CHAPTER 5 Solving First Degree Equations in One Variable
5.1 Solve Equations Using the Subtraction and Addition Properties of Equality
Learning Objectives
By the end of this section, you will be able to:
- Solve equations using the Subtraction and Addition Properties of Equality
- Solve equations that need to be simplified
- Translate an equation and solve
- Translate and solve applications
We are now ready to “get to the good stuff.” You have the basics down and are ready to begin one of the most important topics in algebra: solving equations. The applications are limitless and extend to all careers and fields. Also, the skills and techniques you learn here will help improve your critical thinking and problem-solving skills. This is a great benefit of studying mathematics and will be useful in your life in ways you may not see right now.
Solve Equations Using the Subtraction and Addition Properties of Equality
Solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that make each side of the equation the same. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle.
Solution of an Equation
A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.
The steps to determine if a value is a solution to an equation are listed here.
HOW TO: Determine whether a number is a solution to an equation.
- Substitute the number for the variable in the equation.
- Simplify the expressions on both sides of the equation.
- Determine whether the resulting equation is true.
- If it is true, the number is a solution.
- If it is not true, the number is not a solution.
EXAMPLE 1
Determine whether is a solution for .
Solution
| Multiply. | |
| Add. |
Since results in a true equation, is a solution to the equation .
TRY IT 1.1
Is a solution for
Show answer
no
TRY IT 1.2
Is a solution for
Show answer
no
In that section,we will model how the Subtraction and Addition Properties work and then we will apply them to solve equations.
Subtraction Property of Equality
For all real numbers , and , if , then .
Addition Property of Equality
For all real numbers , and , if , then .
When you add or subtract the same quantity from both sides of an equation, you still have equality.
We will introduce the Subtraction Property of Equality by modeling equations with envelopes and counters. (Figure .1) models the equation .
The goal is to isolate the variable on one side of the equation. So we ‘took away’ from both sides of the equation and found the solution .
Some people picture a balance scale, as in (Figure .2), when they solve equations.
The quantities on both sides of the equal sign in an equation are equal, or balanced. Just as with the balance scale, whatever you do to one side of the equation you must also do to the other to keep it balanced.
Let’s see how to use Subtraction and Addition Properties of Equality to solve equations. We need to isolate the variable on one side of the equation. And we check our solutions by substituting the value into the equation to make sure we have a true statement.
EXAMPLE 2
Solve: .
Solution
To isolate , we undo the addition of by using the Subtraction Property of Equality.
| Subtract 11 from each side to “undo” the addition. | ||
| Simplify. | ||
| Check: | ||
| Substitute . | ||
Since makes a true statement, we know that it is a solution to the equation.
TRY IT 2.1
Solve: .
Show answer
x = −16
TRY IT 2.2
Solve: .
Show answer
x = −20
In the original equation in the previous example, was added to the , so we subtracted to ‘undo’ the addition. In the next example, we will need to ‘undo’ subtraction by using the Addition Property of Equality.
EXAMPLE 3
Solve: .
Solution
| Add 4 to each side to “undo” the subtraction. | ||
| Simplify. | ||
| Check: | ||
| Substitute . | ||
| The solution to is . |
TRY IT 3.1
Solve: .
Show answer
−1
TRY IT 3.2
Solve: .
Show answer
−4
Now let’s solve equations with fractions.
EXAMPLE 4
Solve: .
Solution
| Use the Addition Property of Equality. | ||
| Find the LCD to add the fractions on the right. | ||
| Simplify | ||
| Check: | ||
| Subtract. | ||
| Simplify. | ||
| The solution checks. |
TRY IT 4.1
Solve: .
Show answer
TRY IT 4.2
Solve: .
Show answer
Let’s solve equations that contained decimals.
EXAMPLE 5
Solve .
Solution
| Use the Addition Property of Equality. | ||
| Add. | ||
| Check: | ||
| Substitute . | ||
| Simplify. | ||
| The solution checks. |
TRY IT 5.1
Solve: .
