The writers of the Common Core Standards for Mathematics have provided us with several Progressions Documents that track the progression of specific topics across several grade levels.

In the Progressions Document for 6-8 Expressions and Equations, we can read about a little understood process in middle school math, that is, changing the direction of the inequality when multiplying or dividing by a negative number. Although the rule is easy to memorize, students seldom have a real understanding of why they need to perform this step in solving an inequality. Here is what the Progressions Document has to say about it (bolding added):

Students also recognize one important new consideration in solving inequalities: multiplying or dividing both sides of an inequality by a negative number reverses the order of the comparison it represents. It is useful to present contexts that allows students to make sense of this.

For example:

If the price of a ticket to a school concert is p dollars then the attendance is 1000 – 50p. What range of prices ensures that at least 600 people attend?
Students recognize that the requirement of at least 600 people leads to the inequality 1000 – 50p > 600. Before solving the inequality, they use common sense to anticipate that that answer will be of the form p < ?, since higher prices result in lower attendance.

Another example: Suppose a teacher has been awarded \$250 for classroom supplies. He would like to buy some tape measures for the class to use in measurement projects, but needs to leave at least \$95 to buy some books. If the tape measures cost \$1.75 each, how many tape measures can he buy?

Let t = the number of tape measures he will buy.

Reasoning as above, the more tape measures he buys, the less money he will have left. So it makes sense that the answer will be of the form t < a number.

250 – 1.75t is the amount left after buying t tape measures, and the teacher needs this quantity to be at least 95. So we can write the inequality

250 – 1.75t > 95

Subtracting 250 from both sides, we get

-1.75t > -155

Dividing both sides by -1.75 gives us

t <  88.57

Since t needs to be a whole number in this problem, the number of tape measures needs to be 88 or less. We can check this solution by calculating how much is left after purchasing, for example, 85 tape measures (\$101.25 remaining) , and 90 tape measures (\$92.50 remaining).  Since t = 85 makes the original inequality true, it is a solution, and our inequality has been solved correctly.

Here is a another way to help students develop understanding of this rule:

Consider the solutions to the inequality x< 8.

If we add to both sides (like we can for an equation) will it still be true?

x + 2 < 10

By checking different values for x, we see that the solutions still have to be less than eight.

Can we subtract from both sides and still have the same solutions?

x – 2 < 6

Again, by checking different values of x, we see that the solutions still have to be less than eight.

Does it work for multiplication and division?

2x < 16,  x/2 < 4

Checking different values, we see that x still has to be less than 8 in order for these inequalities to be true (and recall that solutions are the values that make an equation or inequality true).

Does it work when we multiply or divide by a negative number? Multiply both sides of the original inequality by -1:

-x < -8

Let’s try numbers less than 8. If x = 1, we have -1 < -8, which is not true. Checking other values shows that numbers less than 8 do not make this inequality true, and therefore are not solutions. But if the solutions are not the same, then this new inequality (-x < -8) is not equivalent to our original inequality (x < 8) because equivalent equations/inequalities have the same solutions (since we have moved from one form to another with steps that are justified by the equality properties, such as adding the same number to both sides – this concept is developed in the 6th and 7th grade standards of Common Core Math).

Is there anything we can do to the new inequality to make it equivalent to our original?

Since values of x less than 8 were not solutions, let’s check values greater than eight in the new inequality. If x = 9, we have -9 < -8, which is true. Checking values greater than eight shows that all will make this inequality true. We can see that by multiplying by a negative we have flipped to the other side of the number line. While 9 > 8, -9 < -8 since moving to the left on the number line brings us to smaller and smaller numbers (even though they are greater in absolute value).

This is not a proof of why the rule works, but serves as a demonstration that helps students understand why we reverse the direction of the inequality when multiplying/dividing by a negative number.

Students should also be guided in how to check their solutions to ensure that they have chosen the correct inequality symbol or that they’ve reversed the direction appropriately. They should choose values from both their solution and those numbers that are not part of the solution, and substitute those values for the variable in the original inequality to see if the statement is true. For example, in solving -3x + 5 < 7, the solution is x > -2/3. Choose x = -2/3 and numbers on both sides of  -2/3 that will be easy to check, for example x = 0 and x = -1, and substitute them for x in the original inequality.

x = -2/3 :   -3(-2/3) + 5 < 7,    2 + 5 < 7,   7 < 7  this is true, so x= -2/3 is a solution

x = 0:      -3(0) + 5 < 7,    0 + 5 < 7,   5 < 7  this is true, so x=0 is a solution

x = -1:     -3(-1) + 5 < 7,  3 + 5 < 7,   8 < 7  this is not true, so x= -1 is not a solution

Since 0, which is greater than -2/3, is a solution, and -1, which is less than -2/3, is not a solution, then our solution that x must be greater than or equal to -2/3 is correct.