Use proportional equations to find and interpret unknown quantities

Learning goal: Represent a proportional situation with a defined equation, solve for either quantity and check the answer in context. Connect unit-rate, scale-factor and equivalent-ratio methods instead of treating them as unrelated tricks.

A classroom mixing model uses 2/5 liter of concentrate for every one liter of water. Let w represent liters of water and c represent liters of concentrate. The proportional relationship is c = (2/5)w. The constant 2/5 has the interpreted units liters of concentrate per liter of water. The equation compares ingredients; it does not say concentrate is 2/5 of the total mixture.

For three liters of water, c = (2/5) × 3 = 6/5 liters of concentrate. The combined amount is 3 + 6/5 = 21/5 liters if the model treats volumes as additive. The concentrate fraction of the total is (6/5)/(21/5) = 6/21 = 2/7. This distinction between part-to-part and part-to-whole ratios is essential. The supplied two-to-five relationship concerns concentrate and water, while the whole contains seven corresponding parts.

For 3/2 liters of water, c = (2/5) × (3/2) = 3/5 liter. For five liters of water, c = two liters. On a graph with water horizontal and concentrate vertical, (1,2/5) represents the unit-rate pair and (5,2) represents a larger batch with the same composition. The point (0,0) means no water is paired with no concentrate in this mixing rule, not that a zero divided by zero ratio can be calculated.

If the unknown is water, reverse the operation. With 4/5 liter of concentrate, solve 4/5 = (2/5)w. Divide by 2/5 to obtain w = (4/5) ÷ (2/5) = 2 liters. Check by substitution: (2/5) × 2 = 4/5. The answer is not 8/25, which would result from multiplying by 2/5 again. Think about which quantity is being found rather than automatically applying the same operation to every question.

A scale-factor approach reaches the same answer. The pair (water, concentrate) = (1,2/5) can be doubled to (2,4/5). An equivalent-ratio approach writes c/w = 2/5 for nonzero water. Multiplying both sides by w produces c = (2/5)w. These methods express the same multiplicative relationship and should agree when quantities and orientation remain consistent.

Cross multiplication can follow from equivalent fractions, but it should not replace understanding. If c/3 = 2/5, multiply both sides by 15 to get 5c = 6, so c = 6/5. The step is valid because it performs equivalent operations on an equality with nonzero denominators. Randomly crossing nearby numbers in a word problem provides no guarantee that the equation represents the intended relationship.

Use a multistep question to connect the model with available quantities. Suppose there are six liters of water and two liters of concentrate, and all mixture must follow the same ratio. Two liters of concentrate pair with five liters of water, leaving one liter of water unused. Using all six liters would require 12/5 liters of concentrate, more than the available two. The limiting ingredient determines the largest batch under the model; merely adding both available amounts would not preserve the composition.

A second context uses proportional cost: cord costs 5/2 budget units per meter with no fixed charge. Let m be meters and t be total cost; t = (5/2)m. At three meters, t = 15/2 units. With a budget of ten units, m = 10 ÷ (5/2) = four meters, provided fractional lengths can be sold and no other charges apply. State such assumptions because a fixed fee or whole-roll restriction would require a different interpretation.

Check both arithmetic and meaning. Substitute the solution, compare it with an estimate and confirm the unit. A concentrate amount of three liters for 3/2 liters of water would be implausible here because concentrate is less than half the water amount. A correct algebraic result may still require rounding or restricting a domain in another context, such as counting whole packages. Explain that final interpretation instead of presenting a bare number.

Worked batch: Five equal batches each use 3/2 liters of water. Total water is 15/2 liters, and concentrate is (2/5)(15/2) = three liters. Alternatively, one batch uses 3/5 liter concentrate and five batches use three liters. Agreement checks the proportional structure across two methods.

Worked whole: Two liters of concentrate and five liters of water make seven model liters in total. Concentrate is 2/7 of the whole, not 2/5. The five in the original ratio counts water parts, while the seven counts all parts. Identify the whole before writing a fraction of the mixture.

Worked percent step: A batch uses 5 liters of water and 2 liters of concentrate. The water is increased by 20% to 6 liters. Keeping the same ratio requires (2/5) × 6 = 12/5 = 2.4 liters of concentrate, which is also a 20% increase because proportional quantities scale by the same factor. Concentrate stays 2/7 of the whole, about 28.6%, before and after the change.

Worked model limit: A supplier charges four budget units once plus 5/2 per meter. Total cost becomes t = (5/2)m + 4, so the original proportional equation no longer applies. A ten-unit budget leaves six for cord and buys 12/5 meters under a continuous-length rule. The cost change comes from the added fee, not from a failure of the earlier arithmetic.

Guided transfer: Solve the concentrate questions using a table, an equation and one equivalent-ratio calculation. Circle the unit-rate point and explain another graph point. Compare a water-to-concentrate question with a concentrate-to-total question, naming the denominator quantity in each.

Independent performance: Design a proportional recipe or material-use model with a fractional constant. Define variables and units, provide five quantity pairs, draw a scaled graph and solve for an unknown in each direction. Include one multistep availability problem and a part-to-whole question. Check all results by substitution and explain one condition that would make the model inappropriate. A partner tests whether the interpretation of the whole matches the diagram and equations.