Learn on PengiOpenstax Intermediate Algebra 2EChapter 2: Solving Linear Equations

Lesson 2.2: Use a Problem Solving Strategy

In this lesson from OpenStax Intermediate Algebra 2E, students learn a structured seven-step problem-solving strategy for tackling algebraic word problems, including number problems, percent applications, and simple interest problems. The approach guides learners to read, identify, name with a variable, translate into an equation, solve, check, and answer each problem systematically. This skill is foundational for intermediate algebra students who need a reliable method to convert real-world situations into linear equations.

Section 1

πŸ“˜ Use a Problem Solving Strategy

New Concept

Learn a powerful, universal strategy to conquer any word problem. This method breaks down challenges into simple steps, helping you translate real-world scenarios involving numbers, percents, and simple interest into easily solvable equations.

What’s next

This strategy is your new toolkit. Next, you’ll master it step-by-step through worked examples, interactive practice cards, and challenge problems.

Section 2

Problem Solving Strategy

Property

Use a Problem Solving Strategy for word problems.

  • Step 1. Read the problem. Make sure all the words and ideas are understood.
  • Step 2. Identify what you are looking for.
  • Step 3. Name what you are looking for. Choose a variable to represent that quantity.
  • Step 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.
  • Step 5. Solve the equation using proper algebra techniques.
  • Step 6. Check the answer in the problem to make sure it makes sense.
  • Step 7. Answer the question with a complete sentence.

Examples

  • The number of apples in a basket is 5 more than twice the number of oranges. If there are 31 apples, how many oranges are there? Let oo be the number of oranges. The equation is 2o+5=312o + 5 = 31. Solving gives 2o=262o = 26, so o=13o = 13. There are 13 oranges.
  • The sum of five times a number and ten is 55. Find the number. Let nn be the number. The equation is 5n+10=555n + 10 = 55. Solving gives 5n=455n = 45, so n=9n = 9. The number is 9.

Section 3

Consecutive Integers

Property

Consecutive integers are integers that immediately follow each other. If the first integer is nn, the next are n+1,n+2,…n+1, n+2, …

Consecutive even integers are even integers that immediately follow one another. If the first even integer is nn, the next are n+2,n+4,…n+2, n+4, …

Consecutive odd integers are odd integers that immediately follow one another. If the first odd integer is nn, the next are n+2,n+4,…n+2, n+4, …

Section 4

Percent Applications

Property

To solve percent equations, translate English sentences into algebraic equations. Be sure to change the given percent to a decimal before you use it in the equation.

(a) What number is 45% of 84? translates to n=0.45β‹…84n = 0.45 β‹… 84.

(b) 8.5% of what amount is 4.76 dollars? translates to 0.085β‹…n=4.760.085 β‹… n = 4.76.

Section 5

Percent Change

Property

To find the percent change:

  • Step 1. Find the amount of change.
change=newΒ amountβˆ’originalΒ amount\text{change} = \text{new amount} - \text{original amount}
  • Step 2. Find what percent the amount of change is of the original amount. The amount of change is what percent of the original amount?

Examples

  • A shirt's price increased from 20 dollars to 25 dollars. Find the percent change. The change is 25βˆ’20=525 - 20 = 5 dollars. The percent change is β…”520=0.25β…”{5}{20} = 0.25, which is a 25% increase.
  • A car's value decreased from 15,000 dollars to 12,000 dollars. Find the percent change. The change is 12,000βˆ’15,000=βˆ’3,00012,000 - 15,000 = -3,000 dollars. The percent change is β…”βˆ’300015000=βˆ’0.20β…”{-3000}{15000} = -0.20, a 20% decrease.

