# 12.2: Solving Equations Using Addition and Subtraction

**At Grade**Created by: CK-12

## Introduction

*The Bus*

Carl is working on bus costs for his class trip to the amusement park. Since the amusement park is two hours away, Carl’s idea was for his class to take a greyhound bus instead of a school bus. Mrs. Hawk, Carl’s teacher, is alright with the idea as long as it doesn’t cost too much money.

Mrs. Hawk has said that she would like the cost per student to not exceed $20.00.

The amusement park ticket is $14.50.

The bus cost per person is unknown at this point.

Carl wrote the following equation to figure out bus cost per person.

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

Now Carl has two dilemmas. The first is that he needs to figure out how much the bus can cost per person without exceeding the twenty dollar limit. To figure this out, he will need to solve this equation.

Once he has done that, he can multiply the cost per person by 26, since 26 students are going on the trip and this will give him the maximum total cost that the bus can be.

Carl wishes to do all of this research before calling bus companies so that he can have an idea how much the class can afford to pay for a bus.

**To do this, Carl will need some help. First, Carl needs to solve the equation for bus cost per person. Then he needs to use multiplication to figure out the total possible cost for the bus. This lesson will teach you all about solving equations. Pay attention and you will be able to help Carl at the end of the lesson.**

*What You Will Learn*

By the end of this lesson, you will know how to execute the following skills:

- Simplify sums and differences of single-variable expressions
- Solve single-variable addition equations
- Solve single-variable subtraction equations
- Model and solve real-world problems using addition or subtraction equations

*Teaching Time*

I. **Simplify Sums and Differences of Single-Variable Expressions**

In the last lesson you learned how to write single-variable expressions and single variable equations. Now you are going to learn to work with single-variable expressions. The first thing that you are going to learn is how to ** simplify** an expression.

**What does it mean to simplify?**

To simplify means to make smaller or to make simpler. When we simplify in mathematics, we aren’t solving anything, we are just making it smaller.

**How do we simplify expressions?**

Sometimes, you will be given an expression using variables where there is more than one term. A term is a number with a variable. Here is an example of a term.

\begin{align*}4x\end{align*}

This is a term. It is a number and a variable. We haven’t been given a value for \begin{align*}x\end{align*}, so there isn’t anything else we can do with this term. It stays the same. If we have been given a value for \begin{align*}x\end{align*}, then we could evaluate the expression. You have already worked on evaluating expressions.

**When there is more than one LIKE TERM in an expression, we can simplify the expression.**

**What is a like term?**

**A like term means that the terms in question use the same variable.**

**\begin{align*}4x\end{align*} and \begin{align*}5x\end{align*} are like terms. They both have \begin{align*}x\end{align*} as the variable. They are alike.**

**\begin{align*}6x\end{align*} and \begin{align*}2y\end{align*} are not like terms. One has an \begin{align*}x\end{align*} and one has a \begin{align*}y\end{align*}. They are not alike.**

**We can simplify expressions with like terms. We can simplify the sums and differences of expressions with like terms. Let’s start with sums. Here is an example.**

Example

\begin{align*}5x+7x\end{align*}

**First, we look to see if these terms are alike. Both of them have an \begin{align*}x\end{align*}, so they are alike.**

**Next, we can simplify them by adding the numerical part of the terms together. The \begin{align*}x\end{align*} stays the same.**

\begin{align*}&5x+7x\\ &12x\end{align*}

You can think of the \begin{align*}x\end{align*} as a label that lets you know that the terms are alike.

Let’s look at another example.

Example

\begin{align*}7x+2x+5y\end{align*}

**First, we look to see if the terms are alike. Two of the terms have \begin{align*}x\end{align*}’s and one has a \begin{align*}y\end{align*}. The two with the \begin{align*}x\end{align*}’s are alike. The one with the \begin{align*}y\end{align*} is not alike. We can simplify the ones with the \begin{align*}x\end{align*}’s.**

**Next, we simplify the like terms.**

\begin{align*}7x+2x=9x \end{align*}

**We can’t simplify the \begin{align*}5y\end{align*} so it stays the same.**

\begin{align*}9x+5y\end{align*}

**This is our answer.**

**We can also simplify expressions with differences and like terms. Let’s look at an example.**

Example

\begin{align*}9y-2y\end{align*}

**First, you can see that these terms are alike because they both have \begin{align*}y\end{align*}’s. We simplify the expression by subtracting the numerical part of the terms.**

9 - 2 = 7

**Our answer is \begin{align*}7y\end{align*}.**

**Sometimes you can combine like terms that have both sums and differences in the same example.**

Example

\begin{align*}8x-3x+2y+4y\end{align*}

**We begin with the like terms.**

\begin{align*}8x-3x&=5x \\ 2y+4y&=6y\end{align*}

**Next, we put it all together.**

\begin{align*}5x+6y\end{align*}

**This is our answer.**

*Remember that you can only combine terms that are alike!!!*

*Use your notebook and pencil to take some notes on how to identify like terms.*

**Try a few of these on your own. Simplify the expressions by combining like terms.**

- \begin{align*}7z+2z+4z\end{align*}
- \begin{align*}25y-13y\end{align*}
- \begin{align*}7x+2x+4a\end{align*}

*Take a few minutes to check your work with a partner. Are your answers accurate?*

II. **Solve Single-Variable Addition Equations**

In the last lesson you learned how to write an addition equation from a phrase. To do this, you looked for key words that mean addition, words like sum and plus. Let’s look at an example.

