Have you ever been to Harvard Square?

Marc went to lunch with his friend Tony. He bought a day pass for the train and met Tony in Harvard Square for lunch. When he left the house, Marc counted his money and found that he had $15.50 in his pocket. He met Tony in Harvard Square. They had a terrific time walking around and they both loved looking around in the skateboard shop. That was the best.

After walking around for a while, they stopped for lunch and each ate a sandwich. On the way back home later Marc reached into his pocket and found that he only had .35 left from the original $15.50. He began calculating in his head.

That meant that he had spent $15.15 between the day pass and the sandwich. If the day pass was $9.00, how much did Marc spend on lunch?

**The best way to figure this problem out is with an equation. This Concept will teach you all about solving addition and subtraction equations so that by the end of the Concept you will understand how to help Marc figure out the price of his lunch.**

### Guidance

A ** variable** may be used to represent a particular number. For example, in the algebraic equation below, the variable \begin{align*}x\end{align*} represents only one possible number.

\begin{align*}x+3=5\end{align*}

**What number does \begin{align*}x\end{align*} represent?** We can find out by asking ourselves a question. “What number, when added to 3, equals 5?”

**We can solve this one using mental math. Mental math is one strategy for figuring out an unknown variable.**

\begin{align*}2+3=5, x\end{align*} must be equal to 2.

**When we determine the value for a variable in an equation, we are solving that equation.** If the equation involves a simple number fact, such as \begin{align*}2+3=5\end{align*}, then we can solve the equation as we did above.

**However, to solve a more complex equation, such as \begin{align*}x+34=72\end{align*}, we need to use a different strategy besides mental math. Let's take a look at one strategy for solving an equation now.**

When mental math is not an option, one strategy for solving an equation is to ** isolate the variable** in the equation. Isolating the variable means getting the variable by itself on one side of the equal (=) sign.

One way to isolate the variable is to use ** inverse operations**. Inverse operations undo one another. For example, addition is the inverse of subtraction, and vice versa. Multiplication is the inverse of division, and vice versa.

**How do we do that?**

To solve an equation in which a variable is *added* to a number, we can use the inverse operation–– subtraction. We can subtract the number that is being added to the variable from *both* sides of the equation.

We must subtract the number from *both* sides of the equation because of the ** Subtraction Property of Equality**, which states:

if \begin{align*}a=b\end{align*}, then \begin{align*}a-c=b-c\end{align*}.

**If you subtract a number, \begin{align*}c\end{align*}, from one side of an equation, you must subtract that same number, \begin{align*}c\end{align*}, from the other side, too, to keep the values on both sides equal.**

*Solve for* \begin{align*}x. \ x+34=72\end{align*}.

**In the equation, 34 is** *added***to \begin{align*}x\end{align*}. So, we can** *subtract***34 from both sides of the equation to solve for \begin{align*}x\end{align*}.**

\begin{align*}x+34 &= 72\\ x+34-34 &= 72-34\\ x+0 &= 38\\ x &= 38\end{align*}

**The value of \begin{align*}x\end{align*} is 38.**

*Solve for* \begin{align*}b \ 1.5+b=3.5\end{align*}

**In the equation, 1.5 is** *added***to \begin{align*}b\end{align*}. So, we can** *subtract***1.5 from both sides of the equation to solve for \begin{align*}b\end{align*}.**

\begin{align*}1.5+b &= 3.5\\ 1.5-1.5+b &= 3.5-1.5\\ 0+b &= 2.0\\ b &= 2\end{align*}

**The value of \begin{align*}b\end{align*} is 2.**

**The Subtraction Property of Equality states that as long as you subtract the same quantity to both sides of an equation, that the equation will remain equal.**

Each of these properties makes use of an inverse operation. If the operation in the equation is addition, then you use the Subtraction Property of Equality.

Solve each addition equation for the missing variable.

