Learn on PengiOpenstax Elementary Algebra 2EChapter 5: Systems of Linear Equations

Lesson 5: Solve Mixture Applications with Systems of Equations

In this lesson from OpenStax Elementary Algebra 2E, students learn to solve mixture applications and interest problems by setting up and solving systems of two linear equations with two variables. Using organized tables, learners translate real-world scenarios involving tickets, coins, and percent interest into systems and apply methods such as elimination and substitution to find solutions. The lesson builds on earlier single-variable mixture strategies and extends them to a two-variable approach covered in Chapter 5.

Section 1

📘 Solve Mixture Applications with Systems of Equations

New Concept

We'll tackle real-world mixture and interest problems by translating them into systems of two linear equations. This powerful method lets us find unknown quantities, like numbers of coins or investment amounts, using algebra.

What’s next

Next, we'll dive into worked examples. You'll use our interactive tables to set up and solve systems of equations for various mixture scenarios.

Section 2

Ticket and Coin Problems

Property

Use a table to organize items by Number, Value, and Total Value. Two variables represent the number of each item (e.g., nickels and dimes). The system of equations is formed from the total number of items and the total monetary value. For a problem with two types of items, item 1 and item 2, the system is:

{x1+x2=Total Number of Items(value1)x1+(value2)x2=Total Value\begin{cases} x_1 + x_2 = \text{Total Number of Items} \\ (\text{value}_1)x_1 + (\text{value}_2)x_2 = \text{Total Value} \end{cases}

Examples

  • A theater sold 250 tickets, earning 4500 dollars. Adult tickets cost 20 dollars and child tickets cost 10 dollars. Let aa be adult tickets and cc be child tickets. The system is a+c=250a + c = 250 and 20a+10c=450020a + 10c = 4500. Solving gives a=200a=200 and c=50c=50.
  • Dev has a collection of dimes and quarters with a total value of 10.55 dollars. The number of dimes is five more than twice the number of quarters. Let dd be dimes and qq be quarters. The system is 0.10d+0.25q=10.550.10d + 0.25q = 10.55 and d=2q+5d = 2q + 5. Solving gives d=47d=47 and q=21q=21.

Section 3

General Mixture Problems

Property

To solve mixture problems, like trail mix, use a table to track the number of pounds (or other units), the value per unit, and the total value for each component and the final mixture. The system of equations comes from the total quantity and the total value.

{n+c=Total Pounds of Mixture(costn)n+(costc)c=(costmix)(Total Pounds)\begin{cases} n + c = \text{Total Pounds of Mixture} \\ (\text{cost}_n)n + (\text{cost}_c)c = (\text{cost}_{\text{mix}})(\text{Total Pounds}) \end{cases}

Examples

  • Carson wants to make 40 pounds of trail mix costing 6 dollars per pound by mixing nuts at 8 dollars per pound and chocolate chips at 2 dollars per pound. Let nn be pounds of nuts and cc be pounds of chips. The system is n+c=40n+c=40 and 8n+2c=6(40)8n+2c=6(40). He should use 26.67 pounds of nuts and 13.33 pounds of chips.
  • A coffee shop wants to make a 20-pound blend costing 15 dollars per pound using a coffee that costs 12 dollars per pound and another that costs 16 dollars per pound. Let xx and yy be the pounds of each coffee. The system is x+y=20x+y=20 and 12x+16y=15(20)12x+16y=15(20). They should use 5 pounds of the first coffee and 15 pounds of the second.

Section 4

Mixture with Percent Concentration

Property

For mixture problems involving percent concentrations, use a table with columns for the number of units, concentration (as a decimal), and the total amount of the pure substance. The system of equations is derived from the total volume and the total amount of the dissolved substance.

{x+y=Total Volume(conc1)x+(conc2)y=(concfinal)(Total Volume)\begin{cases} x + y = \text{Total Volume} \\ (\text{conc}_1)x + (\text{conc}_2)y = (\text{conc}_{\text{final}})(\text{Total Volume}) \end{cases}

Examples

  • Sasheena needs to make 300 ml of a 50% solution of sulfuric acid. The lab has 20% and 60% solutions. Let xx be the ml of 20% solution and yy be the ml of 60% solution. The system is x+y=300x+y=300 and 0.20x+0.60y=0.50(300)0.20x+0.60y=0.50(300). She should mix 75 ml of the 20% solution and 225 ml of the 60% solution.
  • Jotham needs 50 liters of a 40% alcohol solution. He has a 25% and a 75% solution available. Let xx be liters of the 25% and yy be liters of the 75% solution. The system is x+y=50x+y=50 and 0.25x+0.75y=0.40(50)0.25x+0.75y=0.40(50). He should mix 35 liters of the 25% and 15 liters of the 75% solution.

