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3.15: Solve Inequalities Using Division

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“I think I need new band shoes,” Kayla said to her Mom at breakfast one morning.

“Why is that? Have your feet grown?” her Mom asked taking a sip of coffee.

“I’m not sure, but I know that I have a blister on my toe from them.”

“Alright, must be time for new shoes. You don’t have band practice today, so let’s go after school and get new shoes. That will give you time to break them in before Friday’s game.”

After school, Kayla’s Mom picked her up and they went to the shoe store. Kayla first picked out a very fancy pair of shoes, but they weren’t on sale. Her Mom picked out a nice pair that would work fine, but were a little more economically priced.

“These are half-off. Let’s get them. They will work fine,” her Mom said.

Kayla agreed and they purchased the shoes. Kayla saw her Mom hand over $40.00 and get some change back.

The shoes that Kayla and her Mom bought were half off the original price. They spend less than $40.00 on the shoes. Can you figure out three possible original prices for Kayla’s shoes?

Figuring this out will require you to use inequalities with multiplication and division. Take your time while working on this. At the end of the Concept, you will see this problem again.

Guidance

An inequality is a mathematical statement where a quantity may be equal to or less than or greater than another quantity. You can identify an inequality and solve inequalities using all four operations. Now we are going to look at solving inequalities that involve multiplication.

Let’s look at this more closely.

Just as you sometimes need to multiply or divide both sides of an equation by the same number in order to solve an equation, you may sometimes need to multiply or divide both sides of an inequality by the same number in order to solve it. However, doing so may be a little more complicated.

Do you know how to divide and solve inequalities?

One of the first things you may be wondering is what happens if you divide both sides of an inequality by the same number.

Remember, dividing by a number, c , is the same as multiplying by its reciprocal, \frac{1}{c} . So, the multiplication property of inequality applies to division as well. We can also state this as its own property.

The division property of inequality states that if each side of an inequality is divided by the same positive number, the sense of the inequality stays the same. In other words, the inequality symbol does not change.

If a > b and c > 0 , then \frac{a}{c} > \frac{b}{c} . If a \ge b and c > 0 , then \frac{a}{c} \ge \frac{b}{c} .

If a < b and c > 0 , then \frac{a}{c} < \frac{b}{c} . If a \le b and c > 0 , then \frac{a}{c} \le \frac{b}{c} .

However, if each side of an inequality is divided by the same negative number, the sense of the inequality changes and the inequality symbol must be reversed.

If a > b and d < 0 , then \frac{a}{d} < \frac{b}{d} . If a \ge b and d < 0 , then \frac{a}{d} \le \frac{b}{d} .

If a < b and d < 0 , then \frac{a}{d} > \frac{b}{d} . If a \le b and d < 0 , then \frac{a}{d} \ge \frac{b}{d} .

Remember, that to solve an inequality by using division, you will need to see multiplication in the problem. This is because the inverse of multiplication is division and we use inverse operations to solve inequalities.

Solve this inequality: -3n \ge 12 .

Isolate the variable by using inverse operations. Since -3 is multiplied by n , divide both sides of the inequality by -3 to solve it. Since this involves dividing both sides of the inequality by a negative number, the sense of the inequality will change and you will need to reverse the inequality symbol. This means changing the inequality symbol from a “greater than or equal to” symbol (\ge) to a “less than or equal to” symbol (\le) .

-3n & \ge 12\\\frac{-3n}{-3} & \le \frac{12}{-3}\\1n & \le -4\\n & \le -4

The answer is the n is less than or equal to negative four.

Identify the inequality symbol that goes in each blank.

Example A

s &> t\\s \div (-5) & \ \underline{\;\;\;\;\;} \ t \div (-5)

Solution:  >

Example B

p & \ge q\\p \div 0.5 & \ \underline{\;\;\;\;\;} \ q \div 0.5

Solution: \ge

Example C

-2 & < 2\\\frac{-2}{-1} & \ \underline{\;\;\;\;\;} \ \frac{2}{-1}

Solution: >

Now let's go back to the dilemma from the beginning of the Concept.

