Solving First Degree Equations in One Variable

Teetering high above the floor, this amazing mobile remains aloft thanks to its carefully balanced mass. Any shift in either direction could cause the mobile to become lopsided, or even crash downward. In this chapter, we will solve equations by keeping quantities on both sides of an equal sign in perfect balance.

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.

1. Substitute the number for the variable in the equation.
2. Simplify the expressions on both sides of the equation.
3. 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 .

Figure .1

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.

Figure .2

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.

1. Locate the “equals” word(s). Translate to an equal sign.
2. Translate the words to the left of the “equals” word(s) into an algebraic expression.
3. 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

1. Read the problem. Make sure you understand all the words and ideas.
2. Identify what you are looking for.
3. Name what you are looking for. Choose a variable to represent that quantity.
4. 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.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. 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.
  1. Substitute the number for the variable in the equation.
  2. Simplify the expressions on both sides of the equation.
  3. 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 .
• Translate a word sentence to an algebraic equation.
  1. Locate the “equals” word(s). Translate to an equal sign.
  2. Translate the words to the left of the “equals” word(s) into an algebraic expression.
  3. Translate the words to the right of the “equals” word(s) into an algebraic expression.
• Problem-solving strategy
  1. Read the problem. Make sure you understand all the words and ideas.
  2. Identify what you are looking for.
  3. Name what you are looking for. Choose a variable to represent that quantity.
  4. 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.
  5. Solve the equation using good algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.

Glossary

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.

Solve Equations Using the Division and Multiplication Properties of Equality

You may have noticed that all of the equations we have solved so far have been of the form or . We were able to isolate the variable by adding or subtracting the constant term on the side of the equation with the variable. Now we will see how to solve equations that have a variable multiplied by a constant and so will require division to isolate the variable.

Let’s look at our puzzle again with the envelopes and counters in (Figure 1).

Figure .1

In the illustration there are two identical envelopes that contain the same number of counters. Remember, the left side of the workspace must equal the right side, but the counters on the left side are “hidden” in the envelopes. So how many counters are in each envelope?

How do we determine the number? We have to separate the counters on the right side into two groups of the same size to correspond with the two envelopes on the left side. The 6 counters divided into 2 equal groups gives 3 counters in each group (since ).

What equation models the situation shown in (Figure 2)? There are two envelopes, and each contains counters. Together, the two envelopes must contain a total of 6 counters.

Figure .2

If we divide both sides of the equation by 2, as we did with the envelopes and counters,
we get:

We found that each envelope contains 3 counters. Does this check? We know , so it works! Three counters in each of two envelopes does equal six!

This example leads to the Division Property of Equality.

Division and Multiplication Properties of Equality

Division Property of Equality: For all real numbers , and , if , then .

Multiplication Property of Equality: For all real numbers , if , then .

When you divide or multiply both sides of an equation by the same quantity, you still have equality.

Let’s review how these properties of equality can be applied in order to solve equations. Remember, the goal is to ‘undo’ the operation on the variable. In the example below the variable is multiplied by , so we will divide both sides by to ‘undo’ the multiplication.

EXAMPLE 1

Solve: .

Solution

We use the Division Property of Equality to divide both sides by .

Divide both sides by 4 to undo the multiplication.
Simplify.
Check your answer. Let .

Since this is a true statement, is a solution to .

TRY IT 1.1

Solve: .

Show answer

y = −16

TRY IT 1.2

Solve: .

Show answer

z = −13

In the previous example, to ‘undo’ multiplication, we divided. How do you think we ‘undo’ division?

EXAMPLE 2

Solve: .

Solution

Here is divided by . We can multiply both sides by to isolate .

Multiply both sides by .
Simplify.
Check your answer. Let .

TRY IT 2.1

Solve: .

Show answer

b = 144

TRY IT 2.2

Solve: .

Show answer

c = 128

EXAMPLE 3

Solve: .

Solution

Remember is equivalent to .

Rewrite as .
Divide both sides by .
Check.
Substitute
Simplify.

We see that there are two other ways to solve .

We could multiply both sides by .

We could take the opposite of both sides.

TRY IT 3.1

Solve: .

Show answer

k = −8

TRY IT 3.2

Solve: .

Show answer

g = −3

EXAMPLE 4

Solve: .

Solution

Since the product of a number and its reciprocal is , our strategy will be to isolate by multiplying by the reciprocal of .

Multiply by the reciprocal of .
Reciprocals multiply to one.
Multiply.
Check your answer. Let

Notice that we could have divided both sides of the equation by to isolate . While this would work, multiplying by the reciprocal requires fewer steps.

TRY IT 4.1

Solve: .

Show answer

n = 35

TRY IT 4.2

Solve: .

Show answer

y = 18

Solve Equations That Need to be Simplified

Many equations start out more complicated than the ones we’ve just solved. First, we need to simplify both sides of the equation as much as possible

EXAMPLE 5

Solve: .

Solution

Start by combining like terms to simplify each side.

Combine like terms.
Divide both sides by 12 to isolate x.
Simplify.
Check your answer. Let

TRY IT 5.1

Solve: .

Show answer

x = 2

TRY IT 5.2

Solve: .

Show answer

n = −5

EXAMPLE 6

Solve: .

Solution

Simplify each side by combining like terms.

Simplify each side.
Divide both sides by 3 to isolate y.
Simplify.
Check your answer. Let

Notice that the variable ended up on the right side of the equal sign when we solved the equation. You may prefer to take one more step to write the solution with the variable on the left side of the equal sign.

TRY IT 6.1

Solve: .

Show answer

c = −3

TRY IT 6.2

Solve: .

Show answer

EXAMPLE 7

Solve: .

Solution

Remember—always simplify each side first.

Distribute.
Simplify.
Divide both sides by -3 to isolate n.
Check your answer. Let .

TRY IT 7.1

Solve: .

Show answer

n = −6

TRY IT 7.2

Solve: .

Show answer

n = −5

Key Concepts

• Division and Multiplication Properties of Equality
  • Division Property of Equality: For all real numbers a, b, c, and , if , then .
  • Multiplication Property of Equality: For all real numbers a, b, c, if , then .

Practice Makes Perfect

Solve Equations Using the Division and Multiplication Properties of Equality

In the following exercises, solve each equation for the variable using the Division Property of Equality and check the solution.

