assignment applications 3 6a

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3.6: Applications

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• Page ID 22478

• David Arnold
• College of the Redwoods

Because we’ve increased our fundamental ability to simplify algebraic expressions, we’re now able to tackle a number of more advanced applications. Before we begin, we remind readers of required steps that must accompany solutions of applications.

Requirements for Word Problem Solutions

• Statements such as “Let P represent the perimeter of the rectangle.”
• Labeling unknown values with variables in a table.
• Labeling unknown quantities in a sketch or diagram.
• Set up an Equation. Every solution to a word problem must include a carefully crafted equation that accurately describes the constraints in the problem statement.
• Solve the Equation. You must always solve the equation set up in the previous step.
• Answer the Question. This step is easily overlooked. For example, the problem might ask for Jane’s age, but your equation’s solution gives the age of Jane’s sister Liz. Make sure you answer the original question asked in the problem.
• Look Back. It is important to note that this step does not imply that you should simply check your solution in your equation. After all, it’s possible that your equation incorrectly models the problem’s situation, so you could have a valid solution to an incorrect equation. The important question is: “Does your answer make sense based on the words in the original problem statement.”

Consecutive Integers

The integers are consecutive , in the sense that one follows right after another. For example, 5 and 6 are a pair of consecutive integers. The important relation to notice is the fact that the second integer of this pair is one larger than its predecessor. That is, 6 = 5 + 1.

Definition: Consecutive Integers

Let k represent an integer. The next consecutive integer is the integer k + 1.

Thus, if k is an integer, then k + 1 is the next integer, k + 2 is the next integer after that, and so on.

The three sides of a triangle are consecutive integers and the perimeter is 72 inches. Find the measure of each side of the triangle.

We follow the Requirements for Word Problem Solutions.

1. Set up a Variable Dictionary. In this case, a carefully labeled diagram is the best way to indicate what the unknown variable represents.

In our schematic diagram, we’ve labeled the three sides of the triangle with expressions representing the consecutive integers k, k+ 1, and k+ 2.

2. Set up an Equation. To find the perimeter P of the triangle, sum the three sides.

\[P = k + (k + 1) + (k + 2)\nonumber \]

However, we’re given the fact that the perimeter is 72 inches. Thus,

\[72 = k + (k + 1) + (k + 2)\nonumber \]

3. Solve the Equation. On the right, regroup and combine like terms.

\[72 = 3k + 3\nonumber \]

Now, solve.

\[ \begin{aligned} 72-3 = 3k + 3 - 3 ~ & \textcolor{red}{ \text{ Subtract 3 from both sides.}} \\ 69 = 3k ~ & \textcolor{red}{ \text{ Simplify.}} \\ \frac{69}{3} = \frac{3k}{3} ~ & \textcolor{red}{ \text{ Divide both sides by 3.}} \\ 23 = k ~ & \textcolor{red}{ \text{ Simplify.}} \end{aligned}\nonumber \]

4. Answer the Question. We’ve only found one side, but the question asks for the measure of all three sides. However, the remaining two sides can be found by substituting 23 for k into the expressions k + 1 and k + 2.

k + 1 = 23 + 1 and k + 2 = 23 + 2

Hence, the three sides measure 23 inches, 24 inches, and 25 inches.

5. Look Back. Does our solution make sense? Well, the three sides are certainly consecutive integers, and their sum is 23 inches + 24 inches + 25 inches = 72 inches, which was the given perimeter. Therefore, our solution is correct.

The three sides of a triangle are consecutive integers and the perimeter is 57 centimeters. Find the measure of each side of the triangle.

18, 19, and 20 cm

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Number line.

• simplify\:\frac{2}{3}-\frac{3}{2}+\frac{1}{4}
• simplify\:4+(2+1)^2
• simplify\:\log _{10}(100)
• simplify\:\frac{1}{x+1}\cdot \frac{x^2}{5}
• simplify\:\frac{x^2+4x-45}{x^2+x-30}
• simplify\:\frac{x^2+14x+49}{49-x^2}
• simplify\:\frac{6}{x-1}-\frac{3}{x+1}
• simplify\:\frac{5x}{6}+\frac{3x}{2}

