Have you ever been to an amusement park?

Mrs. Hawk’s sixth grade class finally decided to go to the amusement park. While it will be a lot of fun, the amusement park also incorporates math and science activities into their park, so there will be some educational components to the trip as well. The students are really excited! Not only are they almost seventh graders, but they will get to finish a terrific year with a fun trip.

The amusement park is about two hours away, so Carl offers the class an idea to take a big greyhound bus instead of the typical school bus. The problem is that it costs a lot more to take a greyhound bus than the typical school bus. The students have chosen to attend an amusement park that is more expensive than they originally anticipated.

“How much will that cost?” Sarah asked after Carl presented the idea.

“I don’t know, but it would be a lot more comfortable. The amusement park ticket is $14.50 per person. We could add the cost of the bus to that,” Carl suggested.

“That could get pretty expensive. I would like the total expense for each student to be $20.00,” Mrs. Hawk chimed in. “Why don’t you investigate costs and get back to us?”

Carl agrees to do this. On his paper he makes a few notes.

$14.50 amusement park ticket

Bus cost unknown

Bus cost per person unknown

Total cost per person $20.00

**Carl is puzzled on how to tackle this problem from here. He will need to write an expression and an equation to figure this out. In this Concept, you will begin learning all about expressions and equations and how useful they can be in real-life situations. Pay close attention and you will be able to help Carl at the end of this Concept.**

### Guidance

The last two Concepts focused on writing expressions. Remember that an expression contains some combination of numbers, variables and operations, but does not have an equals sign. When you have an equals sign, you have an ** equation** not an expression.

**An equation has an equal sign. One side of the equation equals the other side of the equation.**

5 + 9 = 14

Here five plus nine is equal to fourteen. The quantity on one side of the equal sign is the same as the quantity on the other side of the equal sign. You have been solving equations for a long time.

**What about equations with a variable in them?**

You can also have equations with variables in them. **When you have a variable in an equation, there is an unknown quantity**. With an expression, there was not an equal sign. With an equation, one side will equal the other side.

Five plus an unknown number is equal to fifteen.

**To write an equation for this phrase, we start by working our way through the problem from the left to the right.**

The first part is 5

"Plus" means addition

"An unknown number" is the variable

"Is equal to" is our equal sign

"Fifteen" is 15

**Let’s write it out.**

\begin{align*}5 + x = 15\end{align*}

Yes you can. You have to pay attention to the key words, but once you have the key words, then you can write a single-variable equation.

Six less than a number is equal to ten.

The first number is 6.

"Less than" means subtraction, but be careful. Since this is “six less than” the order is reversed.

"A number" is the variable

“Is” means equals

"Ten" is 10.

Now let’s put it all together.

\begin{align*}x - 6 = 10\end{align*}

Here is one that uses multiplication.

The product of three and a number is thirty.

"Product" means to multiply

"Three" is 3

"A number" is our variable.

“Is” means equals

Thirty is 30

Put it altogether.

\begin{align*}3y=30\end{align*}

**As long as you walk through each written phrase carefully you will be able to write equations to match. Stay tuned, in another Concept you will learn how to solve equations!**

Practice writing single-variable equations for each phrase.

#### Example A

Fifteen divided by an unknown number is three

**Solution: \begin{align*}\frac{15}{x} = 3\end{align*}**

#### Example B

Six times an unknown number is thirty-six

**Solution: \begin{align*}6y = 36\end{align*}**

#### Example C

Fifteen and twelve is an unknown number.

**Solution: \begin{align*}15 + 12 = x\end{align*}**

Now back to the dilemma of the school trip. Let’s look at the original problem once again.

Mrs. Hawk’s sixth grade class is going on a final class trip to an amusement park. While it will be a lot of fun, the amusement park also incorporates math and science activities into their park, so there will be some educational components to the trip as well. The students are really excited! Not only are they almost seventh graders, but they will get to finish a terrific year with a fun trip.

The amusement park is about two hours away, so Carl offers the class an idea to take a big greyhound bus instead of the typical school bus. The problem is that it costs a lot more to take a greyhound bus than the typical school bus.

“How much will that cost?” Sarah asked after Carl presented the idea.

“I don’t know, but it would be a lot more comfortable. The amusement park ticket is $14.50 per person. We could add the cost of the bus to that,” Carl suggested.

“That could get pretty expensive. I would like the total expense for each student to be $20.00,” Mrs. Hawk chimed in. “Why don’t you investigate costs and get back to us?”

Carl agrees to do this. On his paper he makes a few notes.

$14.50 amusement park ticket

Bus cost unknown

Bus cost per person unknown

Total cost per person $20.00

**First, Carl needs to write an expression to represent the situation. He can use the cost of the amusement park ticket plus the unknown bus cost per person. Because the bus cost per person is unknown, Carl will first need to figure out the total cost of the bus divided by the number of people in his class. There are 26 students in Carl’s class.**

\begin{align*}x=\end{align*} *total cost of bus*

26 students in class

\begin{align*}\frac{x}{26} =\end{align*} ** the cost per person for the bus** \begin{align*}= y\end{align*}

Next, Carl can take the cost per person for the bus, \begin{align*}y\end{align*}, and add that to the price of the amusement park ticket.

\begin{align*}\$14.50 + y\end{align*}

Carl’s teacher has said that she wants the total to be $20.00 per person. Now Carl has enough information to write an equation.

\begin{align*}\$14.50 + y=\$20.00\end{align*}

**Now that Carl has written a couple of equations, he can complete some research on bus costs and figure out the cost of the trip for each person in his class. Stay tuned, Carl will need to learn how to solve equations to accomplish his task. Solving equations is coming up next!!**

### Vocabulary

Here are the vocabulary words in this Concept.

- Expression
- a variable expression has variables or unknown quantities, numbers and operations without an equal sign.

- Equation
- a variable equation has a variable, numbers and operations with an equal sign.

### Guided Practice

Here is one for you to try on your own.

Write an equation for: Six times a number divided by two is equal to four.

**Answer**

There are a couple of different operations here, but we can figure them out.

Six times a number becomes \begin{align*}6x\end{align*}

Next we add the "divided by two" \begin{align*}\frac{6x}{2}\end{align*}

Now we finish the equation.

\begin{align*}\frac{6x}{2} = 4\end{align*}

**This is our answer.**

### Video Review

Here is a video for review.

Khan Academy: Simple Equations

### Practice

Directions: Write each phrase as a single-variable equation.

1. Five less than a number is fifteen.

2. The sum of a number and six is eighteen.

3. Twenty divided by a number is four.

4. Sixteen less than a number is four

5. Twelve and a number is twenty.

6. The product of six and a number is forty-two.

7. Eight times a number is forty.

8. Ten less than a number is twenty-one.

9. A number divided by two is seven.

10. A number times four is forty-eight.

11. An unknown divided by two is fourteen.

12. Twelve times an unknown number is sixty.

13. Fourteen divided by an unknown number is seven.

14. Five and a number is equal to fifty - three.

15. Ten less than a number is seventeen.