1.	2.
3.	4.
5.	6.
7.	8.
9.	10.
11.	12.

In the following exercises, solve each equation for the variable using the Multiplication Property of Equality and check the solution.

13.	14.
15.	16.
17.	18.
19.	20.
21.	22.
23.	24.
25.	26.

Solve Equations That Need to be Simplified

In the following exercises, solve the equation.

27.	28.
29.	30.
31.	32.
33.	34.
35.	36.

Everyday Math

37. Teaching Connie’s kindergarten class has children. She wants them to get into equal groups. Find the number of children in each group, , by solving the equation .	38. Balloons Ramona bought balloons for a party. She wants to make equal bunches. Find the number of balloons in each bunch, , by solving the equation .
39. Unit price Nishant paid for a pack of juice bottles. Find the price of each bottle, , by solving the equation .	40. Ticket price Daria paid for children’s tickets at the ice skating rink. Find the price of each ticket, , by solving the equation .
41. Fabric The drill team used yards of fabric to make flags for one-third of the members. Find how much fabric, , they would need to make flags for the whole team by solving the equation .	42. Fuel economy Tania’s SUV gets half as many miles per gallon (mpg) as her husband’s hybrid car. The SUV gets . Find the miles per gallons, , of the hybrid car, by solving the equation .

Writing Exercises

43. Emiliano thinks is the solution to the equation . Explain why he is wrong.

44. Frida started to solve the equation by adding to both sides. Explain why Frida’s method will result in the correct solution.

Answers

1. 9	3. 3	5. −6
7. 7	9. 15	11. 0
13. 28	15. 36	17. −48
19. 80	21. 25	23. −32
25. 5/2	27. y = −1	29. m = −5
31.	33. q = 24	35. p = 56
37. 6 children	39. $1.08	41. 42 yards
43. Answer will vary.

Attributions

This chapter has been adapted from “Solve Equations Using the Division and Multiplication 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.

Solve an Equation with Constants on Both Sides

You may have noticed that in all the equations we have solved so far, all the variable terms were on only one side of the equation with the constants on the other side. This does not happen all the time—so now we’ll see how to solve equations where the variable terms and/or constant terms are on both sides of the equation.

Our strategy will involve choosing one side of the equation to be the variable side, and the other side of the equation to be the constant side. Then, we will use the Subtraction and Addition Properties of Equality, step by step, to get all the variable terms together on one side of the equation and the constant terms together on the other side.

By doing this, we will transform the equation that started with variables and constants on both sides into the form . We already know how to solve equations of this form by using the Division or Multiplication Properties of Equality.

EXAMPLE 1

Solve: .

Solution

In this equation, the variable is only on the left side. It makes sense to call the left side the variable side. Therefore, the right side will be the constant side. We’ll write the labels above the equation to help us remember what goes where.

Since the left side is the variable side, the 6 is out of place. We must “undo” adding 6 by subtracting 6, and to keep the equality we must subtract 6 from both sides. Use the Subtraction Property of Equality.
Simplify.
Now all the s are on the left and the constant on the right.
Use the Division Property of Equality.
Simplify.
Check:
Let .

TRY IT 1.1

Solve: .

Show answer

x = −4

TRY IT 1.2

Solve: .

Show answer

a = −8

EXAMPLE 1.2

Solve: .

Solution

Notice that the variable is only on the left side of the equation, so this will be the variable side and the right side will be the constant side. Since the left side is the variable side, the is out of place. It is subtracted from the , so to ‘undo’ subtraction, add to both sides.

Add 7 to both sides.
Simplify.
The variables are now on one side and the constants on the other.
Divide both sides by 2.
Simplify.
Check:
Substitute: .

TRY IT 2.1

Solve: .

Show answer

y = 5

TRY IT 2.2

Solve: .

Show answer

m = 9

Solve an Equation with Variables on Both Sides

What if there are variables on both sides of the equation? We will start like we did above—choosing a variable side and a constant side, and then use the Subtraction and Addition Properties of Equality to collect all variables on one side and all constants on the other side. Remember, what you do to the left side of the equation, you must do to the right side too.

EXAMPLE 3

Solve: .

Solution

Here the variable, , is on both sides, but the constants appear only on the right side, so let’s make the right side the “constant” side. Then the left side will be the “variable” side.

We don’t want any variables on the right, so subtract the .
Simplify.
We have all the variables on one side and the constants on the other. We have solved the equation.
Check:
Substitute 7 for .

TRY IT 3.1

Solve: .

Show answer

n = 10

TRY IT 3.2

Solve: .

Show answer

c = 1

EXAMPLE 4

Solve: .

Solution

The only constant, , is on the left side of the equation and variable, , is on both sides. Let’s leave the constant on the left and collect the variables to the right.

Subtract from both sides.
Simplify.
We have the variables on the right and the constants on the left. Divide both sides by 2.
Simplify.
Rewrite with the variable on the left.
Check: Let .

TRY IT 4.1

Solve: .

Show answer

p = −7

TRY IT 4.2

Solve: .

Show answer

m = −3

EXAMPLE 5

Solve: .

Solution

The only constant, , is on the right, so let the left side be the variable side.

Remove the from the right side by adding to both sides.
Simplify.
All the variables are on the left and the constants are on the right. Divide both sides by 8.
Simplify.
Check: Substitute .

TRY IT 5.1

Solve: .

Show answer

j = 2

TRY IT 5.2

Solve: .

Show answer

h = 1

Solve Equations with Variables and Constants on Both Sides

The next example will be the first to have variables and constants on both sides of the equation. As we did before, we’ll collect the variable terms to one side and the constants to the other side.

EXAMPLE 6

Solve: .

Solution

Start by choosing which side will be the variable side and which side will be the constant side. The variable terms are and . Since is greater than , make the left side the variable side and so the right side will be the constant side.

Collect the variable terms to the left side by subtracting from both sides.
Simplify.
Now, collect the constants to the right side by subtracting 5 from both sides.
Simplify.
The solution is .
Check: Let .

TRY IT 6.1

Solve: .

Show answer

x = −1

TRY IT 6.2

Solve: .

Show answer

y = 4

We’ll summarize the steps we took so you can easily refer to them.

