17. Traditional Math: Solving Inequalities with Multiplication and Division—Negative Factors
Seventh Grade
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I usually reserve this topic as a separate one rather than including it in the general “solving inequalities” lesson. Although it is not a difficult concept or procedure, it is sufficiently different from the rest of the general procedure that including it in one lesson usually becomes a last-minute throw-in-the-kitchen-sink type of thing.
Having said that, depending on the ability level of students in particular classrooms, it may make sense to include the topic. If they breeze through the first part with ease, then it probably will do no harm to teach this topic.
In this lesson, examples include all the material covered in the previous lesson with emphasis on multiplying and dividing by negative numbers. Thus, word problems will continue. For that matter, I include word problems all the way through the course, as well as equations and inequalities. Mastering the basic operations of algebra makes it easier for students when they take algebra. Having worked basic equation solving and other aspects of algebra, students see algebra as a continuation of what they started, rather than as a totally foreign subject.
Warm-Ups. In addition to the usual review questions, I like to introduce problems with averages that will become more complex over time. Ultimately they will work up to problems like: “Hal has an average of 90 in math class over the last four tests. What does he need to get on the next test to have an average of 92?” For now, we start with the general concept that if the average of two numbers is x, then the total of the two numbers is 2x, and in general, the total of n numbers whose average is x is nx.
1. The average weight of Steven and Allison is 75 pounds. What is their total weight? Answer: 75 ∙ 2 = 150
2. Solve. 2x – 60 ≤ 90 Answer: 2x ≤ 150; x≤ 75
3. Simplify. -2(x-5) Answer: -2x + 10
4. Solve -2(x-5) = 20 Answer: -2x+10 = 20; -2x = 10; x = -5
5. Solve 2x +10 ≥ 20 Answer: 2x ≥ 10; x ≥ 5
Problem 1 is difficult for students the first time they see it. Prompts that I use to help them are “An average of two weights is the total of two weights divided by two. How do we undo the division to get the total?”
Problem 3 helps set the stage for Problem 4, since the answer to Problem 3 is used in Problem 4.
Multiplying and Dividing by a Negative Number. In Problem 5 of the warm-up, students have solved it in the usual manner. To start the lesson then, I’ll ask them to solve Problem 4 restated as: -2(x-5) ≤ 20. They have already solved it as an equality, so they are likely to solve -2x + 10 ≤ 20, in the same manner that they have before. Namely, once it is in the form -2x ≤ 10 they will divide both sides by -2 and claim that x ≤ -5. That means that if x =-6 the inequality should be true. But plugging in -6 we obtain -2(-6) = 12 which is clearly not less than 10. “What went wrong?”
I’ll then write -3 < 4 and 7 < 9 on the board and ask two tudents to help me graph the two pairs of points:
“What happens when we multiply each side by -1? In order for the new numbers to make sense what inequality sign should we use?”
It becomes obvious that the sign is reversed. I then have two more students plot the new points on a number line.
The graph also makes it obvious what happens; the new numbers obtained by multiplying by -1 are in effect a mirror image of the original points and are reversed.
Demonstrating this for division follows: For 6 > 2, dividing by -2 results in -3 < -1.
Stated simply, multiplying or dividing both sides of an inequality by a negative number reverses the inequality sign. Stated formally it is:
If a<b and c < 0 (c is negative), then a∙c > b∙c and a/c > b/c.
If a>b and c < 0 (c is negative), then a∙c < b∙c and a/c < b/c.
Examples. For these examples I have students write the answers in their notebooks or on mini-whiteboards.
Multiply both sides by -2: 5 < 8 Answer -10> -16
Multiply by -5: -3 > -4; Answer: 15 < 20
Solve for x: -3x > 15 Answer: Dividing both sides by -3 we have x < -5
Solve for y: -(1/2) y ≤ 5; Answer: Multiply by reciprocal of -1/2; y ≥ 10
Solve for x: 6x < -12.
For this last, most students will assume that the inequality sign reverses. I ask “Are we dividing by a negative number?” No, we are not. Therefore, by the rule, the inequality sign remains the same.
Others: 6x -3 > 9 Answer: 6x > 12; x> 2
-7x +2 < -12; Answer: -7x < -14; x > 2
It doesn’t take a lot of examples for students to catch on, but I make sure to mix up problems of multiplication or division by a negative number as well as a positive number to ensure they make the distinction. Some word problems should be included as well:
A cab ride costs $3 per mile. If a ride costs no more than $21, what is the range of miles the ride can be?
Answer: 3x ≤ 21; x ≤ 7. The answer indicates that the range can be up to 7 miles. That is, .a cab ride can be 7 miles or less for the price to stay under $21. Students may be confused by “range”. It is important that students understand that inequalities have many answers making up the solution. This solution comprises a range of numbers. (In previous eras it used to be called a “set” of numbers.)
A number multiplied by -4 is greater than 12. Find all possible values of the number.
Answer: -4x >12; x >-3. All possible values then are all numbers that are greater than -3. Graphing this on a number line helps students understand that all numbers designated by the arrow to the right of -3 (but not including -3) satisfy the inequality.
End of Unit. This section completes the unit on equations and inequalities. Future units will cover more complex equations, and solving equations and inequalities will be interspersed in assignments throughout the year.
Typically, the topic of equations and inequalities require much practice. A homework assignment may sometimes contain 20 problems. I sometimes had contests when time permitted in which two students would stand at the board, facing the classroom. I would write an equation or inequality on the board and when I said “Go” they would turn around and solve the problem. Work had to be shown, and had to be legible; the first one to get the correct answer stayed at the front and a new challenger would come up.
The problems would get more difficult as the competition continued. After about 10 problems, the person remaining at the front would be the winner.