Section 6

Discount and Mark-up

Property

Discount

amountΒ ofΒ discount=discountΒ rateβ‹…originalΒ price\text{amount of discount} = \text{discount rate} β‹… \text{original price}
saleΒ price=originalΒ priceβˆ’discountΒ amount\text{sale price} = \text{original price} - \text{discount amount}

Mark-up

amountΒ ofΒ mark-up=mark-upΒ rateβ‹…originalΒ cost\text{amount of mark-up} = \text{mark-up rate} β‹… \text{original cost}
listΒ price=originalΒ cost+mark-up\text{list price} = \text{original cost} + \text{mark-up}

Examples

  • A coat with an original price of 200 dollars is on sale for 30% off. Find the sale price. The discount is 0.30β‹…200=600.30 β‹… 200 = 60 dollars. The sale price is 200βˆ’60=140200 - 60 = 140 dollars.

Section 7

Simple Interest

Property

If an amount of money, PP, called the principal, is invested or borrowed for a period of tt years at an annual interest rate rr, the amount of interest, II, earned or paid is:

I=PrtI = Prt

where II = interest, PP = principal, rr = rate, and tt = time.

Examples

  • Calculate the simple interest earned on 1,500 dollars at a 5% rate for 4 years. Using I=PrtI = Prt, we get I=(1500)(0.05)(4)=300I = (1500)(0.05)(4) = 300. The interest earned is 300 dollars.
  • Sam borrowed 4,000 dollars and paid 800 dollars in simple interest over 5 years. What was the interest rate? Using I=PrtI = Prt, we have 800=(4000)r(5)βž”800=20000rβž”r=0.04800 = (4000)r(5) βž” 800 = 20000r βž” r = 0.04. The rate was 4%.

Book overview

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Chapter 2: Solving Linear Equations

  1. Lesson 1

    Lesson 2.1: Use a General Strategy to Solve Linear Equations

  2. Lesson 2Current

    Lesson 2.2: Use a Problem Solving Strategy

  3. Lesson 3

    Lesson 3: 2.3 Solve a Formula for a Specific Variable

  4. Lesson 4

    Lesson 2.4: Solve Mixture and Uniform Motion Applications

  5. Lesson 5

    Lesson 2.5: Solve Linear Inequalities

  6. Lesson 6

    Lesson 2.6: Solve Compound Inequalities

  7. Lesson 7

    Lesson 2.7: Solve Absolute Value Inequalities

Lesson overview

Expand to review the lesson summary and core properties.

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Section 1

πŸ“˜ Use a Problem Solving Strategy

New Concept

Learn a powerful, universal strategy to conquer any word problem. This method breaks down challenges into simple steps, helping you translate real-world scenarios involving numbers, percents, and simple interest into easily solvable equations.

What’s next

This strategy is your new toolkit. Next, you’ll master it step-by-step through worked examples, interactive practice cards, and challenge problems.

Section 2

Problem Solving Strategy

Property

Use a Problem Solving Strategy for word problems.

  • Step 1. Read the problem. Make sure all the words and ideas are understood.
  • Step 2. Identify what you are looking for.
  • Step 3. Name what you are looking for. Choose a variable to represent that quantity.
  • Step 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.
  • Step 5. Solve the equation using proper algebra techniques.
  • Step 6. Check the answer in the problem to make sure it makes sense.
  • Step 7. Answer the question with a complete sentence.

Examples

  • The number of apples in a basket is 5 more than twice the number of oranges. If there are 31 apples, how many oranges are there? Let oo be the number of oranges. The equation is 2o+5=312o + 5 = 31. Solving gives 2o=262o = 26, so o=13o = 13. There are 13 oranges.
  • The sum of five times a number and ten is 55. Find the number. Let nn be the number. The equation is 5n+10=555n + 10 = 55. Solving gives 5n=455n = 45, so n=9n = 9. The number is 9.

