12.2: Solving Equations Using Addition and Subtraction
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 singlevariable expressions
 Solve singlevariable addition equations
 Solve singlevariable subtraction equations
 Model and solve realworld problems using addition or subtraction equations
Teaching Time
I. Simplify Sums and Differences of SingleVariable Expressions
In the last lesson you learned how to write singlevariable expressions and single variable equations. Now you are going to learn to work with singlevariable 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*}
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*}
\begin{align*}6x\end{align*}
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*}
Next, we can simplify them by adding the numerical part of the terms together. The \begin{align*}x\end{align*}
\begin{align*}&5x+7x\\
&12x\end{align*}
You can think of the \begin{align*}x\end{align*}
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*}
Next, we simplify the like terms.
\begin{align*}7x+2x=9x \end{align*}
We can’t simplify the \begin{align*}5y\end{align*}
\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*}9y2y\end{align*}
First, you can see that these terms are alike because they both have \begin{align*}y\end{align*}
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*}8x3x+2y+4y\end{align*}
We begin with the like terms.
\begin{align*}8x3x&=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*}
7z+2z+4z 
\begin{align*}25y13y\end{align*}
25y−13y 
\begin{align*}7x+2x+4a\end{align*}
7x+2x+4a
Take a few minutes to check your work with a partner. Are your answers accurate?
II. Solve SingleVariable 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 singlevariable equations, our next step is to learn how to solve them.
What does it mean to solve a singlevariable equation?
To solve a singlevariable 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*}
\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 singlevariable equations.
Example
\begin{align*}x+27=43\end{align*}
The second way of solving a singlevariable 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*}
Is this true? We can check our work by substituting the value for \begin{align*}x\end{align*}
\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*}
Our goal is to get the \begin{align*}x\end{align*}
\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 SingleVariable Subtraction Equations
In the last section, you learned how to solve singlevariable addition equations. Now you are going to learn how to solve singlevariable subtraction equations.
How do we do this?
We can use inverses once again to solve singlevariable 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*}x12=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*}&x12\ = \ \ 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*}5212&=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*}x9=22\end{align*}
 \begin{align*}x3=46\end{align*}
 \begin{align*}x7=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 singlevariable 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*}9xx\end{align*}
7. \begin{align*}12y3y\end{align*}
8. \begin{align*}22x2y\end{align*}
9. \begin{align*}78x10x\end{align*}
10. \begin{align*}22y4y\end{align*}
Directions: Solve each singlevariable 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 singlevariable subtraction problem using the inverse operation. Write your answer in the form: variable = _____.
21. \begin{align*}y5=10\end{align*}
22. \begin{align*}x7=17\end{align*}
23. \begin{align*}a4=12\end{align*}
24. \begin{align*}z6=22\end{align*}
25. \begin{align*}y9=11\end{align*}
25. \begin{align*}b5=12\end{align*}
26. \begin{align*}x8=30\end{align*}
27. \begin{align*}y7=2\end{align*}
28. \begin{align*}x9=1\end{align*}
29. \begin{align*}a6=4\end{align*}
30. \begin{align*}x4=7\end{align*}
My Notes/Highlights Having trouble? Report an issue.
Color  Highlighted Text  Notes 

Show More 