Lesson 8: Systems in Context — Break-Even and Mixtures

Systems earn their keep when the two equations come from a real situation. This lesson covers the two classic families: break-even problems, where a cost line and a revenue (or rival price) line cross, and mixture problems, where amounts and concentrations combine. The algebra is Lesson 7's; the craft is setting the system up so it faithfully matches the story.

Break-even: where cost meets revenue

A small manufacturer has fixed costs of $4,000 (equipment, rent) plus $12 of materials and labour per unit made. Units sell for $20 each. Cost: C = 12x + 4000. Revenue: R = 20x, for x units. The break-even point is where R = C — no profit, no loss: 20x = 12x + 4000 → 8x = 4000 → x = 500 units, at which point both lines equal $10,000. Below 500 units the cost line sits above revenue (a loss); beyond it revenue pulls ahead. At 800 units: R = $16,000, C = 12(800) + 4000 = $13,600, so profit = $2,400. Notice the margin logic: each unit contributes 20 − 12 = $8 toward the fixed costs; 4000 ÷ 8 = 500 units to cover them. That shortcut — fixed cost ÷ per-unit margin — is the same break-even answer in one step.

Break-even between two price plans

Gym A charges $50 to join plus $10 per month; Gym B charges $20 to join plus $15 per month. Which is better? It depends on how long you stay — and the system finds the tipping point: 50 + 10m = 20 + 15m → 30 = 5m → m = 6 months, where both cost $110.

The table narrates the crossing: B starts cheaper (lower joining fee), A catches up at month 6, and A's lower monthly rate wins afterwards. Interpretation is part of the answer: "join B if you'll stay under 6 months, A if longer" uses the solution; merely reporting m = 6 does not.

Mixtures: amounts and concentrations

Mixing problems track two quantities: total amount and *pure substance*. The organising table for blending x litres of a 10% salt solution with y litres of a 30% solution to make 40 litres of an 18% solution:

Amount equation: x + y = 40. Salt equation: 0.10x + 0.30y = 7.2. From the first, x = 40 − y. Substitute: 0.10(40 − y) + 0.30y = 7.2 → 4 + 0.20y = 7.2 → 0.20y = 3.2 → y = 16 L, x = 24 L. Check the salt: 0.10(24) + 0.30(16) = 2.4 + 4.8 = 7.2 ✓, and 7.2/40 = 18% ✓. Sanity: 18% lies closer to 10% than to 30%, so the mix should contain more of the weak solution — and 24 > 16 confirms it. Always ask whether the answer leans the right way.

Coins: counting objects and their value

A jar holds 30 coins, all dimes and quarters, worth $5.10. How many of each? Count: d + q = 30. Value (in dollars): 0.10d + 0.25q = 5.10. From the count, d = 30 − q. Substitute: 0.10(30 − q) + 0.25q = 5.10 → 3 + 0.15q = 5.10 → 0.15q = 2.10 → q = 14 quarters, d = 16 dimes. Check: 16 + 14 = 30 coins ✓; 16 × $0.10 + 14 × $0.25 = $1.60 + $3.50 = $5.10 ✓. Coin problems are mixtures in disguise: count is the "amount" equation, value is the "pure substance" equation, and the coin's denomination plays the role of concentration.

Common mistakes

• Averaging concentrations directly. Mixing equal parts 10% and 30% gives 20%, but mixing 24 L and 16 L does not — the concentration equation must be weighted by amounts (0.10x and 0.30y), because a litre of strong solution brings more salt than a litre of weak.
• Setting cost equal to profit. Break-even is cost = revenue. Profit is the *difference* R − C, which is zero at break-even, not one of the lines.
• Mixing dollars and cents in coin problems. 0.10d + 0.25q = 510 (cents value with dollar coefficients) is a broken equation. Pick one unit and be consistent: $5.10 with 0.10 and 0.25, or 510 with 10 and 25.
• Answering without the lean check. A mixture result that calls for more of the strong solution when the target concentration is near the weak one has failed the reasonableness test — recheck the setup before trusting the arithmetic.

