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12.3: Solving Equations Using Multiplication and Division

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Introduction

Finding a Bus

Carl has been working hard trying to find a coach bus that his sixth grade class can afford. It is more challenging than he expected, so after some frustration he asks his friend Tabitha to help him.

“Here is what I know. The total cost of the bus can’t exceed $143.00 or we can’t afford it. I know this because $143.00 divided by 26 is the cost of the bus per person. Here is the equation I wrote,” Carl explains.

\frac{143}{26}=y is the cost per person

y=\$5.50

“The cost per person can’t exceed $5.50 or the cost of the trip becomes too expensive. I have called three different places and the cheapest bus is $1300,” Carl sighed, looking at Tabitha for any new ideas.

“Wow, that is expensive,” Tabitha said.

“I guess its back to the school bus,” Carl frowned.

At that moment Mrs. Hawk came in with a piece of paper in her hand.

“Carl, Buses are Us just called and they will offer us a deal on a bus. It will be only $998.00,” Mrs. Hawk said.

“I think that is still too much money. Let me see how much that is per person.”

Carl wrote this equation down.

\frac{988}{26}=y

“We have some money in the class account,” Mrs. Hawk said to Carl. “We can take $600 and apply it to the cost of the bus.”

Carl took a pencil to do some quick figuring.

988 – 600 = 388

Now that Carl has this new amount, he is back to the drawing board.

Can you write and solve this new equation with Carl? This lesson is all about solving multiplication and division single-variable equations. Once you have done this work, Carl and Tabitha will be able to figure out the cost of the bus per person.

What You Will Learn

In this lesson you will learn how to complete the following skills.

  • Simplify products and quotients of single-variable expressions.
  • Solve single-variable multiplication equations.
  • Solve single-variable division equations.
  • Model real-world problems using multiplication or division equations.

Teaching Time

I. Simplify Products and Quotients of Single-Variable Expressions

In our last lesson you learned how to simplify expressions. You simplified expressions that had addition and subtraction in them. Now it is time to simplify expressions with multiplication and division in them. Let’s start by reviewing what we mean by the word “simplify.”

To simplify means to make smaller. When simplifying an expression, we don’t evaluate the expression, we just simplify it. To evaluate an expression we must have a given value for the variable. If you have not been given a value for the variable, then you will be simplifying the expression.

Let’s look at an example.

Example

3(5x)

This is a multiplication problem. The number next to the parentheses means that we are going to multiply. However, we haven’t been given a value for x so we are going to simplify this expression. We can’t do anything with the variable, but we can multiply the numerical part of each term.

3 \times 5 = 15

Our answer is 15x.

Let’s look at another one.

Example

4 \cdot (12 \cdot y)

These dots mean to multiply. First, we complete the operation in parentheses.

12 \ \text{times} \ y = 12y

Next we multiply the number parts of the two terms.

4 \times 12 = 48

Our answer is 48y.

Practice a few of these on your own. Simplify each product.

  1. 5 \cdot (4 \cdot x)
  2. 5(6y)
  3. 4(3a)

Check your answers with a friend.

We can also simplify quotients. Remember that a quotient means division.

Example

\frac{4x}{2x}

That is a great question! Do you remember back to fractions? You know that a fraction over itself is equal to one.

\frac{6}{6} = 1

We call that simplifying a fraction. Well, we can simplify the variables too when we divide.

That is the first step. When simplifying a quotient with a variable expression, simplify the variables first.

Example

\frac{4x}{2x} = \frac{4}{2}

x divided by x is equal to one. Since 4 x 1 = 4, and 2 x 1 = 2, those x's really cancel each other out. Then we are left with four divided by two.

4 \div 2 = 2

The answer is 2.

Let’s look at another one.

Example

\frac{8xy}{4x}

Here we have like x’s in both the numerator and denominator. We can simplify those to one and cancel them out. Here is what we are left with.

\frac{8\bcancel{x}y}{4\bcancel{x}} = \frac{8y}{4}

Now we can’t do anything with the y, so we leave it alone.

We can divide four into eight.

8 \div 4 = 2

Don’t forget to add the y in.

Our answer is 2y.

Practice a few of these on your own.

  1. \frac{6y}{3y}
  2. \frac{14xy}{7y}
  3. \frac{25y}{5}

Carefully check your work with a friend. Talk through any differences and correct any errors.

II. Solve Single-Variable Multiplication Equations

Just like you learned how to solve single-variable equations with addition and subtraction, this lesson will teach you how to solve single-variable multiplication and division equations.

Let’s start with solving multiplication equations.

Example

5x=30

Here we need to figure out what the value of x is. We can do this in two ways.

  1. Use mental math
  2. Use the inverse operation

To use mental math we can think to ourselves, “What times five is equal to thirty?”

Using our times tables, we can figure out that 5 times 6 is equal to thirty. The value of x is 6.

To use the inverse operation, we use the opposite operation of multiplication, since this is a multiplication problem. The inverse of multiplication is division.

Once again, we work to get the variable alone on one side of the equation. This time by dividing both sides by the number next to the variable. In this example, we divide both sides by 5.

\frac{5x}{5}= \frac{30}{5}

The fives cancel each other out because five divided by five is one, and "x" times 1 is "x". On the right side, thirty divided by five is six.

\frac{\bcancel{5}x}{\bcancel{5}}&= \frac{30}{5}\\x&=6

You can check your work by substituting the value of x back into the original equation. If both sides are equal, then your work is accurate and correct.

5(6)&=30\\30&=30

Our work is correct.

Let’s look at another example.

Example

7y=49

To do this one, let’s use the inverse operation. We divide both sides by 7 to get the variable alone.

\frac{7y}{7}=\frac{49}{7}

The 7’s cancel each other out, leaving y alone. Forty-nine divided by seven is seven.