Example

Five plus an unknown number is ten.

**Here is our equation.**

\begin{align*}5+x=10\end{align*}

Now that you have learned how to write single-variable equations, our next step is to learn how to solve them.

**What does it mean to solve a single-variable equation?**

To solve a single-variable equation means that you are going to figure out the value of the variable or the unknown number.

**We can do this in a couple of ways.**

The first way is to use mental math. Let’s look at the equation that we just wrote.

\begin{align*}5+x=10\end{align*}

Using mental math you can ask yourself, “Five plus what number is equal to 10?”

**The answer is 5.**

**You can check your answer too.** To do this, simply substitute the value for \begin{align*}x\end{align*} into the equation and see if it forms a true statement.

\begin{align*}5+5&=10\\ 10 &= 10\end{align*}

**This is a true statement. Our answer is correct.**

**Sometimes, it may seem difficult to figure out the value of the variable using mental math. What do we do then?** Let’s look at an example and work on the next way to solve single-variable equations.

Example

\begin{align*}x+27=43\end{align*}

**The second way of solving a single-variable equation involves “using the inverse operation.”**

**That is an excellent question! An** *inverse operation***is the opposite of the given operation.** In the example above, the given operation is addition, so we can use the opposite of addition (subtraction) to solve the problem.

**How do we do this?**

Let’s sort it out with the example.

\begin{align*}x+27=43\end{align*}

Since 27 is being added, we can subtract 27 from both sides of the equation. That will help us get the variable on one side of the equation. We need to get the variable by itself to figure out what the value of it is. Let’s subtract 27 from both sides and see what we end up with as an answer.

\begin{align*}& x+27 \ = \ 43\\ & \underline{ \;\;\; -27 \quad -27}\\ & \ x+0 \ = \ 16\\ & \qquad x \ = \ 16\end{align*}

**Where did the 0 come from?**

Well, if you look at the inverse, +27 – 27 is equal to 0. Once we have a \begin{align*}O\end{align*} next to variable, we have succeeded in getting the variable alone. Then all we have on the left side of the equal sign is the variable. On the right side of the equals, we subtract 27 and we end up with an answer of 16.

**Is this true?** We can check our work by substituting the value for \begin{align*}x\end{align*} back into the original equation. If it is true, then one side of the equation will equal the other side.

\begin{align*}16+27&=43\\ 43&=43\end{align*}

**Our answer is correct.**

Let’s look at another example.

Example

\begin{align*}45+x=67\end{align*}

**This one may seem a little trickier because the \begin{align*}x\end{align*} is in the middle of the equation and not at the start of it. We can still use an inverse to sort it out.**

Our goal is to get the \begin{align*}x\end{align*} alone. To do this, we need to do something with the 45. Notice that it is a positive 45. We can use the inverse of a positive 45 which is a negative 45 and subtract 45 from both sides of the equation.

\begin{align*}&\ \ 45 \ + \ x = \ 67\\ & \underline{-45 \qquad \quad -45}\\ & \qquad \quad \ x \ = \ 22\end{align*}

**Notice here again that 45 – 45 is equal to 0. On the left side of the equals we succeeded in getting the variable by itself. On right side, we subtracted 45 and got an answer of 22.**

**Is this true?** We can figure out if this is true by substituting our answer for \begin{align*}x\end{align*} into the original equation. If it is true, then one side of the equation will equal the other side of the equation.

\begin{align*}45+22&=67\\ 67&=67\end{align*}

**Our answer checks out.**

**Practice a few of these on your own. Solve each equation then check your answer. Write your answers in the form variable = _____.**

- \begin{align*}x+16=22\end{align*}
- \begin{align*}y+15=30\end{align*}
- \begin{align*}12+x=18\end{align*}

*Check your answers with a peer.*

III. **Solve Single-Variable Subtraction Equations**

In the last section, you learned how to solve single-variable addition equations. Now you are going to learn how to solve single-variable subtraction equations.

**How do we do this?**

We can use inverses once again to solve single-variable subtraction equations. The inverse of subtraction is addition, so we can use the inverse operation to help us in solving each problem.

Let’s look at an example.

