Find solutions to a two-equation system with constraints on variables representing product quantities and prices | Step-by-Step Solution

Let's Tackle This System of Equations Together! 🎯

1. What We're Solving:

We have a business scenario with three products where we know:

• The total quantity of all products is 100 units
• The total cost of all products is also 100 (dollars/currency units)
• We need to find how many of each product (x, y, z) to buy, given their prices (a, b, c)

This gives us the system:

• x + y + z = 100 (quantity constraint)
• ax + by + cz = 100 (budget constraint)

2. The Approach:

This is a classic underdetermined system - we have 2 equations but 3 unknowns! This means we'll have infinitely many solutions unless we add more constraints. The key insight is learning how to:

• Express solutions in terms of parameters
• Apply the constraint that x, y, z must be positive integers
• See how different price scenarios affect our solutions

3. Step-by-Step Solution:

Step 1: Set up the parameter approach Since we have 2 equations and 3 unknowns, we can express two variables in terms of the third. Let's solve for x and y in terms of z.

Step 2: Solve the system algebraically From equation 1: x + y = 100 - z From equation 2: ax + by = 100 - cz

Now we have a 2×2 system in x and y:

• x + y = 100 - z
• ax + by = 100 - cz

Step 3: Apply Cramer's rule or substitution Using substitution: x = 100 - z - y Substituting into equation 2: a(100 - z - y) + by = 100 - cz Simplifying: 100a - az - ay + by = 100 - cz Solving for y: y(b - a) = 100 - cz - 100a + az = 100(1 - a) + z(a - c)

Therefore: y = [100(1 - a) + z(a - c)]/(b - a) And: x = 100 - z - y

Step 4: Apply constraints For valid solutions, we need:

• x, y, z > 0 (positive integers)
• The denominators can't be zero (b ≠ a)
• z can range from some minimum to maximum value

4. The Framework:

General Solution:

• z = any valid positive integer
• y = [100(1 - a) + z(a - c)]/(b - a)
• x = 100 - z - y

For Case 1 (decimal prices): You'd substitute specific values for a, b, c and find the range of valid z values.

For Case 2 (integer prices): The constraint that a, b, c are integers might make it easier to find integer solutions for x, y, z.

Analysis approach:

• 1. Choose specific values for a, b, c
• 2. Determine the valid range for z
• 3. Check which values give positive integer solutions
• 4. Verify your solutions in both original equations

5. Memory Tip: 💡

Think of this as a "budget shopping problem" - you're trying to spend exactly $100 on exactly 100 items with three different prices. The magic happens when you realize that once you decide how many of one item to buy (z), the math tells you exactly how many of the other two items you need!

Remember: Fewer equations than unknowns = parameter family of solutions!

You've got this! Try plugging in some specific price values and see what patterns emerge. What happens when the prices are very different vs. very similar?