which happens to be: -5 < x < 5.
“So this inequality represents all numbers that are between -5 and 5, but do not include either endpoint.”
“How about |n-15| < 3?”
They have worked similar problems with absolute value equations so this isn’t totally foreign. They will get n < 18 for the first solution and for the second solution –(n-15) < 3, n – 15 > -3, n >12.
Examples.
There will be some who get this immediately—there is no solution because the absolute value of any number cannot be negative.
A Shortcut and “Or” Situations. Now that they have done several of these problems, I will introduce a quicker way of solving these based on a suggestion from a student in one of my algebra classes. Looking at, |x -6| < 12, we see immediately that the first solution is x < 18. For the second solution my student said “Since we’re going to end up dividing by a negative, can’t we skip making the left hand side negative, and just make the right hand side negative and flip the inequality sign?”
I said we could and called this short-cut the “Eli Theorem” after the student who discovered it.
Now I introduce problems for which the solutions are not common to both inequalities. These are associated with inequalities in the form |x + c| > k in which c and k are constants. I write on the board:
The first solution is x ≥ 8 The second solution is x ≤ -2. “Let’s graph it,” I say. The graph clearly shows that the solutions are not common to any particular interval.
“In this situation we cannot write this as a single inequality because the numbers satisfying are either in one interval or the other. So this is an ‘or’ situation and we write it like this:”
To summarize this as a rule I say that when an inequality has a greater than sign, it’s an “or” situation. When the inequality has a less than sign, it’s an “and” situation.
I now announce a “challenge problem”. I’ll have them work this in their notebooks and as I give hints and prompts, I say them loud enough for the whole class to hear—a technique that has benefits for all.
First prompt: “Is it an ‘and’ or an ‘or’ inequality?”
Second prompt: “We have to isolate the absolute value term and its coefficient.”
They now should have:
This simplifies to 2x – 5 > 7 and x > 6 for the first solution. For the second solution, it is 2x -5 < -7, x < -1.
I have them graph it, so they can clearly see why it is an “or” situation.
Homework. Other examples follow, some of which are from the homework. The homework consists of a combination of making single inequality statements from two inequalities and solving absolute value inequalities.