Example

\begin{align*}x-12=40\end{align*}

**If you think this through, it means “Some number minus twelve is equal to 40.”**

**To figure this out, you can use the inverse of subtraction (addition) and add 12 to both sides of the equation. That will help to get the variable alone and solve the problem.**

\begin{align*}& x-12\ = \ \ 40\\ & \underline{\quad + 12 \quad \ +12}\\ & \qquad x \ \ = \ \ 52\end{align*}

**Notice that -12 + 12 is equal to 0. That got the variable alone on the left side of the equals. On the right side, we added 12 and got an answer of 52.**

**To check this answer, we can substitute it back into the original problem and see if we have a true statement.**

\begin{align*}52-12&=40\\ 40&=40\end{align*}

**Our answer is true so our work is accurate.**

**Try and solve a few of these on your own.**

- \begin{align*}x-9=22\end{align*}
- \begin{align*}x-3=46\end{align*}
- \begin{align*}x-7=23\end{align*}

*Take a few minutes to check your work.*

## Real Life Example Completed

*The Bus*

**Now that you have learned all about solving single-variable addition and subtraction equations, you are ready to help Carl. Reread the original problem once again and underline any important information.**

Carl is working on bus costs for his class trip to the amusement park. Since the amusement park is two hours away, Carl’s idea was for his class to take a greyhound bus instead of a school bus. Mrs. Hawk, Carl’s teacher, is alright with the idea as long as it doesn’t cost too much money.

Mrs. Hawk has said that she would like the cost per student to not exceed $20.00.

The amusement park ticket is $14.50.

The bus cost per person is unknown at this point.

Carl wrote the following equation to figure out bus cost per person.

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

Now Carl has two dilemmas. The first is that he needs to figure out how much the bus can cost per person without exceeding the twenty dollar limit. To figure this out, he will need to solve this equation.

Once he has done that, he can multiply the cost per person by 26, since 26 students are going on the trip and this will give him the maximum total cost that the bus can be.

Carl wishes to do all of this research before calling bus companies so that he can have an idea how much the class can afford to pay for a bus.

**First, Carl needs to solve the equation for the unknown quantity \begin{align*}x\end{align*}.**

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

We use the inverse of addition and subtract $14.50 from both sides of the equation.

\begin{align*}& \quad \$14.50 + \ x = \ \$20.00\\ &\ \underline{-\$14.50 \qquad \ -\$14.50}\\ & \qquad \quad \ 0 + \ x \ = \ \$5.50\\ & \qquad \qquad \quad \ x \ = \ \$5.50\end{align*}

**Each person can pay $5.50 for the bus.**

**What is the total amount that the bus can cost?**

To figure this out, you can multiply the number of people going by the price each person can pay for the bus. In this case, that is $5.50.

\begin{align*}& \ 5.50\\ &\underline{\times \; 26\;\;\;\;}\\ &\$143.00\end{align*}

**The students can afford to pay $143.00 for a coach or greyhound bus.**

**Carl is not very optimistic after looking at his arithmetic. He doesn’t think that they will be able to get a bus for that little amount of money. Carl decides to call a few places anyway, maybe they will get lucky and find a coach bus for $143.00.**

## Vocabulary

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

- Expression
- a combination of variables, numbers and operations without an equal sign.

- Simplify
- to make smaller

- Inverse
- the opposite. An inverse operation is the opposite operation.

- Sum
- the answer to an addition problem

- Difference
- the answer to a subtraction problem

## Technology Integration

James Sousa, Solve One Step Equations by Adding and Subtracting Whole Numbers

Other Videos:

- http://www.mathplayground.com/howto_solvevariable.html – This is a video on how to solve a variable equation. It goes through each type of equation step by step.

## Time to Practice

Directions: Simplify the following expressions by combining like terms. If the expression is already in simplest form please write “already in simplest form.”

1. \begin{align*}4x+6x\end{align*}

2. \begin{align*}8y+5y\end{align*}

3. \begin{align*}9z+2z\end{align*}

4. \begin{align*}8x+2y\end{align*}

5. \begin{align*}7y+3y+2x\end{align*}

6. \begin{align*}9x-x\end{align*}

7. \begin{align*}12y-3y\end{align*}

8. \begin{align*}22x-2y\end{align*}

9. \begin{align*}78x-10x\end{align*}

10. \begin{align*}22y-4y\end{align*}

Directions: Solve each single-variable addition equation. Write your answer in the form: variable = _____. For example, \begin{align*}x = 3\end{align*}

11. \begin{align*}x+4=11\end{align*}

12. \begin{align*}x+11=22\end{align*}

13. \begin{align*}x+3=8\end{align*}

14. \begin{align*}x+12=20\end{align*}

15. \begin{align*}x+9=11\end{align*}

16. \begin{align*}x+8=30\end{align*}

17. \begin{align*}22+x=29\end{align*}

18. \begin{align*}18+x=25\end{align*}

19. \begin{align*}15+x=20\end{align*}

20. \begin{align*}13+x=24\end{align*}

Directions: Solve each single-variable subtraction problem using the inverse operation. Write your answer in the form: variable = _____.

21. \begin{align*}y-5=10\end{align*}

22. \begin{align*}x-7=17\end{align*}

23. \begin{align*}a-4=12\end{align*}

24. \begin{align*}z-6=22\end{align*}

25. \begin{align*}y-9=11\end{align*}

25. \begin{align*}b-5=12\end{align*}

26. \begin{align*}x-8=30\end{align*}

27. \begin{align*}y-7=2\end{align*}

28. \begin{align*}x-9=1\end{align*}

29. \begin{align*}a-6=4\end{align*}

30. \begin{align*}x-4=7\end{align*}

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