Learn on PengiYoshiwara Elementary AlgebraChapter 4: Applications of Linear Equations

Lesson 4: Problem Solving with Systems

New Concept Learn to translate complex word problems into solvable systems of equations. We'll model real world scenarios involving interest, mixtures, and motion by setting up and solving two equations with two variables.

Section 1

📘 Problem Solving with Systems

New Concept

Learn to translate complex word problems into solvable systems of equations. We'll model real-world scenarios involving interest, mixtures, and motion by setting up and solving two equations with two variables.

What’s next

Next, we'll walk through examples for interest, mixture, and motion problems. Then you'll solve them yourself in interactive practice cards.

Section 2

Solving Interest Problems

Property

To solve problems involving two investments, use the formula for simple interest, I=PrtI = Prt, where II is the interest, PP is the principal, rr is the annual interest rate, and tt is the time in years. A system of equations is created where one equation represents the sum of the principals, and the second equation represents the sum of the interests earned from each investment.

Total Principal: P1+P2=Total PP_1 + P_2 = \text{Total P}
Total Interest: I1+I2=Total II_1 + I_2 = \text{Total I}

Examples

  • Maria invests a total of 12,000 dollars in two separate accounts. One account pays 5% annual interest, and the other pays 7%. Her total interest for the year is 700 dollars. If xx is the amount at 5% and yy is the amount at 7%, the system is x+y=12000x + y = 12000 and 0.05x+0.07y=7000.05x + 0.07y = 700.

Section 3

Solving Mixture Problems

Property

The formula for mixture problems is P=rWP = rW, where PP is the part (amount of pure ingredient), rr is the rate (percent concentration), and WW is the whole (total amount of the solution). A system is formed by one equation for the total amount of the mixture and a second for the total amount of the key ingredient. Note: We cannot add percents unless they are percents of the same whole amount.

Total Amount: W1+W2=WtotalW_1 + W_2 = W_{\text{total}}
Amount of Ingredient: r1W1+r2W2=rtotalWtotalr_1W_1 + r_2W_2 = r_{\text{total}}W_{\text{total}}

Examples

  • A chemist mixes a 15% saline solution with a 40% saline solution to get 100 liters of a 25% solution. Let xx be the amount of the 15% solution and yy be the amount of the 40% solution. The system is x+y=100x + y = 100 and 0.15x+0.40y=0.25(100)0.15x + 0.40y = 0.25(100).

Section 4

Solving Motion Problems

Property

Motion problems use the formula D=RTD = RT, where DD is distance, RR is rate (speed), and TT is time. A system of equations arises when there are two objects in motion or two parts of a journey. The system is built by applying the D=RTD = RT formula to each moving object or travel segment.

Examples

  • A boat travels 48 miles upstream in 4 hours and makes the return trip downstream in 3 hours. Let bb be the boat's speed in still water and cc be the current's speed. The system is 4(bc)=484(b - c) = 48 and 3(b+c)=483(b + c) = 48.
  • Two trains leave from the same station at the same time, traveling in opposite directions. One train is 20 mph faster than the other. After 3 hours, they are 420 miles apart. Let r1r_1 and r2r_2 be their speeds. The system is r1=r2+20r_1 = r_2 + 20 and 3r1+3r2=4203r_1 + 3r_2 = 420.

Book overview

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Continue this chapter

Chapter 4: Applications of Linear Equations

  1. Lesson 1

    Lesson 1: The Distributive Law

  2. Lesson 2

    Lesson 2: Systems of Linear Equations

  3. Lesson 3

    Lesson 3: Algebraic Solution of Systems

  4. Lesson 4Current

    Lesson 4: Problem Solving with Systems

  5. Lesson 5

    Lesson 5: Point-Slope Formula

  6. Lesson 6

    Lesson 4.6: Chapter Summary and Review

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Problem Solving with Systems

New Concept

Learn to translate complex word problems into solvable systems of equations. We'll model real-world scenarios involving interest, mixtures, and motion by setting up and solving two equations with two variables.

What’s next

Next, we'll walk through examples for interest, mixture, and motion problems. Then you'll solve them yourself in interactive practice cards.

Section 2

Solving Interest Problems

Property

To solve problems involving two investments, use the formula for simple interest, I=PrtI = Prt, where II is the interest, PP is the principal, rr is the annual interest rate, and tt is the time in years. A system of equations is created where one equation represents the sum of the principals, and the second equation represents the sum of the interests earned from each investment.

Total Principal: P1+P2=Total PP_1 + P_2 = \text{Total P}
Total Interest: I1+I2=Total II_1 + I_2 = \text{Total I}

Examples

  • Maria invests a total of 12,000 dollars in two separate accounts. One account pays 5% annual interest, and the other pays 7%. Her total interest for the year is 700 dollars. If xx is the amount at 5% and yy is the amount at 7%, the system is x+y=12000x + y = 12000 and 0.05x+0.07y=7000.05x + 0.07y = 700.

Section 3

Solving Mixture Problems

Property

The formula for mixture problems is P=rWP = rW, where PP is the part (amount of pure ingredient), rr is the rate (percent concentration), and WW is the whole (total amount of the solution). A system is formed by one equation for the total amount of the mixture and a second for the total amount of the key ingredient. Note: We cannot add percents unless they are percents of the same whole amount.

Total Amount: W1+W2=WtotalW_1 + W_2 = W_{\text{total}}
Amount of Ingredient: r1W1+r2W2=rtotalWtotalr_1W_1 + r_2W_2 = r_{\text{total}}W_{\text{total}}

Examples

  • A chemist mixes a 15% saline solution with a 40% saline solution to get 100 liters of a 25% solution. Let xx be the amount of the 15% solution and yy be the amount of the 40% solution. The system is x+y=100x + y = 100 and 0.15x+0.40y=0.25(100)0.15x + 0.40y = 0.25(100).

Section 4

Solving Motion Problems

Property

Motion problems use the formula D=RTD = RT, where DD is distance, RR is rate (speed), and TT is time. A system of equations arises when there are two objects in motion or two parts of a journey. The system is built by applying the D=RTD = RT formula to each moving object or travel segment.

Examples

  • A boat travels 48 miles upstream in 4 hours and makes the return trip downstream in 3 hours. Let bb be the boat's speed in still water and cc be the current's speed. The system is 4(bc)=484(b - c) = 48 and 3(b+c)=483(b + c) = 48.
  • Two trains leave from the same station at the same time, traveling in opposite directions. One train is 20 mph faster than the other. After 3 hours, they are 420 miles apart. Let r1r_1 and r2r_2 be their speeds. The system is r1=r2+20r_1 = r_2 + 20 and 3r1+3r2=4203r_1 + 3r_2 = 420.

Book overview

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

Continue this chapter

Chapter 4: Applications of Linear Equations

  1. Lesson 1

    Lesson 1: The Distributive Law

  2. Lesson 2

    Lesson 2: Systems of Linear Equations

  3. Lesson 3

    Lesson 3: Algebraic Solution of Systems

  4. Lesson 4Current

    Lesson 4: Problem Solving with Systems

  5. Lesson 5

    Lesson 5: Point-Slope Formula

  6. Lesson 6

    Lesson 4.6: Chapter Summary and Review