HOW TO: Solve an Equation with Variables and Constants on Both Sides

1. Choose one side to be the variable side and then the other will be the constant side.
2. Collect the variable terms to the variable side, using the Addition or Subtraction Property of Equality.
3. Collect the constants to the other side, using the Addition or Subtraction Property of Equality.
4. Make the coefficient of the variable , using the Multiplication or Division Property of Equality.
5. Check the solution by substituting it into the original equation.

It is a good idea to make the variable side the one in which the variable has the larger coefficient. This usually makes the arithmetic easier.

EXAMPLE 7

Solve: .

Solution

We have on the left and on the right. Since > , make the left side the “variable” side.

We don’t want variables on the right side—add to both sides to leave only constants on the right.
Combine like terms.
We don’t want any constants on the left side, so add 2 to both sides.
Simplify.
The variable term is on the left and the constant term is on the right. To get the coefficient of to be one, divide both sides by 9.
Simplify.
Check: Substitute 1 for .

TRY IT 7.1

Solve: .

Show answer

q = 1

TRY IT 7.2

Solve: .

Show answer

n = 1

EXAMPLE 8

Solve: .

Solution

This equation has on the left and on the right. Since > , make the right side the variable side and the left side the constant side.

Subtract from both sides to remove the variable term from the left.
Combine like terms.
Subtract 8 from both sides to remove the constant from the right.
Simplify.
Divide both sides by 3 to make 1 the coefficient of .
Simplify.
Check: Let .

Note that we could have made the left side the variable side instead of the right side, but it would have led to a negative coefficient on the variable term. While we could work with the negative, there is less chance of error when working with positives. The strategy outlined above helps avoid the negatives!

TRY IT 8.1

Solve: .

Show answer

a = −5

TRY IT 8.2

Solve: .

Show answer

k = −6

To solve an equation with fractions, we still follow the same steps to get the solution.

EXAMPLE 9

Solve: .

Solution

Since > , make the left side the variable side and the right side the constant side.

Subtract from both sides.
Combine like terms.
Subtract 5 from both sides.
Simplify.
Check: Let .

TRY IT 9.1

Solve: .

Show answer

x = 10

TRY IT 9.2

Solve: .

Show answer

y = −3

We follow the same steps when the equation has decimals, too.

EXAMPLE 10

Solve: .

Solution

Since > , make the left side the variable side and the right side the constant side.

Subtract from both sides.
Combine like terms.
Subtract 4 from both sides.
Simplify.
Use the Division Property of Equality.
Simplify.
Check: Let .

TRY IT 10.1

Solve: .

Show answer

x = −5

TRY IT 10.2

Solve: .

Show answer

y = −5

Solve Equations Using a General Strategy

Each of the first few sections of this chapter has dealt with solving one specific form of a linear equation. It’s time now to lay out an overall strategy that can be used to solve any linear equation. We call this the general strategy. Some equations won’t require all the steps to solve, but many will. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.

HOW TO: Use a General Strategy for Solving Linear Equations

1. Simplify each side of the equation as much as possible. Use the Distributive Property to remove any parentheses. Combine like terms.
2. Collect all the variable terms to one side of the equation. Use the Addition or Subtraction Property of Equality.
3. Collect all the constant terms to the other side of the equation. Use the Addition or Subtraction Property of Equality.
4. Make the coefficient of the variable term to equal to . Use the Multiplication or Division Property of Equality. State the solution to the equation.
5. Check the solution. Substitute the solution into the original equation to make sure the result is a true statement.

EXAMPLE 11

Solve: .

Solution

Simplify each side of the equation as much as possible. Use the Distributive Property.
Collect all variable terms on one side of the equation—all s are already on the left side.
Collect constant terms on the other side of the equation. Subtract 6 from each side
Simplify.
Make the coefficient of the variable term equal to 1. Divide each side by 3.
Simplify.
Check: Let .

TRY IT 11.1

Solve: .

Show answer

x = 4

TRY IT 11.2

Solve: .

Show answer

y = 1

EXAMPLE 12

Solve: .

Solution

Simplify each side of the equation as much as possible by distributing. The only term is on the left side, so all variable terms are on the left side of the equation.
Add 5 to both sides to get all constant terms on the right side of the equation.
Simplify.
Make the coefficient of the variable term equal to 1 by multiplying both sides by -1.
Simplify.
Check: Let .

TRY IT 12.1

Solve: .

Show answer

y = −6

TRY IT 12.2

Solve: .

Show answer

z = 8

EXAMPLE 13

Solve: .

Solution

Simplify each side of the equation as much as possible. Distribute.
Combine like terms
The only is on the left side, so all variable terms are on one side of the equation.
Add 3 to both sides to get all constant terms on the other side of the equation.
Simplify.
Make the coefficient of the variable term equal to 1 by dividing both sides by 4.
Simplify.
Check: Let .

TRY IT 13.1

Solve: .

Show answer

a = 2

TRY IT 13.2

Solve: .

Show answer

n = 2

EXAMPLE 14

Solve: .

Solution

Be careful when distributing the negative.

Simplify—use the Distributive Property.
Combine like terms.
Add 2 to both sides to collect constants on the right.
Simplify.
Divide both sides by −6.
Simplify.
Check: Let .

TRY IT 14.1

Solve: .

Show answer

TRY IT 14.2

Solve: .

Show answer

EXAMPLE 15

Solve: .

Solution

Distribute.
Combine like terms.
Subtract to get all the variables on the right since > .
Simplify.
Subtract 9 to get the constants on the left.
Simplify.
Divide by 5.
Simplify.
Check: Substitute: .

TRY IT 14.1

Solve: .

Show answer

p = −2

TRY IT 14.2

Solve: .

Show answer

q = −8

EXAMPLE 15

Solve: .

Solution

Distribute.
Add to get all the variables on the left.
Simplify.
Add 1 to get constants on the right.
Simplify.
Divide by 4.
Simplify.
Check: Let .

TRY IT 15.1

Solve: .

Show answer

u = 2

TRY IT 15.2

Solve: .

Show answer

x = 4

In many applications, we will have to solve equations with decimals. The same general strategy will work for these equations.

EXAMPLE 16

Solve: .

Solution

Distribute.
Subtract to get all the s to the left.
Simplify.
Subtract 1.2 to get the constants to the right.
Simplify.
Divide.
Simplify.
Check: Let .

TRY IT 16.1

Solve: .

Show answer

1

TRY IT 16.2

Solve: .