\frac{\bcancel{7}y}{\bcancel{7}}&=\frac{49}{7}\\ y&=7

Check your work. Substitute 7 back into the original problem for y.

7(7)&=49\\49&=49

Our work is accurate.

Practice solving a few equations on your own. Write your answer is the form variable = _____.

  1. 8x=64
  2. 2a=26
  3. 6y=42

Check your answers with a friend.

III. Solve Single-Variable Division Equations

Division equations are a bit tricky because you have to multiply to solve them. We often think in terms of multiplication, but we don’t think in terms of division. When you have a division equation, you have to use multiplication to solve for the variable. Remember that multiplication is the inverse operation for division.

There are two different types of division equations that we will be solving. Let’s look at an example of the first type of division equation.

Example

\frac{x}{3}=12

This type of division problem has a missing numerator. We don’t know the value of the numerator so we use a variable in place of the unknown number.

To figure out the numerator, we multiply the denominator with the value on the right side of the equals.

To check the answer, substitute it back into the original equation for the variable. If one side equals the other side, then your work is accurate and correct.

\frac{36}{3}&=12\\12 & = 12

Our work is correct.

Now let’s look at an example of the second type of division equation.

Example

\frac{4}{x}=2

To solve this equation we need to multiply the denominator with the value on the right side of the equals. In this case, the denominator is a variable. We multiply it by two and rewrite the problem.

Now we have a multiplication problem to solve. We solve it by using division. To get the variable alone, we divide both sides of the equation by 2.

\frac{2x}{2}&=\frac{4}{2}\\x&=2

Is this accurate? Let’s substitute it back into the original problem to check.

\frac{4}{2} & = 2\\2 &= 2

Our work is accurate and correct.

Practice solving these equations. Write your answer in the form variable = _____.

  1. \frac{x}{5}=7
  2. \frac{x}{2} = 3
  3. \frac{12}{x}=6

Check your answers with a partner. Correct any errors before continuing.

Real Life Example Completed

Finding a Bus

Here is the original problem once again. Take a few minutes to reread the problem and underline any important information.

Carl has been working hard trying to find a coach bus that his sixth grade class can afford. It is more challenging than he expected, and after some frustration he asks his friend Tabitha to help him.

“Here is what I know. The total cost of the bus can’t exceed $143.00 or we can’t afford it. I know this because $143.00 divided by 26 is the cost of the bus per person. Here is the equation I wrote,” Carl explains.

\frac{143}{26}=y is the cost per person

y=\$5.50

“The cost per person can’t exceed $5.50 or the cost of the trip becomes too expensive. I have called three different places and the cheapest bus is $1300,” Carl sighed, looking at Tabitha for any new ideas.

“Wow, that is expensive,” Tabitha said.

“I guess its back to the school bus,” Carl frowned.

At that moment Mrs. Hawk came in with a piece of paper in her hand.

“Carl, Buses are Us just called and they will offer us a deal on a bus. It will be only $998.00,” Mrs. Hawk said.

“I think that is still too much money. Let me see how much that is per person.”

Carl wrote this equation down.

\frac{988}{26}=y

“We have some money in the class account,” Mrs. Hawk said to Carl. “We can take $600 and apply it to the cost of the bus.”

Carl took a pencil to do some quick figuring.

988 – 600 = 388

Now that Carl has this new amount, he is back to the drawing board.

Can you write and solve this new equation with Carl?

Mrs. Hawk has said that the students can apply class funds to the cost of the bus. Carl needs to write an equation and solve for the bus cost per person with these new figures.

The cost of the bus – class funds = new cost

988 – 600 = 388

There are 26 students in the class.

y is the cost of the bus per person

Let’s write the equation. This is a division problem.

\frac{388}{26} & = y\\y & = \$14.92 \ \text{per person}

Carl knows that his original figure was $5.50 per person for the bus. To find the difference he subtracts $5.50 from $14.92.

14.92 – 5.50 = $9.42

There is a difference of $9.42 that each person would have to pay to take the coach bus.

Carl and Tabitha show their work to Mrs. Hawk. Mrs. Hawk suggests that they ask the class to have a car wash to raise the additional funds. The students think this is a terrific idea!!

Vocabulary

Here are the vocabulary words that are found in this lesson.

Product
the answer to a multiplication problem
Quotient
the answer to a division problem
Inverse Operation
the opposite operation

Technology Integration

Khan Academy, Simple Equations

James Sousa, Solving One Step Equation by Multiplication and Division

Other Videos:

  1. http://www.mathplayground.com/howto_solvevariable.html – This is a video that teaches students how to solve single-variable equations using inverse operations.
  2. http://www.onlinemathlearning.com/solving-equations.html – This video goes through how to solve a single step equation. It does incorporate integers into the equations.

Time to Practice

Directions: Simplify the products and quotients of the following single-variable expressions.

1. 5x(4)

2. 3(5x)

3. 4y(2)

4. 2 \cdot (8 \cdot a)

5. 4 \cdot (7x)

6. \frac{9y}{3y}

7. \frac{64x}{8x}

8. \frac{12xy}{4}

9. \frac{10ab}{5a}

10. \frac{18xy}{6y}

Directions: Solve each single-variable multiplication equation.

11. 4x=16

12. 3y=12

13. 8a=72

14. 12v=36

15. 9x=45

16. 10y=100

17. 7x=21

18. 9a=99

19. 16x=32

20. 14y=28

Directions: Solve each single-variable division equation.

21. \frac{x}{4}=8

22. \frac{6}{x}=3

23. \frac{x}{9}=9

24. \frac{x}{5}=3

25. \frac{20}{y}=4

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CK.MAT.ENG.SE.1.Math-Grade-6.12.3

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