Exam Experience

Question 1. A product costs C = 8x + 240 to make and sells for revenue R = 14x. The break-even point is: A. 20 units B. 40 units C. 30 units D. 240 units Answer: B. 14x = 8x + 240 → 6x = 240 → x = 40. (Equivalently, fixed cost 240 ÷ margin 6 = 40.) Option D just quotes the fixed cost; A divides by the wrong margin.

Question 2. Plan A costs $30 plus $0.05 per minute; Plan B costs $20 plus $0.10 per minute. The plans cost the same at: A. 100 minutes B. 50 minutes C. 200 minutes D. 500 minutes Answer: C. 30 + 0.05m = 20 + 0.10m → 10 = 0.05m → m = 200. Check: A: 30 + 10 = $40; B: 20 + 20 = $40 ✓.

Question 3. Which system models mixing x litres of a 20% solution with y litres of a 50% solution to obtain 30 litres of a 30% solution? A. x + y = 30 and 20x + 50y = 30 B. x + y = 30 and 0.20x + 0.50y = 9 C. x + y = 30 and 0.20x + 0.50y = 30 D. x − y = 30 and 0.20x + 0.50y = 0.30 Answer: B. The amount equation totals 30 litres; the substance equation must total 30% *of 30 litres* = 9 litres of pure substance. (Solving it gives x = 20, y = 10.) A forgets the decimals, C targets 30 litres of substance — more than the whole mixture could hold unless it were pure.

Question 4 (constructed response). A roaster blends two coffees: a $9-per-pound bean and a $13-per-pound bean. The blend must weigh 50 pounds and sell for $10.60 per pound. (a) Write a system of equations for the pounds x of the $9 bean and y of the $13 bean. (b) Solve the system. (c) Verify your blend has the correct total value.

Scoring guide (4 points).

• 1 point: the system x + y = 50 and 9x + 13y = 530 (the total value 50 × $10.60).
• 2 points: correct solving, e.g. x = 50 − y substituted: 9(50 − y) + 13y = 530 → 450 + 4y = 530 → y = 20, so 20 lb of the $13 bean and 30 lb of the $9 bean (1 point for the method, 1 for both amounts).
• 1 point: value check — 30 × $9 + 20 × $13 = $270 + $260 = $530 = 50 × $10.60 ✓, ideally with the lean observation that $10.60 is nearer $9, matching the larger share of the cheaper bean.

Deeper still: interest splits, and mixtures where the total changes

An investment system. $5,000 is split between an account paying 3% and one paying 5% simple annual interest. Total interest after one year is $190. How much is in each? Let x = amount at 3%, y = amount at 5%: x + y = 5000 and 0.03x + 0.05y = 190. Substitute x = 5000 − y: 0.03(5000 − y) + 0.05y = 190 → 150 + 0.02y = 190 → 0.02y = 40 → y = $2,000 at 5%, x = $3,000 at 3%. Check: interest 0.03(3000) + 0.05(2000) = 90 + 100 = $190 ✓. Structurally this is a mixture problem — the "concentration" is the interest rate and the "pure substance" is dollars of interest — one more costume for the same two equations. A mixture where the total changes. Start with 10 L of a 12% salt solution (so 1.2 L of salt) and boil off 2 L of *pure water*. The salt doesn't leave: the new concentration is 1.2/(10 − 2) = 1.2/8 = 15%. No system is needed — one quantity (salt) is conserved while the total moves — but students routinely try to solve it as a standard mix. The diagnostic question for any mixture problem: *what stays the same?* Standard mixing conserves both total and substance separately; evaporation conserves only the substance. Ask first, and the right equation writes itself.

One more idea: the profit function, and reading its slope

Subtract the two lines of the break-even model and a third function appears: profit P(x) = R(x) − C(x) = 20x − (12x + 4000) = 8x − 4000. This one line tells the whole commercial story. Its zero is the break-even point (8x = 4000 → x = 500 ✓). Its slope, 8, is the per-unit margin — every additional unit sold adds exactly $8 of profit, which is why the fixed cost divided by the margin (4000/8) gave the break-even in one step. Evaluating: at 650 units, P = 8(650) − 4000 = $1,200 profit; at 300 units, P = 2400 − 4000 = −$1,600, a loss stated as a negative profit. Framing any two-plan comparison this way — build the *difference* function and read its zero and slope — converts "which is better?" into "where does the difference change sign?", a question with a graph and a clean algebraic answer.