Show answer
b = 6.4
TRY IT 5.2
Solve: .
Show answer
c = 14
Solve Equations That Need to Be Simplified
In the examples up to this point, we have been able to isolate the variable with just one operation. Many of the equations we encounter in algebra will take more steps to solve. Usually, we will need to simplify one or both sides of an equation before using the Subtraction or Addition Properties of Equality. You should always simplify as much as possible before trying to isolate the variable.
EXAMPLE 6
Solve: .
Solution
The left side of the equation has an expression that we should simplify before trying to isolate the variable.
| Rearrange the terms, using the Commutative Property of Addition. | |
| Combine like terms. | |
| Add 11 to both sides to isolate . | |
| Simplify. | |
| Check. Substitute into the original equation. |
The solution checks.
TRY IT 6.1
Solve: .
Show answer
y = 15
TRY IT 6.2
Solve: .
Show answer
z = 2
EXAMPLE 7
Solve: .
Solution
The left side of the equation has an expression that we should simplify.
| Distribute on the left. | |
| Use the Commutative Property to rearrange terms. | |
| Combine like terms. | |
| Isolate n using the Addition Property of Equality. | |
| Simplify. | |
| Check. Substitute into the original equation. The solution checks. |
TRY IT 7.1
Solve: .
Show answer
p = 5
TRY IT 7.2
Solve: .
Show answer
q = −16
EXAMPLE 8
Solve: .
Solution
Both sides of the equation have expressions that we should simplify before we isolate the variable.
| Distribute on the left, subtract on the right. | |
| Use the Commutative Property of Addition. | |
| Combine like terms. | |
| Undo subtraction by using the Addition Property of Equality. | |
| Simplify. | |
| Check. Let . | |
| The solution checks. |
TRY IT 8.1
Solve: .
Show answer
h = −1
TRY IT 8.2
Solve: .
Show answer
x = 1
Translate an Equation and Solve
Previously, we translated word sentences into equations. The first step is to look for the word (or words) that translate(s) to the equal sign. The list below reminds us of some of the words that translate to the equal sign (=):
- is
- is equal to
- is the same as
- the result is
- gives
- was
- will be
Let’s review the steps we used to translate a sentence into an equation.
HOW TO: Translate a word sentence to an algebraic equation.
- Locate the “equals” word(s). Translate to an equal sign.
- Translate the words to the left of the “equals” word(s) into an algebraic expression.
- Translate the words to the right of the “equals” word(s) into an algebraic expression.
Now we are ready to try an example.
EXAMPLE 9
Translate and solve: five more than is equal to .
Solution
| Translate. | |
| Subtract 5 from both sides. | |
| Simplify. | |
| Check: Is five more than ? | The solution checks. |
TRY IT 9.1
Translate and solve: Eleven more than is equal to .
Show answer
x + 11 = 41; x = 30
TRY IT 9.2
Translate and solve: Twelve less than is equal to .
Show answer
y − 12 = 51; y = 63
EXAMPLE 10
Translate and solve: The difference of and is .
Solution
| Translate. | |
| Simplify. | |
| Check. |
|
| The solution checks. |
TRY IT 10.1
Translate and solve: The difference of and is .
Show answer
4x − 3x = 14; x = 14
TRY IT 10.2
Translate and solve: The difference of and is .
Show answer
7a − 6a = −8; a = −8
Translate and Solve Applications
In most of the application problems we solved earlier, we were able to find the quantity we were looking for by simplifying an algebraic expression. Now we will be using equations to solve application problems. We’ll start by restating the problem in just one sentence, assign a variable, and then translate the sentence into an equation to solve. When assigning a variable, choose a letter that reminds you of what you are looking for.
EXAMPLE 11
The Robles family has two dogs, Buster and Chandler. Together, they weigh pounds.
Chandler weighs pounds. How much does Buster weigh?