Section 5

Solve Interest Applications

Property

The formula to model interest applications is I=PrtI = Prt. Interest, II, is the product of the principal, PP, the rate, rr, and the time, tt. For one year, t=1t=1. The system of equations comes from the Principal column and the Interest column.

{s+b=Total Principal(rates)s+(rateb)b=Total Interest\begin{cases} s + b = \text{Total Principal} \\ (\text{rate}_s)s + (\text{rate}_b)b = \text{Total Interest} \end{cases}

Examples

  • Adnan has 60,000 dollars to invest and hopes to earn 6% interest per year. He puts some money into a stock fund that earns 8% per year and the rest into bonds that earn 3% per year. Let ss be the amount in stocks and bb be the amount in bonds. The system is s+b=60000s+b=60000 and 0.08s+0.03b=0.06(60000)0.08s+0.03b=0.06(60000). He should put 36,000 dollars in stocks and 24,000 dollars in bonds.
  • Rosie owes 50,000 dollars on two student loans. The interest rate on her bank loan is 8.5% and the interest rate on the federal loan is 4.5%. The total interest she paid last year was 3050 dollars. Let bb be the bank loan and ff be the federal loan. The system is b+f=50000b+f=50000 and 0.085b+0.045f=30500.085b+0.045f=3050. The bank loan is 20,000 dollars and the federal loan is 30,000 dollars.

Book overview

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Chapter 5: Systems of Linear Equations

  1. Lesson 1

    Lesson 1: Solve Systems of Equations by Graphing

  2. Lesson 2

    Lesson 2: Solving Systems of Equations by Substitution

  3. Lesson 3

    Lesson 3: Solve Systems of Equations by Elimination

  4. Lesson 4

    Lesson 4: Solve Applications with Systems of Equations

  5. Lesson 5Current

    Lesson 5: Solve Mixture Applications with Systems of Equations

  6. Lesson 6

    Lesson 6: Graphing Systems of Linear Inequalities

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Solve Mixture Applications with Systems of Equations

New Concept

We'll tackle real-world mixture and interest problems by translating them into systems of two linear equations. This powerful method lets us find unknown quantities, like numbers of coins or investment amounts, using algebra.

What’s next

Next, we'll dive into worked examples. You'll use our interactive tables to set up and solve systems of equations for various mixture scenarios.

Section 2

Ticket and Coin Problems

Property

Use a table to organize items by Number, Value, and Total Value. Two variables represent the number of each item (e.g., nickels and dimes). The system of equations is formed from the total number of items and the total monetary value. For a problem with two types of items, item 1 and item 2, the system is:

{x1+x2=Total Number of Items(value1)x1+(value2)x2=Total Value\begin{cases} x_1 + x_2 = \text{Total Number of Items} \\ (\text{value}_1)x_1 + (\text{value}_2)x_2 = \text{Total Value} \end{cases}

Examples

  • A theater sold 250 tickets, earning 4500 dollars. Adult tickets cost 20 dollars and child tickets cost 10 dollars. Let aa be adult tickets and cc be child tickets. The system is a+c=250a + c = 250 and 20a+10c=450020a + 10c = 4500. Solving gives a=200a=200 and c=50c=50.
  • Dev has a collection of dimes and quarters with a total value of 10.55 dollars. The number of dimes is five more than twice the number of quarters. Let dd be dimes and qq be quarters. The system is 0.10d+0.25q=10.550.10d + 0.25q = 10.55 and d=2q+5d = 2q + 5. Solving gives d=47d=47 and q=21q=21.

Section 3

General Mixture Problems

Property

To solve mixture problems, like trail mix, use a table to track the number of pounds (or other units), the value per unit, and the total value for each component and the final mixture. The system of equations comes from the total quantity and the total value.