Show answer

−1

Key Concepts

• Solve an equation with variables and constants on both sides
  1. Choose one side to be the variable side and then the other will be the constant side.
  2. Collect the variable terms to the variable side, using the Addition or Subtraction Property of Equality.
  3. Collect the constants to the other side, using the Addition or Subtraction Property of Equality.
  4. Make the coefficient of the variable 1, using the Multiplication or Division Property of Equality.
  5. Check the solution by substituting into the original equation.
• General strategy for solving linear equations
  1. Simplify each side of the equation as much as possible. Use the Distributive Property to remove any parentheses. Combine like terms.
  2. Collect all the variable terms to one side of the equation. Use the Addition or Subtraction Property of Equality.
  3. Collect all the constant terms to the other side of the equation. Use the Addition or Subtraction Property of Equality.
  4. Make the coefficient of the variable term to equal to 1. Use the Multiplication or Division Property of Equality. State the solution to the equation.
  5. Check the solution. Substitute the solution into the original equation to make sure the result is a true statement.

Practice Makes Perfect

Solve an Equation with Constants on Both Sides

In the following exercises, solve the equation for the variable.

1.	2.
3.	4.
5.	6.
7.	8.
9.	10.
11.	12.

Solve an Equation with Variables on Both Sides

In the following exercises, solve the equation for the variable.

13.	14.
15.	16.
17.	18.
19.	20.
21.	22.
23.	24.

Solve an Equation with Variables and Constants on Both Sides

In the following exercises, solve the equations for the variable.

25.	26.
27.	28.
29.	30.
31.	32.
33.	34.
35.	36.
37.	38.
39.	40.
41.	42.
43.	44.
45.	46.
47.	48.
49.	50.

Solve an Equation Using the General Strategy

In the following exercises, solve the linear equation using the general strategy.

51.	52.
53.	54.
55.	56.
57.	58.
59.	60.
61.	62.
63.	64.
65.	66.
67.	68.
69.	70.
71.	72.
73.	74.
75.	76.
77.	78.
79.	80.
81.	82.
83.	84.
85.	86.

Everyday Math

Making a fence 87.  Jovani has a fence around the rectangular garden in his backyard. The perimeter of the fence is feet. The length is feet more than the width. Find the width, , by solving the equation .	Concert tickets  88. At a school concert, the total value of tickets sold was Student tickets sold for and adult tickets sold for The number of adult tickets sold was less than times the number of student tickets. Find the number of student tickets sold, , by solving the equation .
Coins 89. Rhonda has in nickels and dimes. The number of dimes is one less than twice the number of nickels. Find the number of nickels, , by solving the equation .	Fencing 90. Micah has feet of fencing to make a rectangular dog pen in his yard. He wants the length to be feet more than the width. Find the length, , by solving the equation .

Writing Exercises

91. When solving an equation with variables on both sides, why is it usually better to choose the side with the larger coefficient as the variable side?	92. Solve the equation , explaining all the steps of your solution.
93. What is the first step you take when solving the equation Explain why this is your first step.	94. Solve the equation explaining all the steps of your solution as in the examples in this section.
95. Using your own words, list the steps in the General Strategy for Solving Linear Equations.	96. Explain why you should simplify both sides of an equation as much as possible before collecting the variable terms to one side and the constant terms to the other side.

Answers

1. 6	3.6	5. -8
7. -8	9. -4	11. -2
13. -11	15. 9	17. -3
19. 3	21. -3/4	25. 19
27. 7	29. -5	31. -4
33. 2	35. 4	37. -6
39. 7	41. -40	43. 15
45. 3.46	47. 60	49. 23
51. 9	53. 6	55. 3
57. −2	59. −1	61. 5
63. 0.52	65. 0.25	67. −9
69. 2	71. 6	73. 3/2
75. 3	77. −4	79. 2
81. 34	83. 10	85. 2
87. 30 feet	89. 8 nickels	91. Answers will vary.
93. Answers will vary.	95. Answers will vary.

Attributions

This chapter has been adapted from “Solve Equations with Variables and Constants on Both Sides” 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.

Solve Equations with Fraction Coefficients

Let’s use the General Strategy for Solving Linear Equations introduced earlier to solve the equation .

To isolate the term, subtract from both sides.
Simplify the left side.
Change the constants to equivalent fractions with the LCD.
Subtract.
Multiply both sides by the reciprocal of .
Simplify.

This method worked fine, but many students don’t feel very confident when they see all those fractions. So we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.

We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but with no fractions. This process is called clearing the equation of fractions. Let’s solve the same equation again, but this time use the method that clears the fractions.

EXAMPLE 1

Solve: .

Solution

Find the least common denominator of all the fractions in the equation.
Multiply both sides of the equation by that LCD, 8. This clears the fractions.
Use the Distributive Property.
Simplify — and notice, no more fractions!
Solve using the General Strategy for Solving Linear Equations.
Simplify.
Check: Let

TRY IT 1.1

Solve: .

Show answer

TRY IT 1.2

Solve: .

Show answer

y = 3

Notice in (Figure) that once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve! We then used the General Strategy for Solving Linear Equations.

HOW TO: Solve Equations with Fraction Coefficients by Clearing the Fractions

1. Find the least common denominator of all the fractions in the equation.
2. Multiply both sides of the equation by that LCD. This clears the fractions.
3. Solve using the General Strategy for Solving Linear Equations.

EXAMPLE 2

Solve: .

Solution

We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.

Find the least common denominator of all the fractions in the equation.
Multiply both sides of the equation by 12.
Distribute.
Simplify — and notice, no more fractions!
Combine like terms.
Divide by 7.
Simplify.
Check: Let .

TRY IT 2.1

Solve: .

Show answer

v = 40

TRY IT 2.2

Solve: .

Show answer

u = −12

In the next example, we’ll have variables and fractions on both sides of the equation.

EXAMPLE 3

Solve: .

Solution

Find the LCD of all the fractions in the equation.
Multiply both sides by the LCD.
Distribute.
Simplify — no more fractions!
Subtract from both sides.
Simplify.
Subtract 2 from both sides.
Simplify.
Divide by 5.
Simplify.
Check: Substitute .

TRY IT 3.1

Solve: .

Show answer

a = −2

TRY IT 3.2

Solve: .