Solution
| Read the problem carefully. | |
| Identify what you are asked to find, and choose a variable to represent it. | How much does Buster weigh? Let Buster’s weight |
| Write a sentence that gives the information to find it. | Buster’s weight plus Chandler’s weight equals 71 pounds. |
| We will restate the problem, and then include the given information. | Buster’s weight plus 28 equals 71. |
| Translate the sentence into an equation, using the variable . | |
| Solve the equation using good algebraic techniques. | |
| Check the answer in the problem and make sure it makes sense. | Is 43 pounds a reasonable weight for a dog? Yes. Does Buster’s weight plus Chandler’s weight equal 71 pounds? |
| ✓ | |
| Write a complete sentence that answers the question, “How much does Buster weigh?” | Buster weighs 43 pounds |
TRY IT 11.1
Translate into an algebraic equation and solve: The Pappas family has two cats, Zeus and Athena. Together, they weigh pounds. Zeus weighs pounds. How much does Athena weigh?
Show answer
a + 6 = 13; Athena weighs 7 pounds.
TRY IT 11.2
Translate into an algebraic equation and solve: Sam and Henry are roommates. Together, they have books. Sam has books. How many books does Henry have?
Show answer
26 + h = 68; Henry has 42 books.
Devise a Problem-Solving Strategy
- Read the problem. Make sure you understand all the words and ideas.
- Identify what you are looking for.
- Name what you are looking for. Choose a variable to represent that quantity.
- Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
- Solve the equation using good algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.
EXAMPLE 12
Shayla paid for her new car. This was less than the sticker price. What was the sticker price of the car?
Solution
| What are you asked to find? | “What was the sticker price of the car?” |
| Assign a variable. | Let the sticker price of the car. |
| Write a sentence that gives the information to find it. | $24,575 is $875 less than the sticker price $24,575 is $875 less than |
| Translate into an equation. | |
| Solve. | |
| Check: | Is $875 less than $25,450 equal to $24,575?
✓ |
| Write a sentence that answers the question. | The sticker price was $25,450. |
TRY IT 12.1
Translate into an algebraic equation and solve: Eddie paid for his new car. This was less than the sticker price. What was the sticker price of the car?
Show answer
19,875 = s − 1025; the sticker price is $20,900.
TRY IT 12.2
Translate into an algebraic equation and solve: The admission price for the movies during the day is . This is less than the price at night. How much does the movie cost at night?
Show answer
7.75 = n − 3.25; the price at night is $11.00.
Key Concepts
- Determine whether a number is a solution to an equation.
- Substitute the number for the variable in the equation.
- Simplify the expressions on both sides of the equation.
- Determine whether the resulting equation is true.
If it is true, the number is a solution.
If it is not true, the number is not a solution. - Subtraction and Addition Properties of Equality
- Subtraction Property of Equality
For all real numbers a, b, and c,
if a = b then . - Addition Property of Equality
For all real numbers a, b, and c,
if a = b then .
- Subtraction Property of Equality
- Translate a word sentence to an algebraic equation.
- Locate the “equals” word(s). Translate to an equal sign.
- Translate the words to the left of the “equals” word(s) into an algebraic expression.
- Translate the words to the right of the “equals” word(s) into an algebraic expression.
- Problem-solving strategy
- Read the problem. Make sure you understand all the words and ideas.
- Identify what you are looking for.
- Name what you are looking for. Choose a variable to represent that quantity.
- Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
- Solve the equation using good algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.
Glossary
- solution of an equation
- A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.