{n+c=Total Pounds of Mixture(costn)n+(costc)c=(costmix)(Total Pounds)\begin{cases} n + c = \text{Total Pounds of Mixture} \\ (\text{cost}_n)n + (\text{cost}_c)c = (\text{cost}_{\text{mix}})(\text{Total Pounds}) \end{cases}

Examples

  • Carson wants to make 40 pounds of trail mix costing 6 dollars per pound by mixing nuts at 8 dollars per pound and chocolate chips at 2 dollars per pound. Let nn be pounds of nuts and cc be pounds of chips. The system is n+c=40n+c=40 and 8n+2c=6(40)8n+2c=6(40). He should use 26.67 pounds of nuts and 13.33 pounds of chips.
  • A coffee shop wants to make a 20-pound blend costing 15 dollars per pound using a coffee that costs 12 dollars per pound and another that costs 16 dollars per pound. Let xx and yy be the pounds of each coffee. The system is x+y=20x+y=20 and 12x+16y=15(20)12x+16y=15(20). They should use 5 pounds of the first coffee and 15 pounds of the second.

Section 4

Mixture with Percent Concentration

Property

For mixture problems involving percent concentrations, use a table with columns for the number of units, concentration (as a decimal), and the total amount of the pure substance. The system of equations is derived from the total volume and the total amount of the dissolved substance.

{x+y=Total Volume(conc1)x+(conc2)y=(concfinal)(Total Volume)\begin{cases} x + y = \text{Total Volume} \\ (\text{conc}_1)x + (\text{conc}_2)y = (\text{conc}_{\text{final}})(\text{Total Volume}) \end{cases}

Examples

  • Sasheena needs to make 300 ml of a 50% solution of sulfuric acid. The lab has 20% and 60% solutions. Let xx be the ml of 20% solution and yy be the ml of 60% solution. The system is x+y=300x+y=300 and 0.20x+0.60y=0.50(300)0.20x+0.60y=0.50(300). She should mix 75 ml of the 20% solution and 225 ml of the 60% solution.
  • Jotham needs 50 liters of a 40% alcohol solution. He has a 25% and a 75% solution available. Let xx be liters of the 25% and yy be liters of the 75% solution. The system is x+y=50x+y=50 and 0.25x+0.75y=0.40(50)0.25x+0.75y=0.40(50). He should mix 35 liters of the 25% and 15 liters of the 75% solution.

Section 5

Solve Interest Applications

Property

The formula to model interest applications is I=PrtI = Prt. Interest, II, is the product of the principal, PP, the rate, rr, and the time, tt. For one year, t=1t=1. The system of equations comes from the Principal column and the Interest column.

{s+b=Total Principal(rates)s+(rateb)b=Total Interest\begin{cases} s + b = \text{Total Principal} \\ (\text{rate}_s)s + (\text{rate}_b)b = \text{Total Interest} \end{cases}

Examples

  • Adnan has 60,000 dollars to invest and hopes to earn 6% interest per year. He puts some money into a stock fund that earns 8% per year and the rest into bonds that earn 3% per year. Let ss be the amount in stocks and bb be the amount in bonds. The system is s+b=60000s+b=60000 and 0.08s+0.03b=0.06(60000)0.08s+0.03b=0.06(60000). He should put 36,000 dollars in stocks and 24,000 dollars in bonds.
  • Rosie owes 50,000 dollars on two student loans. The interest rate on her bank loan is 8.5% and the interest rate on the federal loan is 4.5%. The total interest she paid last year was 3050 dollars. Let bb be the bank loan and ff be the federal loan. The system is b+f=50000b+f=50000 and 0.085b+0.045f=30500.085b+0.045f=3050. The bank loan is 20,000 dollars and the federal loan is 30,000 dollars.

Book overview

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

Continue this chapter

Chapter 5: Systems of Linear Equations

  1. Lesson 1

    Lesson 1: Solve Systems of Equations by Graphing

  2. Lesson 2

    Lesson 2: Solving Systems of Equations by Substitution

  3. Lesson 3

    Lesson 3: Solve Systems of Equations by Elimination

  4. Lesson 4

    Lesson 4: Solve Applications with Systems of Equations

  5. Lesson 5Current

    Lesson 5: Solve Mixture Applications with Systems of Equations

  6. Lesson 6

    Lesson 6: Graphing Systems of Linear Inequalities