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# 7.3: Solving Addition and Subtraction Equations

Difficulty Level: At Grade Created by: CK-12

## Introduction

The Lunch Dilemma

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. 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$0.35 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 lesson will teach you all about solving addition and subtraction equations so that by the end of the lesson you will understand how to help Marc figure out the price of his lunch. What You Will Learn By the end of this lesson, you will be able to demonstrate the following skills: • Solve single-variable addition equations. • Solve single-variable subtraction equations. • Recognize and apply the addition and subtraction properties of equality and the inverse relationship of addition and subtraction. • Model and solve real-world problems using addition and subtraction equations. Teaching Time I. Solve Single-Variable Addition Equations A variable may be used to represent a particular number. For example, in the algebraic equation below, the variable x\begin{align*}x\end{align*} represents only one possible number. x+3=5\begin{align*}x+3=5\end{align*} What number does x\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. If x+3=5,x\begin{align*}x+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 2+3=5\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 x+34=72\begin{align*}x+34=72\end{align*}, we may 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 which states that if two things are the same and you take the same amount from each thing, they will still be the same afterward. In math terms, this looks like: if a=b\begin{align*}a=b\end{align*}, then ac=bc\begin{align*}a-c=b-c\end{align*}. If you subtract a number, c\begin{align*}c\end{align*}, from one side of an equation, you must subtract that same number, c\begin{align*}c\end{align*}, from the other side, too, to keep the values on both sides equal. Example Solve for x: x+34=72\begin{align*} \ x+34=72\end{align*}. In the equation, 34 is added to x\begin{align*}x\end{align*}. So, we can subtract 34 from both sides of the equation to solve for x\begin{align*}x\end{align*}. x+34x+3434x+0x=72=7234=38=38\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. Example Solve for b: \begin{align*}\ 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. 7G. Lesson Exercises Solve each addition equation for the missing variable. 1. \begin{align*}x+36=90\end{align*} 2. \begin{align*}x+27=35\end{align*} 3. \begin{align*}y+1.7=6.5\end{align*} Take a few minutes to check your answers. Is your work accurate? II. Solve Single-Variable Subtraction Equations To solve an equation in which a number is subtracted from a variable, we can use the inverse of subtraction––addition. We can add that number to both sides of the equation to solve it. You can think about this as working backwards from the operation. If we have a problem with addition, we subtract. If we have a problem with subtraction, we add. We must add the number to both sides of the equation because of the Addition Property of Equality, which is very much like the Subtraction Property of Equality, and states: if \begin{align*}a=b\end{align*}, then \begin{align*}a+c=b+c\end{align*}. So, if you add a number, \begin{align*}c\end{align*}, to one side of an equation, you must add that same number, \begin{align*}c\end{align*}, to the other side, too, to keep the values on both sides equal. Example Solve for \begin{align*}a \ a-15=18\end{align*}. In the equation, 15 is subtracted from \begin{align*}a\end{align*}. So, we can add 15 to both sides of the equation to solve for \begin{align*}a\end{align*}. \begin{align*}a-15 &= 18\\ a-15+15 &= 18+15\\ a+(-15+15) &= 33\\ a+0 &= 33\\ a &= 33\end{align*} Notice how we rewrote the subtraction as adding a negative integer. The value of \begin{align*}a\end{align*} is 33. Example Solve for k: \begin{align*}\ k-\frac{1}{3}=\frac{2}{3}\end{align*}. In the equation, \begin{align*}\frac{1}{3}\end{align*} is subtracted from \begin{align*}k\end{align*}. So, we can add \begin{align*}\frac{1}{3}\end{align*} to both sides of the equation to solve for \begin{align*}k\end{align*}. \begin{align*}k-\frac{1}{3} &= \frac{2}{3}\\ k-\frac{1}{3}+\frac{1}{3} &= \frac{2}{3}+\frac{1}{3}\\ k+\left(-\frac{1}{3}+\frac{1}{3}\right) &= \frac{3}{3}\\ k+0 &= \frac{3}{3}\\ k &= \frac{3}{3}=1\end{align*} The value of \begin{align*}k\end{align*} is 1. 7H. Lesson Exercises Solve each equation. 1. \begin{align*}x-44=22\end{align*} 2. \begin{align*}x-1.3=5.6\end{align*} 3. \begin{align*}y-\frac{1}{4}=\frac{2}{4}\end{align*} Take a few minutes to check your work with a peer. III. Using Properties In the last two sections, you learned how to solve single-variable addition and subtraction equations. When solving those equations, you utilized two properties that have already been mentioned. Let’s just review them here. The Addition Property of Equality states that as long as you add the same quantity from both sides of an equation, the equation will remain equal. The Subtraction Property of Equality states that as long as you subtract the same quantity to both sides of an equation, 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. If the operation in the equation is subtraction, then you use the Addition Property of Equality. You can look back at the previous examples to review how we did this. Write both of these properties and their definitions down in your notebook before continuing. IV. Model and Solve Real-World Problems Using Addition and Subtraction Equations We can use equations to solve real-world problems. In fact, sometimes an equation can help us to organize the given information so that we can solve it. Writing and solving equations is often the best way to work through a real-world problem. Example 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. a. Write an equation to represent \begin{align*}b\end{align*}, the number of blue tiles in the bag. b. Determine how many blue tiles are in the bag. Consider part \begin{align*}a\end{align*} first. 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. Next, consider part \begin{align*}b\end{align*}. 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. Example Harry earned19.50 this week. That is 6.50 less than he earned last week. a. Write an equation to represent \begin{align*}m\end{align*}, the amount of money, in dollars, that he earned last week. b. Determine how much money Harry earned last week. Consider part \begin{align*}a\end{align*} first. Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem. \begin{align*}& \text{Harry earned} \ \underline{\19.50 \ \text{this week}}. \ \text{That} \ \underline{\text{is}} \ \underline{\6.50} \ \underline{\text{less than}} \ldots \underline{\text{last week}.}\\ & \qquad \qquad \qquad \qquad \quad \downarrow \qquad \qquad \qquad \ \ \downarrow \quad \Box \quad \quad \downarrow \qquad \qquad \qquad \Box\\ & \qquad \qquad \qquad \qquad \quad \downarrow \qquad \qquad \qquad \ \ \downarrow \quad \Box \quad \quad \downarrow \qquad \qquad \qquad \Box\\ & \qquad \qquad \qquad \qquad \quad \downarrow \qquad \qquad \qquad \ \ \downarrow \quad \Box \quad \quad \downarrow \qquad \qquad \qquad \Box\\ & \qquad \qquad \qquad \qquad \ 19.50 \qquad \qquad \quad \ \ = \quad m \quad \quad - \qquad \qquad \quad 6.50\end{align*} This equation, \begin{align*}19.50=m-6.50\end{align*}, represents \begin{align*}m\end{align*}, the number of dollars earned last week. Next, consider part \begin{align*}b\end{align*}. Solve the equation to find the number of blue tiles in the bag. \begin{align*}19.50 &= m-6.50\\ 19.50+6.50 &= m-6.50+6.50\\ 26.00 &= m+(-6.50+6.50)\\ 26 &= m+0\\ 26 &= m\end{align*} Harry earned26.00 last week.