#### Example A

\begin{align*}x+36=90\end{align*}

**Solution: \begin{align*}x = 54\end{align*}**

#### Example B

\begin{align*}x+27=35\end{align*}

**Solution: \begin{align*}x = 8\end{align*}**

#### Example C

\begin{align*}y+1.7=6.5\end{align*}

**Solution: \begin{align*}y = 4.8\end{align*}**

Here is the original problem once again.

Marc went to lunch with his friend Tony. He bought a day pass for the train and met Tony in Harvard Square for lunch. When he left the house, Marc counted his money and found that he had $15.50 in his pocket. He met Tony in Harvard Square. They had a terrific time walking around and they both loved looking around in the skateboard shop. That was the best.

After walking around for a while, they stopped for lunch and each ate a sandwich. On the way back home later Marc reached into his pocket and found that he only had .35 left from the original $15.50. He began calculating in his head.

That meant that he had spent $15.15 between the day pass and the sandwich. If the day pass was $9.00, how much did Marc spend on lunch?

To figure out the price of lunch, we need to write an equation.

Lunch is our unknown quantity. We can use \begin{align*}l\end{align*} for the price of lunch.

We know that Marc spent $9.00 on a day pass.

We also know that the total spent was $15.15.

Now we can write an equation to solve for the price of lunch.

\begin{align*}l+9=15.15\end{align*}

We can solve this by subtracting 9 from both sides of the equation.

\begin{align*}l+9-9=15.15-9\end{align*}

\begin{align*}l=\$6.15\end{align*}

**The cost of Marc’s lunch is $6.15.**

### Vocabulary

- Isolate the variable
- an explanation used to describe the action of getting the variable alone on one side of the equal sign.

- Inverse Operation
- the opposite operation.

- Subtraction Property of Equality
- states that you can subtract the same quantity from both sides of an equation and have the equation still balance.

### Guided Practice

Here is the original problem once again.

The number of gray tiles in a bag is 4 more than the number of blue tiles in the bag. There are 11 gray tiles in the bag.

**Answer**

Write an equation to represent \begin{align*}b\end{align*}, the number of blue tiles in the bag.

Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem. Since there are 11 gray tiles in the bag, represent the number of gray tiles as 11.

\begin{align*}& The \ \underline{number \ of \ gray \ tiles} \ldots \underline{is} \ \underline{4} \ \underline{more \ than} \ the \ \underline{number \ of \ blue \ tiles} \ldots\\ & \qquad \qquad \downarrow \qquad \qquad \qquad \qquad \ \downarrow \ \ \downarrow \qquad \ \ \downarrow \qquad \qquad \qquad \qquad \ \downarrow\\ & \qquad \qquad 11 \qquad \qquad \qquad \quad \ \ = \ 4 \qquad \ + \qquad \qquad \qquad \qquad b\end{align*}

This equation, \begin{align*}11=4+b\end{align*}, represents \begin{align*}b\end{align*}, the number of blue tiles in the bag.Solve the equation to find the number of blue tiles in the bag.

\begin{align*}11 &= 4+b\\ 11-4 &= 4-4+b\\ 7 &= 0+b\\ 7 &= b\end{align*}

**There are 7 blue tiles in the bag.**

### Video Review

- This is a James Sousa video on solving single variable addition equations.

### Practice

Directions: Solve each single-variable addition equation.

1. \begin{align*}x+7=14\end{align*}

2. \begin{align*}y+17=34\end{align*}

3. \begin{align*}a+27=34\end{align*}

4. \begin{align*}x+30=47\end{align*}

5. \begin{align*}x+45=53\end{align*}

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

7. \begin{align*}a+38=74\end{align*}

8. \begin{align*}b+45=80\end{align*}

9. \begin{align*}c+54=75\end{align*}

10. \begin{align*}y+197=423\end{align*}

11. \begin{align*}y+297=523\end{align*}

12. \begin{align*}y+397=603\end{align*}

13. \begin{align*}y+97=405\end{align*}

14. \begin{align*}y+94=102\end{align*}

15. \begin{align*}y+87=323\end{align*}