Have you ever tried to figure out the cost of a field trip? Take a look at this dilemma.

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 a dilemma. 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.

**In this Concept, you will learn how to solve single variable equations like this one. Pay attention and you can help Carl at the end of the Concept.**

### Guidance

In the last Concept 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.

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 the next way to solve single-variable equations.

\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?**

\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.**

\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 = _____.

#### Example A

\begin{align*}x+16=22\end{align*}

**Solution: \begin{align*}x = 6\end{align*}**

#### Example B

\begin{align*}y+15=30\end{align*}

**Solution: \begin{align*}y = 15\end{align*}**

#### Example C

\begin{align*}12+x=18\end{align*}

**Solution: \begin{align*}x = 6\end{align*}**

Remember Carl and the bus dilemma?

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.**

### Vocabulary

Here are the vocabulary words in this Concept.

- 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

### Guided Practice

Here is one for you to try on your own.

Solve this equation.

\begin{align*}x + 15 = 32\end{align*}

**Answer**

We can use mental math to solve this problem.

\begin{align*}15 + 15 = 30 + 2 = 32\end{align*}

**Our answer is \begin{align*}x = 17\end{align*}.**

### Video Review

Here are videos for review.

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

### Practice

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

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

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

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

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

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

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

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

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

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

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

11. \begin{align*}x + 18=24\end{align*}

12. \begin{align*}23+x=33\end{align*}

13. \begin{align*}y+18=31\end{align*}

14. \begin{align*}21+x=54\end{align*}

15. \begin{align*}88+x=91\end{align*}