## Real Life Example Completed

The Lunch Dilemma

Here is the original problem once again. Reread it and underline any important information.

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. 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$0.35 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 spent9.00 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 was6.15.

## Vocabulary

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

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".
states that you can add the same quantity to both sides of an equation and have the equation still "balance".

## Technology Integration

Other Videos:

1. http://www.mathplayground.com/howto_solvevariable.html – This is a video on how to solve a variable equation.
2. http://www.mathplayground.com/mv_solving_single_step_equations.html – This is a Brightstorm video on solving single step equations.

## Time to 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+97=123\end{align*}

Directions: Solve each single-variable subtraction equation.

11. \begin{align*}x-8=9\end{align*}

12. \begin{align*}x-18=29\end{align*}

13. \begin{align*}a-9=29\end{align*}

14. \begin{align*}a-4=30\end{align*}

15. \begin{align*}b-14=27\end{align*}

16. \begin{align*}b-13=50\end{align*}

17. \begin{align*}y-23=57\end{align*}

18. \begin{align*}y-15=27\end{align*}

19. \begin{align*}x-9=32\end{align*}

20. \begin{align*}c-19=32\end{align*}

Directions: Solve each equation.

21. \begin{align*}x+\frac{3}{7}=\frac{5}{7}\end{align*}

22. \begin{align*}h-2=\frac{1}{2}\end{align*}

23. \begin{align*}m-3.4=7.1\end{align*}

24. \begin{align*}s+1.4=3.6\end{align*}

Directions: Write an equation for each problem and then solve it.

25. Christina has 40 coins in her collection. She has 21 fewer coins in her collection than Shaya has in hers.

a. Write an equation to represent \begin{align*}s\end{align*}, the number of coins in Shaya's collection.

b. Determine how many coins are in Shaya's collection.

26. At a book store, each hardcover book costs $19.99. Each hardcover book costs$12.50 more than each paperback book.

a. Write an equation to represent \begin{align*}p\end{align*}, the cost, in dollars, of a paperback book.

b. Determine the cost, in dollars, of a paperback book.

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