Show answer

c = −2

In (Figure), we’ll start by using the Distributive Property. This step will clear the fractions right away!

EXAMPLE 4

Solve: .

Solution

Distribute.
Simplify. Now there are no fractions to clear!
Subtract 1 from both sides.
Simplify.
Divide by 2.
Simplify.
Check: Let .

TRY IT 4.1

Solve: .

Show answer

p = −4

TRY IT 4.2

Solve: .

Show answer

q = 2

Many times, there will still be fractions, even after distributing.

EXAMPLE 5

Solve: .

Solution

Distribute.
Simplify.
Multiply by the LCD, 4.
Distribute.
Simplify.
Collect the terms to the left.
Simplify.
Collect the constants to the right.
Simplify.
Check: Substitute for .

TRY IT 5.1

Solve: .

Show answer

n = 2

TRY IT 5.2

Solve: .

Show answer

m = −1

Solve Equations with Decimal Coefficients

Some equations have decimals in them. This kind of equation will occur when we solve problems dealing with money and percent. But decimals are really another way to represent fractions. For example, and . So, when we have an equation with decimals, we can use the same process we used to clear fractions—multiply both sides of the equation by the least common denominator.

EXAMPLE 6

Solve: .

Solution

The only decimal in the equation is . Since , the LCD is . We can multiply both sides by to clear the decimal.

Multiply both sides by the LCD.
Distribute.
Multiply, and notice, no more decimals!
Add 50 to get all constants to the right.
Simplify.
Divide both sides by 8.
Simplify.
Check: Let .

TRY IT 6.1

Solve: .

Show answer

x = 20

TRY IT 6.2

Solve: .

Show answer

x = 10

EXAMPLE 7

Solve: .

Solution

Look at the decimals and think of the equivalent fractions.

Notice, the LCD is .

By multiplying by the LCD we will clear the decimals.

Multiply both sides by 100.
Distribute.
Multiply, and now no more decimals.
Collect the variables to the right.
Simplify.
Collect the constants to the left.
Simplify.
Divide by 19.
Simplify.
Check: Let .

TRY IT 7.1

Solve: .

Show answer

h = 12

TRY IT 7.2

Solve: .

Show answer

k = −1

The next example uses an equation that is typical of the ones we will see in the money applications in the next chapter. Notice that we will distribute the decimal first before we clear all decimals in the equation.

EXAMPLE 8

Solve: .

Solution

Distribute first.
Combine like terms.
To clear decimals, multiply by 100.
Distribute.
Subtract 15 from both sides.
Simplify.
Divide by 30.
Simplify.
Check: Let .

TRY IT 8.1

Solve: .

Show answer

n = 9

TRY IT 8.2

Solve: .

Show answer

d = 16

Key Concepts

• Solve equations with fraction coefficients by clearing the fractions.
  1. Find the least common denominator of all the fractions in the equation.
  2. Multiply both sides of the equation by that LCD. This clears the fractions.
  3. Solve using the General Strategy for Solving Linear Equations.

Practices Makes Perfect

Solve equations with fraction coefficients

In the following exercises, solve the equation by clearing the fractions.

1.	2.
3.	4.
5.	6.
7.	8.
9.	10.
11.	12.
13.	14.
15.	16.
17.	18.
19.	20.
21.	22.
23.	24.

Solve Equations with Decimal Coefficients

In the following exercises, solve the equation by clearing the decimals.

25.	26.
27.	28.
29.	30.
31.	32.
33.	34.
35.	36.
37.	38.
39.	40.

Everyday Math

Coins  41. Taylor has in dimes and pennies. The number of pennies is more than the number of dimes. Solve the equation for , the number of dimes.

Stamps 42.  Travis bought worth of stamps and stamps. The number of stamps was less than the number of stamps. Solve the equation for , to find the number of stamps Travis bought.

Writing Exercises

43. Explain how to find the least common denominator of .	44. If an equation has several fractions, how does multiplying both sides by the LCD make it easier to solve?
45. If an equation has fractions only on one side, why do you have to multiply both sides of the equation by the LCD?	46. In the equation , what is the LCD? How do you know?

Answers

1. x = -1	3. y = -1	5.
7. x = 4	9. m = 20	11. x = -3
13.	15. x = 1	17. b = 12
19. x = 1	21. p = -41	23.
25. y = 10	27. j = 2	29. x = 18
31. x = 18	33. x = 20	35. n = 9
37. d = 8	39. q = 11	41 d = 18
43. Answers will vary.	45.Answers will vary.

Attributions

This chapter has been adapted from “Solve Equations with Fraction or Decimal Coefficients” 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.

Solve Equations Using the General Strategy

Until now we have dealt with solving one specific form of a linear equation. It is time now to lay out one overall strategy that can be used to solve any linear equation. Some equations we solve will not require all these steps to solve, but many will.

Beginning by simplifying each side of the equation makes the remaining steps easier.

EXAMPLE 1. How to Solve Linear Equations Using the General Strategy

Solve: .

Solution

EXAMPLE 2

Solve: .

Solution

Simplify each side of the equation as much as possible by distributing.
The only term is on the left side, so all variable terms are on the left side of the equation.
Add to both sides to get all constant terms on the right side of the equation.
Simplify.
Rewrite as .
Make the coefficient of the variable term to equal to by dividing both sides by .
Simplify.
Check: Let .

TRY IT 2.1

Solve: .

Show answer

TRY IT 2.2

Solve: .

Show answer

EXAMPLE 3

Solve: .

Solution

Simplify each side of the equation as much as possible.
Distribute.
Combine like terms.
The only term is on the left side, so all variable terms are on one side of the equation.
Add to both sides to get all constant terms on the other side of the equation.
Simplify.
Make the coefficient of the variable term to equal to by dividing both sides by .
Simplify.
Check:
Let .

TRY IT 3.1

Solve: .

Show answer

TRY IT 3.2

Solve: .

Show answer

EXAMPLE 4

Solve: .

Solution

Distribute.
Add to get the variables only to the left.
Simplify.
Add to get constants only on the right.
Simplify.
Divide by .
Simplify.
Check:
Let .

TRY IT 4.1

Solve: .

Show answer

TRY IT 4.2

Solve: .

Show answer

EXAMPLE 5

Solve: .

Solution

Simplify—use the Distributive Property.
Combine like terms.
Add to both sides to collect constants on the right.
Simplify.
Divide both sides by .
Simplify.
Check: Let .