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89 100 89 100

72 100 72 100

62 100 , 62% 62 100 , 62%

41 100 , 41% 41 100 , 41%

• ⓐ 12 25 12 25
• ⓑ 11 10 11 10
• ⓐ 16 25 16 25
• ⓐ 161 250 161 250
• ⓐ 113 250 113 250
• ⓑ 7 80 7 80
• ⓐ 6 25 6 25
• ⓑ 0.24 0.24
• ⓐ 22 50 22 50
• ⓑ 0.44 0.44
• ⓐ 3 10 3 10
• ⓐ 12.5 100 12.5 100
• ⓑ 0.125 0.125

11. 1 – %, or 11 1 9 % 11. 1 – %, or 11 1 9 %

16 . 6 – %, or 16 2 3 % 16 . 6 – %, or 16 2 3 %

• ⓐ 5 9 = 20 36 5 9 = 20 36
• ⓑ 7 11 = 28 44 7 11 = 28 44
• ⓒ 2.50 8 = 3.75 12 2.50 8 = 3.75 12
• ⓐ 6 7 = 36 42 6 7 = 36 42
• ⓑ 8 36 = 12 54 8 36 = 12 54
• ⓒ 3.75 6 = 2.50 4 3.75 6 = 2.50 4

n 105 = 60 100 n 105 = 60 100

n 85 = 40 100 n 85 = 40 100

36 n = 25 100 36 n = 25 100

27 n = 36 100 27 n = 36 100

n 100 = 39 52 n 100 = 39 52

n 100 = 23 92 n 100 = 23 92

Section 6.1 Exercises

6 100 6 100

32 1000 32 1000

• ⓐ 57 100 57 100
• ⓐ 42 100 42 100

17 100 17 100

13 25 13 25

23 125 23 125

19 200 19 200

• ⓐ 3 200 3 200
• ⓑ 0.015 0.015
• ⓐ 7023 10,000 7023 10,000
• ⓑ 0.7023 0.7023
• ⓑ 0.25 0.25

266 . 6 – % 266 . 6 – %

• ⓐ 1 10 1 10
• ⓑ approximately $9.50 approximately $9.50

17 20 ; 0.85 17 20 ; 0.85

1 4 , 1 2 , 3 4 , 1 . 1 4 , 1 2 , 3 4 , 1 .

The Szetos sold their home for five times what they paid 30 years ago.

Section 6.2 Exercises

2,407 grams

The original number should be greater than 44.80% is less than 100%, so when 80% is converted to a decimal and multiplied to the base in the percent equation, the resulting amount of 44 is less. 44 is only the larger number in cases where the percent is greater than 100%.

Alex should have packed half as many shorts and twice as many shirts.

Section 6.3 Exercises

• ⓑ $1,260.45
• ⓒ 25% off first, then $20 off
• ⓐ Priam is correct. The original price is 100%. Since the discount rate was 40%, the sale price was 60% of the original price.

Section 6.4 Exercises

Answers will vary.

Section 6.5 Exercises

4 15 = 36 135 4 15 = 36 135

12 5 = 96 40 12 5 = 96 40

5 7 = 115 161 5 7 = 115 161

8 1 = 48 6 8 1 = 48 6

9.36 18 = 2.60 5 9.36 18 = 2.60 5

18.04 11 = 4.92 3 18.04 11 = 4.92 3

3 4 cup 3 4 cup

48 quarters

n 250 = 35 100 n 250 = 35 100

n 47 = 110 100 n 47 = 110 100

45 n = 30 100 45 n = 30 100

90 n = 150 100 90 n = 150 100

17 85 = p 100 17 85 = p 100

340 260 = p 100 340 260 = p 100

Review Exercises

32 100 32 100

13 100 , 13% 13 100 , 13%

12 25 12 25

641 1000 641 1000

• ⓐ 27 200 27 200
• ⓑ 0.135 0.135

3 8 = 12 32 3 8 = 12 32

1 18 = 23 414 1 18 = 23 414

n 395 = 62 100 n 395 = 62 100

15 1000 = p 100 15 1000 = p 100

Practice Test

0.24 , 6 25 0.24 , 6 25

3.5 , 3 1 2 3.5 , 3 1 2

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• Authors: Lynn Marecek, MaryAnne Anthony-Smith
• Publisher/website: OpenStax
• Book title: Prealgebra
• Publication date: Sep 25, 2015
• Location: Houston, Texas
• Book URL: https://openstax.org/books/prealgebra/pages/1-introduction
• Section URL: https://openstax.org/books/prealgebra/pages/chapter-6

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