First, write an inequality to describe the values in the problem.

x is the unknown original price of the shoes. This is what we are looking to figure out.

\div 2 is the sale price since the shoes were half off.

 < less than $40.00

Since Kayla’s Mom gave the clerk $40.00 and received change, we know that the shoes were less than $40.00.

Here is our inequality.

\frac{x}{2} < \$ 40.00

Now we can solve it for possible prices.

2 \cdot \frac{x}{2} & < \$ 40.00 \cdot 2\\x &< \$ 80.00

Three possible original prices could be $75.00, $60.00 or $55.00. But there are many possible options.

Vocabulary

Inequality
a mathematical statement where two quantities may or may not be equal. There are many possible answers to an inequality.
Multiplication Property of Inequality
multiplication of a positive number = inequality is the same. Multiplication of a negative number, the sense of the inequality changes and the sign is reversed.
Division Property of Inequality
Division by a positive number = inequality is the same. Division by a negative number, the sense of the inequality changes and the sign is reversed.

Guided Practice

Here is one for you to try on your own.

Maddie is buying bottles of water for a school event. The store only sells bottles of water in six-packs. That is, bottles of water are only sold in packages of 6 bottles each. Maddie wants to make sure there are at least 72 bottles of water available at the event.

a. Write an inequality to represent, p , the number of six-packs of water she could buy.

b. If Maddie buys 11 six-packs of water for the event, will that be enough? Explain.

Solution

Consider part a first.

Use a number, an operation sign, a variable, or an inequality symbol to represent each part of the problem. Since bottles of water are sold in packages of 6, or six-packs, you can find the total number of bottles purchased by multiplying 6 by the number of packages purchased. The key words “at least” indicate that you should use a \ge symbol.

& \ldots \underline{sells \ bottles \ of \ water \ in \ six-packs} \ldots make \ sure \ there \ are \ \underline{at \ least} \ \underline{72 \ bottles \ of \ water} \ldots\\& \qquad \qquad \qquad \qquad \quad \downarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \downarrow \qquad \qquad \quad \quad \downarrow\\& \qquad \qquad \quad \qquad \quad \ 6 \times p \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \ge \qquad \qquad \quad \ 72

The equation 6 \times p \ge 72 or 6p \ge 72 represents this problem.

You may also want to consider that the value of p must be an integer greater than or equal to 0. Think about why that is for a moment.

The reason that the value of p must be greater than zero is because Maddie cannot buy a negative number of six-packs of water. The value of p must be an integer because the problem also states that the store only sells bottled water in six-packs. That means it is not possible to buy a fraction of a six-pack at that particular store. When using inequalities to represent real-life situations, you should always think about which values would make sense for the variable and which values would not make sense. Sometimes a negative value or a fractional value make sense. Other times, they do not.

Next, consider part b .

Solve the inequality to help you.

6p & \ge 72\\\frac{6p}{6} & \ge \frac{72}{6}\\1p & \ge 12\\p & \ge 12

Now, let's consider if Maddie could buy 11 six-packs of water and have enough for the event. According to the inequality above, the number of six-packs, p , that she should buy must be greater than or equal to 12. Since 11 is less than 12, it is not a solution for this inequality.

That means that if Maddie bought only 11 six-packs of bottled water, she would not have enough water for the event.

Video Review

Khan Academy Inequalities Using Multiplication and Division

Practice

Directions: Solve each inequality using division.

  1. 3x<12
  2. 5x>60
  3. 6x<72
  4. 9x>81
  5. 11x>121
  6. 12x \le 48
  7. 13x \ge 39
  8. 15x \le 60
  9. 22x>66
  10. 48x \le 192

Directions: Solve each inequality.

  1. 4x > -36
  2. -9n \ge 63
  3. 7n \ge 63
  4. -19x \le 38
  5. -9y < 18

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Date Created:

Dec 19, 2012

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Dec 19, 2014
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