TRY IT 5.1

Solve: .

Show answer

TRY IT 5.2

Solve: .

Show answer

EXAMPLE 6

Solve: .

Solution

Distribute.
Combine like terms.
Subtract to get the variables only on the right side since > .
Simplify.
Subtract to get the constants on left.
Simplify.
Divide by 6.
Simplify.
Check:
Let .

EXAMPLE 7

Solve: .

Solution

Simplify from the innermost parentheses first.
Combine like terms in the brackets.
Distribute.
Add to get the s’s to the right.
Simplify.
Subtract 600 to get the constants to the left.
Simplify.
Divide.
Simplify.
Check:
Substitute .

EXAMPLE 8

Solve: .

Solution

Distribute.
Subtract to get the variables to the left.
Simplify.
Subtract to get the constants to the right.
Simplify.
Divide.
Simplify.
Check:
Let .

Classify Equations

Consider the equation we solved at the start of the last section, . The solution we found was . This means the equation is true when we replace the variable, x, with the value . We showed this when we checked the solution and evaluated for .

If we evaluate for a different value of x, the left side will not be .

The equation is true when we replace the variable, x, with the value , but not true when we replace x with any other value. Whether or not the equation is true depends on the value of the variable. Equations like this are called conditional equations.

All the equations we have solved so far are conditional equations.

Now let’s consider the equation . Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for y.

Distribute.
Subtract to get the ’s to one side.
Simplify—the ’s are gone!

But is true.

This means that the equation is true for any value of y. We say the solution to the equation is all of the real numbers. An equation that is true for any value of the variable like this is called an identity.

What happens when we solve the equation ?

Subtract to get the constant alone on the right.
Simplify—the ’s are gone!

But .

Solving the equation led to the false statement . The equation will not be true for any value of z. It has no solution. An equation that has no solution, or that is false for all values of the variable, is called a contradiction.

EXAMPLE 9

Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.

Solution

Distribute.
Combine like terms.
Subtract to get the ’s to one side.
Simplify.
This is a true statement.	The equation is an identity. The solution is every real number.

EXAMPLE 10

Classify as a conditional equation, an identity, or a contradiction. Then state the solution.

Solution

Distribute.
Combine like terms.
Add to both sides.
Simplify.
Divide.
Simplify.
The equation is true when .	This is a conditional equation. The solution is .

EXAMPLE 11

Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.

Solution

Distribute.
Combine like terms.
Subtract from both sides.
Simplify.
But .	The equation is a contradiction. It has no solution.

Key Concepts

• General Strategy for Solving Linear Equations
  1. Simplify each side of the equation as much as possible.
     Use the Distributive Property to remove any parentheses.
     Combine like terms.
  2. Collect all the variable terms on one side of the equation.
     Use the Addition or Subtraction Property of Equality.
  3. Collect all the constant terms on the other side of the equation.
     Use the Addition or Subtraction Property of Equality.
  4. Make the coefficient of the variable term to equal to 1.
     Use the Multiplication or Division Property of Equality.
     State the solution to the equation.
  5. Check the solution.
     Substitute the solution into the original equation.

Glossary

Practice Makes Perfect

Solve Equations Using the General Strategy for Solving Linear Equations

In the following exercises, solve each linear equation.

1.	2.
3.	4.
5.	6.
7.	8.
9.	10.
11.	12.
13.	14.
15.	16.
17.	18.
19.	20.
21.	22.
23.	24.
25.	26.
27.	28.
29.	30.
31.	32.
33.	34.
35.	36.
37.	38.
39.	40.
41.	42.
43.	44.
45.	46.
47.	48.
49.	50.
51.	52.
53.	54.
55.	56.
57.	58.

Classify Equations

In the following exercises, classify each equation as a conditional equation, an identity, or a contradiction and then state the solution.

59.	60.
61.	62.
63.	64.
65.	66.
67.	68.
69.	70.
71.	72.
73.	74.
75.	76.
77.	78.

Everyday Math

79. Coins. Rhonda has $1.90 in nickels and dimes. The number of dimes is one less than twice the number of nickels. Find the number of nickels, n, by solving the equation .

80. Fencing. Micah has 44 feet of fencing to make a dog run in his yard. He wants the length to be 2.5 feet more than the width. Find the length, L, by solving the equation .

Writing Exercises

81. Explain why you should simplify both sides of an equation as much as possible before collecting the variable terms to one side and the constant terms to the other side.	82. Using your own words, list the steps in the general strategy for solving linear equations.
83. Solve the equation explaining all the steps of your solution as in the examples in this section.	84. What is the first step you take when solving the equation ? Why is this your first step?

Answers

1.	3.	5.
7.	9.	11.
13.	15.	17.
19.	21.	23.
25.	27.	29.
31.	33.	35.
37.	39.	41.
43.	45.	47.
49.	51.	53.
55.	57.	59. identity; all real numbers
61. identity; all real numbers	63. conditional equation;	65. conditional equation;
67. contradiction; no solution	69. contradiction; no solution	71. conditional equation;
73. contradiction; no solution	75. identity; all real numbers	77. identity; all real numbers
79. 8 nickels	81. Answers will vary.	83. Answers will vary.

Attributions

This chapter has been adapted from “Use a General Strategy to Solve Linear Equations” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

Use the Distance, Rate, and Time Formula

One formula you will use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant rate. Rate is an equivalent word for “speed.” The basic idea of rate may already familiar to you. Do you know what distance you travel if you drive at a steady rate of 60 miles per hour for 2 hours? (This might happen if you use your car’s cruise control while driving on the highway.) If you said 120 miles, you already know how to use this formula!

Distance, Rate, and Time

For an object moving at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula:

We will use the Strategy for Solving Applications that we used earlier in this chapter. When our problem requires a formula, we change Step 4. In place of writing a sentence, we write the appropriate formula. We write the revised steps here for reference.

You may want to create a mini-chart to summarize the information in the problem. See the chart in this first example.

EXAMPLE 1

Jamal rides his bike at a uniform rate of 12 miles per hour for hours. What distance has he traveled?

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.		distance traveled
Step 3. Name. Choose a variable to represent it.		Let d = distance.
Step 4. Translate: Write the appropriate formula.
Substitute in the given information.
Step 5. Solve the equation.		miles
Step 6. Check
Does 42 miles make sense?
Jamal rides:
Step 7. Answer the question with a complete sentence.		Jamal rode 42 miles.