Practice Makes Perfect
Solve Equations Using the Subtraction and Addition Properties of Equality
In the following exercises, determine whether the given value is a solution to the equation.
| 1. Is a solution of | 2. Is a solution of |
| 3. Is a solution of | 4. Is a solution of |
In the following exercises, solve each equation.
| 5. | 6. |
| 7. | 8. |
| 9. | 10. |
| 11. | 12. |
| 13. | 14. |
| 15. | 16. |
| 17. | 18. |
| 19. | 20. |
Solve Equations that Need to be Simplified
In the following exercises, solve each equation.
| 21. | 22. |
| 23. | 24. |
| 25. | 26. |
| 27. | 28. |
| 29. | 30. |
| 31. | 32. |
| 33. | 34. |
| 35. | 36. |
| 37. | 38. |
Translate to an Equation and Solve
In the following exercises, translate to an equation and then solve.
| 39. The sum of and is . | 40.Five more than is equal to . |
| 41.Three less than is . | 42. Ten less than is . |
| 43. Eight more than is equal to . | 44. The sum of and is . |
| 45. The difference of and is . | 46. The difference of and is . |
| 47. The difference of and is . | 48. The difference of and is . |
| 49. The sum of and is . | 50. The sum of and is . |
Translate and Solve Applications
In the following exercises, translate into an equation and solve.
| 51.Jeff read a total of pages in his English and Psychology textbooks. He read pages in his English textbook. How many pages did he read in his Psychology textbook? | 52. Pilar drove from home to school and then to her aunt’s house, a total of miles. The distance from Pilar’s house to school is miles. What is the distance from school to her aunt’s house? |
| 53. Eva’s daughter is years younger than her son. Eva’s son is years old. How old is her daughter? | 54. Pablo’s father is years older than his mother. Pablo’s mother is years old. How old is his father? |
| 55. For a family birthday dinner, Celeste bought a turkey that weighed pounds less than the one she bought for Thanksgiving. The birthday dinner turkey weighed pounds. How much did the Thanksgiving turkey weigh? | 56. Allie weighs pounds less than her twin sister Lorrie. Allie weighs pounds. How much does Lorrie weigh? |
| 57. Connor’s temperature was degrees higher this morning than it had been last night. His temperature this morning was degrees. What was his temperature last night? | 58. The nurse reported that Tricia’s daughter had gained pounds since her last checkup and now weighs pounds. How much did Tricia’s daughter weigh at her last checkup? |
| 59. Ron’s paycheck this week was less than his paycheck last week. His paycheck this week was . How much was Ron’s paycheck last week? | 60. Melissa’s math book cost less than her art book cost. Her math book cost . How much did her art book cost? |
Everyday Math
| 61.Construction Miguel wants to drill a hole for a screw. The screw should be inch larger than the hole. Let equal the size of the hole he should drill. Solve the equation to see what size the hole should be. | Baking 62. Kelsey needs cup of sugar for the cookie recipe she wants to make. She only has cup of sugar and will borrow the rest from her neighbour. Let equal the amount of sugar she will borrow. Solve the equation to find the amount of sugar she should ask to borrow. |
Writing Exercises
| 63. Write a word sentence that translates the equation and then make up an application that uses this equation in its solution. | 64. Is a solution to the equation How do you know? |
Answers
| 1. yes | 3. no | 5. x = 5 |
| 7. | 9. p = −11.7 | 11. a = 10 |
| 13. | 15. y = 13.8 | 17. x = −27 |
| 19. | 21. 17 | 23. 8 |
| 25. −20 | 27. 2 | 29. −1.7 |
| 31. −2 | 33. −4 | 35. 6 |
| 37. −41 | 39. x + (−5) = 33; x = 38 | 41.y − 3 = −19; y = −16 |
| 43. p + 8 = 52; p = 44 | 45. 5c − 4c = 60; 60 | 47. |
| 49. −9m + 10m = −25; m = −25 | 51. Let p equal the number of pages read in the Psychology book 41 + p = 54. Jeff read pages in his Psychology book. | 53. Let d equal the daughter’s age. d = 12 − 5. Eva’s daughter’s age is 7 years old. |
| 55. 21 pounds | 57. 100.5 degrees | 59. $121.19 |
| 61. | 63. Answers will vary. |
Attributions
This chapter has been adapted from “Solve Equations Using the Subtraction and Addition Properties of Equality” in Prealgebra (OpenStax) by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.