Eighth Grade Mathematics: Mixture, Distance-Rate-Time, and Cost Systems
Systems of linear equations are indispensable tools in professional fields including chemistry, economics, navigation, and finance. Whenever a real-world scenario involves two interconnected unknown quantities bounded by two distinct conditions—such as blending solutions of different concentrations, traveling with and against wind currents, or managing inventory revenue—a 2x2 system provides an exact mathematical resolution. In this chapter, you will master setting up, structuring, and solving classical mixture, distance-rate-time, and cost systems.
The Structural Anatomy of Word Problem Systems
Every applied system problem contains two distinct unknowns and two independent relationships. To set up a system successfully, follow a structured protocol:
The Systems Modeling Protocol:
1. Define Variables: Assign two letters with explicit units (e.g., Let x = pounds of cashews).
2. Write Equation 1: Represents the "Quantity" or "Physical Total" (e.g., total weight).
3. Write Equation 2: Represents the "Value", "Rate", or "Concentration" (e.g., total cost).
4. Choose Method: Solve using substitution or elimination based on coefficient structure.
5. Answer Contextually: State answers with appropriate physical units and verify.
Category 1: Cost and Quantity Systems
In cost systems, Equation 1 tracks the count of items purchased, while Equation 2 tracks the monetary value of those items.
Example: A theater sold 300 total tickets for $2,100.
Student tickets cost $5 each; Adult tickets cost $9 each.
Step 1: Variables:
Let s = number of student tickets sold
Let a = number of adult tickets sold
Step 2: Quantity Equation (Total tickets):
s + a = 300
Step 3: Value Equation (Total revenue):
5s + 9a = 2100
Step 4: Solve by Substitution:
From Eq 1: s = 300 - a
Substitute: 5*(300 - a) + 9a = 2100
1500 - 5a + 9a = 2100
4a = 600 ---> a = 150 adult tickets
Find s: s = 300 - 150 = 150 student tickets
Category 2: Distance, Rate, and Time (Wind and Current) Systems
When a vehicle travels in a moving medium (like a boat in a river current or an airplane in wind), the speed of the medium either assists or opposes the vehicle's speed:
With Current / Tailwind: Effective Speed = Vehicle Speed + Medium Speed (v + c)
Against Current / Headwind: Effective Speed = Vehicle Speed - Medium Speed (v - c)
Distance Formula: Distance = Speed * Time ---> Rate = Distance / Time
Example: A boat travels 48 miles downstream (with current) in 3 hours.
The return trip upstream (against current) takes 4 hours.
Downstream Rate: 48 / 3 = 16 mph ---> b + c = 16
Upstream Rate: 48 / 4 = 12 mph ---> b - c = 12
Solve by Elimination:
b + c = 16
+ b - c = 12
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2b = 28 ---> b = 14 mph (Boat speed in still water)
14 + c = 16 ---> c = 2 mph (Current speed of river)
Adding the two equations immediately eliminates the current speed c, isolating the boat's speed.
Category 3: Mixture and Concentration Systems
In chemistry and agriculture, solutions of different percentages are blended to achieve a desired concentration:
Mixture Formula:
(Amount 1 * Conc 1) + (Amount 2 * Conc 2) = (Total Amount * Final Conc)
Always convert percentages to decimals (e.g., 20% becomes 0.20) before setting up the value equation.
Chapter Exercises
Exercise 1: A school concession stand sells hot dogs for 3 dollars each and pretzels for 2 dollars each. During a basketball game, the stand sold 180 total food items and collected 440 dollars in revenue. Write a system of linear equations modeling this situation and solve for the number of hot dogs and pretzels sold.
Exercise 2: An airplane travels 1800 miles from City A to City B with a tailwind in 3 hours. On the return trip against the same wind, the flight takes 3 hours and 36 minutes (which is 3.6 hours). Find the speed of the airplane in calm air and the speed of the wind.
Exercise 3: A laboratory scientist needs to create 50 liters of an 18 percent acid solution by mixing a 10 percent acid solution with a 30 percent acid solution. Set up a system of equations and calculate how many liters of each acid solution must be combined.
Solutions and Explanations
Solution 1: Let h represent the number of hot dogs sold, and let p represent the number of pretzels sold. The first equation represents the total quantity of items sold: h + p = 180. The second equation represents the total dollar revenue generated: 3h + 2p = 440. To solve using substitution, isolate h in the first equation: h = 180 - p. Substitute this expression into the second equation: 3*(180 - p) + 2p = 440. Distribute the 3: 540 - 3p + 2p = 440. Combine like terms: 540 - p = 440. Subtract 540 from both sides: -p = -100, which yields p = 100 pretzels. Now find h using h = 180 - 100 = 80 hot dogs. Verifying the revenue: 80 hot dogs at 3 dollars is 240 dollars, and 100 pretzels at 2 dollars is 200 dollars; 240 + 200 = 440 dollars. The stand sold 80 hot dogs and 100 pretzels.
Solution 2: Let a represent the speed of the airplane in calm air in miles per hour, and let w represent the speed of the wind in miles per hour. When flying with the tailwind, the effective ground speed is a + w. Using Rate = Distance / Time, the tailwind rate is 1800 / 3 = 600 miles per hour, giving the first equation: a + w = 600. When flying against the headwind, the effective ground speed is a - w. The headwind rate is 1800 / 3.6 = 500 miles per hour, giving the second equation: a - w = 500. Now solve the system using elimination by adding the two equations vertically: (a + a) + (w - w) = 600 + 500, which yields 2a = 1100. Dividing by 2 gives a = 550 miles per hour for the airplane's speed in calm air. Back-substitute into the first equation: 550 + w = 600, which yields w = 50 miles per hour for the wind speed. The plane travels at 550 mph in still air and the wind speed is 50 mph.
Solution 3: Let x represent the volume in liters of the 10 percent acid solution, and let y represent the volume in liters of the 30 percent acid solution. The first equation represents the total volume of liquid: x + y = 50. The second equation represents the total volume of pure acid: 0.10x + 0.30y = 0.18*(50). Calculating 0.18 * 50 gives 9 liters of pure acid, so the second equation is 0.10x + 0.30y = 9. Clear decimals by multiplying the second equation by 100: 10x + 30y = 900, which simplifies by dividing by 10 to x + 3y = 90. Now subtract the first equation (x + y = 50) from this simplified equation: (x - x) + (3y - y) = 90 - 50, which gives 2y = 40. Dividing by 2 yields y = 20 liters of the 30 percent solution. Now substitute into x + y = 50: x + 20 = 50, which gives x = 30 liters of the 10 percent solution. To verify: 30 liters of 10% acid provides 3 liters of acid, and 20 liters of 30% acid provides 6 liters of acid; 3 + 6 = 9 liters of pure acid in a total of 50 liters, which is exactly 9 / 50 = 18%. The scientist must combine 30 liters of the 10% solution and 20 liters of the 30% solution.