TRY IT 1.1

Lindsay drove for hours at 60 miles per hour. How much distance did she travel?

Show answer

330 miles

TRY IT 1.2

Trinh walked for hours at 3 miles per hour. How far did she walk?

Show answer

7 miles

EXAMPLE 2

Rey is planning to drive from his house in Saskatoon to visit his grandmother in Winnipeg, a distance of 520 miles. If he can drive at a steady rate of 65 miles per hour, how many hours will the trip take?

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.		How many hours (time)
Step 3. Name. Choose a variable to represent it.		Let t = time.
		d = 600 km r = 75 km/h t = ? hours
Step 4. Translate. Write the appropriate formula.
Substitute in the given information.
Step 5. Solve the equation.
Step 6. Check. Substitute the numbers into the formula and make sure the result is a true statement.
Step 7. Answer the question with a complete sentence. Rey’s trip will take 8 hours.

Solve a Formula for a Specific Variable

You are probably familiar with some geometry formulas. A formula is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. It is important to be familiar with formulas and be able to manipulate them easily.

In (Example 1) and (Example 2), we used the formula . This formula gives the value of , distance, when you substitute in the values of , the rate and time. But in (Example 2), we had to find the value of . We substituted in values of and then used algebra to solve for . If you had to do this often, you might wonder why there is not a formula that gives the value of when you substitute in the values of . We can make a formula like this by solving the formula for .

To solve a formula for a specific variable means to isolate that variable on one side of the equals sign with a coefficient of 1. All other variables and constants are on the other side of the equals sign. To see how to solve a formula for a specific variable, we will start with the distance, rate and time formula.

EXAMPLE 3

Solve the formula for :

1. when and
2. in general

Solution

We will write the solutions side-by-side to demonstrate that solving a formula in general uses the same steps as when we have numbers to substitute.

a) when and		b) in general
Write the formula.		Write the formula.
Substitute.
Divide, to isolate .		Divide, to isolate .
Simplify.		Simplify.

We say the formula is solved for .

TRY IT 3.1

Solve the formula for :

a) when b) in general

Show answer

a) b)

TRY IT 3.2

Solve the formula for :

a) when b) in general

Show answer

a) b)

EXAMPLE 4

Solve the formula for :

a) when and b) in general

Solution

a) when and		b) in general
Write the formula.		Write the formula.
Substitute.
Clear the fractions.		Clear the fractions.
Simplify.		Simplify.
Solve for .		Solve for .

We can now find the height of a triangle, if we know the area and the base, by using the formula .

The formula is used to calculate simple interest, I, for a principal, P, invested at rate, r, for t years.

EXAMPLE 5

Solve the formula to find the principal, :

a) when , , b) in general

Solution

a), ,		b) in general
Write the formula.		Write the formula.
Substitute.
Simplify.		Simplify.
Divide, to isolate P.		Divide, to isolate P.
Simplify.		Simplify.
The principal is

Later in this class, and in future algebra classes, you’ll encounter equations that relate two variables, usually x and y. You might be given an equation that is solved for y and need to solve it for x, or vice versa. In the following example, we’re given an equation with both x and y on the same side and we’ll solve it for y.

EXAMPLE 6

Solve the formula for y:

a) when b) in general

Solution

a) when		b) in general
Substitute.
Subtract to isolate the -term.		Subtract to isolate the -term.
Divide.		Divide.
Simplify.		Simplify.

Now we will solve a formula in general without using numbers as a guide.

EXAMPLE 7

Solve the formula for .

Solution

We will isolate on one side of the equation.
Both and are added to , so we subtract them from both sides of the equation.
Simplify.

EXAMPLE 8

Solve the formula for y.

Solution

Subtract from both sides to isolate the term with .
Simplify.
Divide by 5 to make the coefficient 1.
Simplify.

The fraction is simplified. We cannot divide by 5

Key Concepts

• To Solve an Application (with a formula)
  1. Read the problem. Make sure all the words and ideas are understood.
  2. Identify what we are looking for.
  3. Name what we are looking for. Choose a variable to represent that quantity.
  4. Translate into an equation. Write the appropriate formula for the situation. Substitute in the given information.
  5. Solve the equation using good algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.
• Distance, Rate and Time
  For an object moving at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula: where d = distance, r = rate, t = time.
• To solve a formula for a specific variable means to get that variable by itself with a coefficient of 1 on one side of the equation and all other variables and constants on the other side.

Practice Makes Perfect

Use the Distance, Rate, and Time Formula

In the following exercises, solve.

1. Socorro drove for hours at 60 miles per hour. How much distance did she travel?	2. Steve drove for hours at 72 miles per hour. How much distance did he travel?
3. Francie rode her bike for hours at 12 miles per hour. How far did she ride?	4. Yuki walked for hours at 4 miles per hour. How far did she walk?
5. Marta is taking the bus from Abbotsford to Cranbrook. The distance is 774 km and the bus travels at a steady rate of 86 miles per hour. How long will the bus ride be?	6. Connor wants to drive from Vancouver to the Nakusp, a distance of 630 km. If he drives at a steady rate of 90 km/h, how many hours will the trip take?
7. Kareem wants to ride his bike from Golden, BC to Banff, AB. The distance is 140 km. If he rides at a steady rate of 20 km/h, how many hours will the trip take?	8. Aurelia is driving from Calgary to Edmonton at a rate of 85 km/h. The distance is 300 km. To the nearest tenth of an hour, how long will the trip take?
9. Alejandra is driving to Prince George, 450 km away. If she wants to be there in 6 hours, at what rate does she need to drive?	10. Javier is driving to Vernon, 240 km away. If he needs to be in Vernon in 3 hours, at what rate does he need to drive?
11. Philip got a ride with a friend from Calgary to Kelowna, a distance of 890 km. If the trip took 10 hours, how fast was the friend driving?	12. Aisha took the train from Spokane to Seattle. The distance is 280 miles and the trip took 3.5 hours. What was the speed of the train?

Solve a Formula for a Specific Variable

In the following exercises, use the formula .

