In this section, I address how to solve equations of the form |x+c| = n in which x is the variable and c and n are constants. Students have by now learned what absolute value is. They have learned the definition that |x| represents the distance from x to 0. Distances are always positive. An informal definition from which most students operate is that the absolute value of any number is positive regardless of sign.
In either Algebra 2 or pre-calculus, they may encounter another definition that is more formal:
For any real number x, the absolute value of x is denoted by |x| and is defined as:
Students get confused by this until they’ve had a bit more experience. Nevertheless, I provide a discussion of this definition and prime the pump in the Warm-Ups with a question on what -(-6) equals. The definition provides a background for why |x| = p is equivalent to x = p and –p. Using numbers as I describe in the lesson, the value of -6 if it is inside the absolute value brackets is its opposite—that is, -(-6) which is 6. In formal terms, using a variable instead of a number, if x is a negative number (like -6), then its absolute value is its opposite, or -x.
For equations, the same result is obtained whether the right hand side or left hand side is made negative for the second solution. For inequalities, which are discussed in the subsequent lesson, I show that making the right hand side negative results in an inequality that is false.
Having said this, I would assess the ability level of the class carefully. If it appears that including this discussion will confuse more than enlighten and possibly interfere with understanding the basic procedure, I would omit it. In my experience, most students in algebra 1—particularly in eighth grade—will not understand the definition. An indicator is if students keep saying “But how could the absolute value of x be -x? I thought the absolute value of any number is positive?”
Stronger students tend to understand more of it. Students will see it again in either algebra 2 or pre-calculus should they take that route in high school. In answer to the question of whether the formal definition will be on the test, I tell them that it will be an extra credit question—therefore no penalty if they can’t answer it.
My overall approach, whether the formal definition is included in the lesson or not, is on the procedure for solving absolute value equations, and subsequently absolute value inequalities. The procedure is what most will hang on to—even the stronger students whose understanding may be in the “kind of understand it” category. The procedural approach is entirely appropriate for students in eighth grade.
Warm-Ups.
1. Simplify. 2x²·x/4x Answer: x²/2 or ½( x²)
2. Solve for z. xz = t/3 Answer:z = t/3x
3. The width of a rectangle is 7 cm less than the length. If the perimeter is 410 cm, what are the dimensions of the rectangle? Answer: Let x = width, then x – 7 is the length. 2x + 2(x -7) = 410; 2x + 2x – 14 = 410; 4x =424;
x = 106 cm, x – 7 = 99 cm
4. Solve. 20x + 30 + 10x -20 = -20 Answer: 30x +10=-20; 30x = – 30; x = – 1
5. -(-6) = ? Answer: 6
Problem 1 is a review of exponent multiplication and division rules. Students may not remember what to do with the numbers. Problem 2 is a literal equation. To solve, both sides are divided by x which is the same as multiplying both sides by its reciprocal 1/x. Students may be confused about this so I go over this one carefully. Problem 3 is a rectangle perimeter problem that will be repeated throughout the year, with variations. In the spirit of formative assessment, I make a note of which students repeatedly forget the formula for the perimeter of a rectangle—they usually have more difficulty as the class progresses.
Problem 5 is asking what is the opposite number of -6? Students may be confused by this, so I remind them what an opposite number is by asking what is the opposite number of positive 6? It’s negative 6, so by the same logic, the opposite of negative 6 is positive 6. This topic is discussed as part of the lesson to follow.
Absolute Value. “What is the absolute value of – 5,280?” I ask. By this time they know what absolute value is, so I tell them to raise their hands and don’t shout out the answer. I pick someone at random, usually someone who is reluctant to participate and I will always hear the correct answer: 5,280[1].
Then I ask “If I have |x| = 5,280, what is x?”
The answers vary. The first answers I hear are usually positive 5,280 or negative 5,280 both of which I respond with “Correct!” I then elaborate. “The answer is both positive and negative 5,280, since the absolute value of a negative number is always positive.”
So far, they are following. Now I want to up the ante and present the formal definition of absolute value. Stronger students tend to follow this discussion more than others, but like many explanations of the conceptual understanding, students will glom on to the procedure(s) that follow in the general spirit that my daughter expressed when I helped with math when she was younger: “Just tell me what to do!”
“In the Warm-Ups, we talked about What is –(–6)? Remember, the opposite of 6 is –6. And the opposite of -6 is therefore 6. This is important in understanding the formal definition of absolute value. When you first learned about it, absolute value of a number was defined as the ‘distance of that number from zero on the number line’. That’s one definition but now I’m going to show you the algebraic definition.”
I have them write this down in their notebooks:
The nice part of having them write it in their notebooks is their too busy to evince expressions of despair, frustration, or confusion.
“So let’s see what this means. If I have |x| = 6 there are two answers that will work: 6 and –6. That’s because the absolute value of whatever is inside the brackets, whether negative or positive, is always positive.”
“The formal definition of absolute value says this.”
I always hope someone will say “It does?” so I can respond “Yes, and here’s why!” But that has never occurred.