Section 3

Consecutive Integers

Property

Consecutive integers are integers that immediately follow each other. If the first integer is nn, the next are n+1,n+2,…n+1, n+2, …

Consecutive even integers are even integers that immediately follow one another. If the first even integer is nn, the next are n+2,n+4,…n+2, n+4, …

Consecutive odd integers are odd integers that immediately follow one another. If the first odd integer is nn, the next are n+2,n+4,…n+2, n+4, …

Section 4

Percent Applications

Property

To solve percent equations, translate English sentences into algebraic equations. Be sure to change the given percent to a decimal before you use it in the equation.

(a) What number is 45% of 84? translates to n=0.45β‹…84n = 0.45 β‹… 84.

(b) 8.5% of what amount is 4.76 dollars? translates to 0.085β‹…n=4.760.085 β‹… n = 4.76.

Section 5

Percent Change

Property

To find the percent change:

  • Step 1. Find the amount of change.
change=newΒ amountβˆ’originalΒ amount\text{change} = \text{new amount} - \text{original amount}
  • Step 2. Find what percent the amount of change is of the original amount. The amount of change is what percent of the original amount?

Examples

  • A shirt's price increased from 20 dollars to 25 dollars. Find the percent change. The change is 25βˆ’20=525 - 20 = 5 dollars. The percent change is β…”520=0.25β…”{5}{20} = 0.25, which is a 25% increase.
  • A car's value decreased from 15,000 dollars to 12,000 dollars. Find the percent change. The change is 12,000βˆ’15,000=βˆ’3,00012,000 - 15,000 = -3,000 dollars. The percent change is β…”βˆ’300015000=βˆ’0.20β…”{-3000}{15000} = -0.20, a 20% decrease.

Section 6

Discount and Mark-up

Property

Discount

amountΒ ofΒ discount=discountΒ rateβ‹…originalΒ price\text{amount of discount} = \text{discount rate} β‹… \text{original price}
saleΒ price=originalΒ priceβˆ’discountΒ amount\text{sale price} = \text{original price} - \text{discount amount}

Mark-up

amountΒ ofΒ mark-up=mark-upΒ rateβ‹…originalΒ cost\text{amount of mark-up} = \text{mark-up rate} β‹… \text{original cost}
listΒ price=originalΒ cost+mark-up\text{list price} = \text{original cost} + \text{mark-up}

Examples

  • A coat with an original price of 200 dollars is on sale for 30% off. Find the sale price. The discount is 0.30β‹…200=600.30 β‹… 200 = 60 dollars. The sale price is 200βˆ’60=140200 - 60 = 140 dollars.

Section 7

Simple Interest

Property

If an amount of money, PP, called the principal, is invested or borrowed for a period of tt years at an annual interest rate rr, the amount of interest, II, earned or paid is:

I=PrtI = Prt

where II = interest, PP = principal, rr = rate, and tt = time.

Examples

  • Calculate the simple interest earned on 1,500 dollars at a 5% rate for 4 years. Using I=PrtI = Prt, we get I=(1500)(0.05)(4)=300I = (1500)(0.05)(4) = 300. The interest earned is 300 dollars.
  • Sam borrowed 4,000 dollars and paid 800 dollars in simple interest over 5 years. What was the interest rate? Using I=PrtI = Prt, we have 800=(4000)r(5)βž”800=20000rβž”r=0.04800 = (4000)r(5) βž” 800 = 20000r βž” r = 0.04. The rate was 4%.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 2: Solving Linear Equations

  1. Lesson 1

    Lesson 2.1: Use a General Strategy to Solve Linear Equations

  2. Lesson 2Current

    Lesson 2.2: Use a Problem Solving Strategy

  3. Lesson 3

    Lesson 3: 2.3 Solve a Formula for a Specific Variable

  4. Lesson 4

    Lesson 2.4: Solve Mixture and Uniform Motion Applications

  5. Lesson 5

    Lesson 2.5: Solve Linear Inequalities

  6. Lesson 6

    Lesson 2.6: Solve Compound Inequalities

  7. Lesson 7

    Lesson 2.7: Solve Absolute Value Inequalities