13. Solve for a) when b) in general	14. Solve for a) when and b) in general
15. Solve for a) when b) in general	16. Solve for a) when b) in general
17. Solve for a) when b) in general	18. Solve for a) when b) in general
19. Solve for a) when b) in general In the following exercises, use the formula .	20. Solve for a) when b) in general
21. Solve for a) when b) in general	22. Solve for a) when b) in general
23. Solve for the principal, P for a) ,, b) in general	24. Solve for a) when b) in general In the following exercises, use the formula I = Prt.
25. Solve for the time, t for a) , , b) in general	26. Solve for the principal, P for a) ,, b) in general
27. Solve the formula for y a) when b) in general	28. Solve for the time, t for a) ,, b) in general In the following exercises, solve.
29. Solve the formula for y a) when b) in general	30. Solve the formula for y a) when b) in general
31. Solve for .	32. Solve the formula for y a) when b) in general
33. Solve for .	34. Solve for .
35. Solve the formula for y.	36. Solve for .
37. Solve the formula for y.	38. Solve the formula for y.
39. Solve the formula for y.	40. Solve the formula for y.
41. Solve the formula for y.	42. Solve the formula for y.
43. Solve the formula for .	44. Solve the formula for y.
45. Solve the formula for .	46. Solve the formula for .
47. Solve the formula for .	48. Solve the formula for .
49. Solve the formula for .

Everyday Math

50. Converting temperature. Yon was visiting the United States and he saw that the temperature in Seattle one day was 50o Fahrenheit. Solve for C in the formula to find the Celsius temperature.

51. Converting temperature. While on a tour in Greece, Tatyana saw that the temperature was 40o Celsius. Solve for F in the formula to find the Fahrenheit temperature.

Writing Exercises

52. Solve the equation for a) when b) in general c) Which solution is easier for you, a) or b)? Why?

53. Solve the equation for a) when b) in general c) Which solution is easier for you, a) or b)? Why?

Answers

1. 290 miles	3. 30 miles	5. 9 hours.
7. 75 km/h	9. 3.5 hours	11. 7 hours
13. 7	15. 89 km/h	17. a) b)
19. a) b)	21. a) b)	23. a) b)
25. a) b)	27. a) b)	29. a) b)
31. a) years b)	33. a) b)	35. a) b)
37.	39.	41.
43.	45.	47.
49.	51.	53.
55. 10°C	57. Answers will vary.

Attributions

This chapter has been adapted from “Solve a Formula for a Specific Variable” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

Approach Word Problems with a Positive Attitude

“If you think you can… or think you can’t… you’re right.”—Henry Ford

The world is full of word problems! Will my income qualify me to rent that apartment? How much punch do I need to make for the party? What size diamond can I afford to buy my girlfriend? Should I fly or drive to my family reunion?

How much money do I need to fill the car with gas? How much tip should I leave at a restaurant? How many socks should I pack for vacation? What size turkey do I need to buy for Thanksgiving dinner, and then what time do I need to put it in the oven? If my sister and I buy our mother a present, how much does each of us pay?

Now that we can solve equations, we are ready to apply our new skills to word problems. Do you know anyone who has had negative experiences in the past with word problems? Have you ever had thoughts like the student below?

Negative thoughts can be barriers to success.

Figure .1

When we feel we have no control, and continue repeating negative thoughts, we set up barriers to success. We need to calm our fears and change our negative feelings.

Start with a fresh slate and begin to think positive thoughts. If we take control and believe we can be successful, we will be able to master word problems! Read the positive thoughts in (Figure 2) and say them out loud.

Thinking positive thoughts is a first step towards success.

Figure .2

Think of something, outside of school, that you can do now but couldn’t do 3 years ago. Is it driving a car? Snowboarding? Cooking a gourmet meal? Speaking a new language? Your past experiences with word problems happened when you were younger—now you’re older and ready to succeed!

Use a Problem-Solving Strategy for Word Problems

We have reviewed translating English phrases into algebraic expressions, using some basic mathematical vocabulary and symbols. We have also translated English sentences into algebraic equations and solved some word problems. The word problems applied math to everyday situations. We restated the situation in one sentence, assigned a variable, and then wrote an equation to solve the problem. This method works as long as the situation is familiar and the math is not too complicated.

Now, we’ll expand our strategy so we can use it to successfully solve any word problem. We’ll list the strategy here, and then we’ll use it to solve some problems. We summarize below an effective strategy for problem solving.

EXAMPLE 1

Pilar bought a purse on sale for $18, which is one-half of the original price. What was the original price of the purse?

Solution

Step 1. Read the problem. Read the problem two or more times if necessary. Look up any unfamiliar words in a dictionary or on the internet.

• In this problem, is it clear what is being discussed? Is every word familiar?

Step 2. Identify what you are looking for. Did you ever go into your bedroom to get something and then forget what you were looking for? It’s hard to find something if you are not sure what it is! Read the problem again and look for words that tell you what you are looking for!

• In this problem, the words “what was the original price of the purse” tell us what we need to find.

Step 3. Name what we are looking for. Choose a variable to represent that quantity. We can use any letter for the variable, but choose one that makes it easy to remember what it represents.

• Let the original price of the purse.

Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Translate the English sentence into an algebraic equation.

Reread the problem carefully to see how the given information is related. Often, there is one sentence that gives this information, or it may help to write one sentence with all the important information. Look for clue words to help translate the sentence into algebra. Translate the sentence into an equation.

Restate the problem in one sentence with all the important information.
Translate into an equation.

Step 5. Solve the equation using good algebraic techniques. Even if you know the solution right away, using good algebraic techniques here will better prepare you to solve problems that do not have obvious answers.

Solve the equation.
Multiply both sides by 2.
Simplify.

Step 6. Check the answer in the problem to make sure it makes sense. We solved the equation and found that , which means “the original price” was $36

• Does $36 make sense in the problem? Yes, because 18 is one-half of 36, and the purse was on sale at half the original price.

Step 7. Answer the question with a complete sentence. The problem asked “What was the original price of the purse?”

• The answer to the question is: “The original price of the purse was $36.”

If this were a homework exercise, our work might look like this:

Pilar bought a purse on sale for $18, which is one-half the original price. What was the original price of the purse?

	Let the original price.
	18 is one-half the original price.
Multiply both sides by 2.
Simplify.
Check. Is $36 a reasonable price for a purse?	Yes.
Is 18 one half of 36?
	✓
	The original price of the purse was $36.