“Let’s see why. If 6 is inside the brackets, we know 6 > 0, the definition tells us if x > 0, then |x| = x. Well, x is 6 in this case so |6| = 6. BUT…” (dramatic pause), “if –6 is inside the brackets, what does the definition say? Is –6 less than zero?”
General nods of agreement are seen and an occasional “Yeah” can be heard.
“So by the definition |-6| is –(whatever is in the brackets). So it’s –(–6), which we know equals positive 6.”
Solving Absolute Value Equations. This is the lead in to the method for solving absolute value equations. I write on the board:
“The first step is to remove the absolute value bracket, so we have x = 25. That’s one answer—the positive value.”
“The next step is to remove the bracket once more, but this time make the left hand side negative and solve. What does the equation look like now?” I have them show their white boards. I should see -x = 25.”
“Remember, -x = -1x, we just don’t write the -1. How do we solve the equation?” Most know to divide both sides by -1, resulting in x = 25.”
“Now, here’s some good news. For absolute value equations we can also make the right hand side negative and it will work the same way. Is that easier?”
There is general agreement and maybe even a few “Thank God” exclamation.
“For equations we can make the right hand or left hand sides negative. You probably like to use the right hand side because it eliminates a step, right? That’s fine, but as we’ll see tomorrow, for absolute value inequalities it won’t work. But let’s worry about that tomorrow. For today, either side will do.”
I then demonstrate the right hand method with the example.
“If we make the right hand side negative we have x = -25. We can see this is correct because plugging in either answer into the absolute value brackets gets me 25.”
I now write on the board, and have them write in their notebooks:
“The first step is remove the brackets. What do you get?”
They should get x-5 = 4. I am on the lookout, however, for students who write x + 5 = 4, because they think that any negative value inside absolute value brackets becomes positive. I quickly disabuse people of this popular misconception.
“Yes, if I had |-5|, removing the brackets would result in 5, you are correct. “But if I have, say |1 – 5|. What does this equal?”
The answer comes back “4”.
“Yes, 4 is correct. But 6 is not. |1 – 5| is not equal to 1 + 5. The –5 inside the bracket stays as is, because it is part of the numerical expression 1 – 5.”
“You can also see that |1 – 5| does not equal |1| - |5|, because that would equal –4. So you can’t split up numbers or variables within an expression inside the absolute value brackets. But now let’s move to the next step. We have x–5 = 4.We need to solve that, which is easy. So after you solve it, I want you to do the next step following what we did with the first example, and do it in your notebooks.”
I look to see that they have x = 9 for the first answer, and for the second part they have x – 5 = – 4, and x = 1.
“Any idea how to check if –4 and 1 are correct?” Generally they know by now to plug it back in the original equation and yes, |9–5| = |– 1– 4| = 5.
Other Examples.
1. |2x + 9| = 5 There are usually questions about this, but once brackets are removed for positive and negative case, both equations are solved like they are used to solving them. Answer: x = –2, x = –7.
2. 2|x – 5|= 10 Possible prompt is “Can you divide first?” to get them to divide both sides by 2 to obtain |x–5|=5. Answer: x = 10, x = 0
3. |x| = – 345 This has no answer. Students generally get angry that I’ve given them a problem with no answer, but they are also forgiving. The point is that such an equation is impossible; there is no solution, because absolute values are always positive.
Graphing Solutions. Graphing the solution to such equations amounts to placing a dot on a number line in the appropriate spot. It is not hard, and may seem superfluous but it is in anticipation of the next lesson which is on inequalities in general, and eventually absolute value inequalities.
“Let’s say I have the equation |x– 3| = 5. Solve it first and then we’ll graph the solutions. It’s very simple. You just plot the two solutions on the number line.”
I have them work in their notebooks. The solutions are x = 8, –2. The solutions on the number line are:
We try a few more examples and then I pose a challenge question. “We’ve seen that for absolute value equations we will get two solutions. Can you find an equation for which there is only one solution? Think about it and tell me when you find it or think you’ve found it.”
In the meantime, I get them started on homework. The answer, if you’re curious, is any equation for which the right hand side is zero.
Homework. Problems should be as shown here, with some requiring making the left hand side negative. Some should be graphed, but not all. I monitor their work as they start on homework and look for students who may solve only the first equation after removing the absolute value brackets.
[1] I pick this number for several reasons. One, it is a large negative number that is unusual and different than the usual ones asked about. Second, it is the number of feet in a mile, which increasing numbers of students do not know. I continually provide reminders of this and other measurement information they should know like number of ounces in a pound, pounds in a ton, feet in a yard and so forth. I wouldn’t mind this alarming diminishing of information as much as I do if we were on our way to converting to the metric system, but it doesn’t look like that is happening any time soon. And the fact that one can just Google the conversion does not provide me any comfort.
[2] Some textbooks such as Dolciani’s warn against having students say they are “distributing the negative”. I’ve considered this advice, and for me it is like the word “cancel”. I want them to know when they cancel they are either adding an additive inverse or dividing, but I don’t mind the word. Similarly I want students to know they are distributing –1 but I accept that they will say “distributing the negative” and I don’t lose sleep over it.