One pump fills a tank in 6 hours; another takes 9. How long together? Their rates add, not their times: the answer is 3 hours 36 minutes. For the equation to describe this situation, assume constant flows, an initially empty tank, simultaneous starts, no leaks and no interference between pumps.
We use fractions and percentages and equivalent transformations. Every number here belongs to an invented teaching example.
Name the unknown together with its unit
Let t be the numerical value of their joint time in hours, t > 0. Take the tank’s capacity as one complete job. The first pump completes 1/6 of a job each hour, the second 1/9. Rate is measured in tanks per hour, not hours.
In t hours, their contributions are t/6 and t/9 tanks. The table translates words into expressions; “together they filled one tank” supplies the equality.
| Pump | Alone, h | Tank/h | In t h |
|---|---|---|---|
| A | 6 | 1/6 | t/6 |
| B | 9 | 1/9 | t/9 |
(t)/(6) + (t)/(9) = 1.
Multiply by 18: 3t + 2t = 18, so t = 18/5 = 3.6 hours. The decimal part, 0.6 hours, equals 0.6 ⋅ 60 = 36 minutes, not 60 or 6 minutes.
Check the equation and the situation
The first pump fills 3.6/6 = 0.6 tanks; the second fills 3.6/9 = 0.4. Together that is exactly one. The answer is positive and below 6 hours: adding a positive flow should beat the faster pump working alone. Joint work, OpenStax §8.8.
The average (6 + 9)/2 = 7.5 hours cannot work: it is even slower than the first pump alone. If flow changes with water level or the pumps share a bottleneck, we need another model. Algebra cannot verify those assumptions for us.
Motion: relate the times
For constant speed, s = vt: distance equals speed times time. Travelling the same distance d km at 60 rather than 80 km/h adds half an hour. Therefore:
(d)/(60) − (d)/(80) = (1)/(2).
Both quotients are hours. Multiply by 240: 4d − 3d = 120, giving d = 120 km. Check: 2 hours versus 1.5 hours, a difference of 0.5 hours. This model includes no extra stops. Motion tables and units, OpenStax §8.8.
Mixtures: identify what is conserved
We have 20 kg of syrup containing 10% sugar by mass. How many kilograms of 40% syrup should we add for a 20% mixture? Let x ≥ 0 be the added mass in kilograms. With no losses or evaporation, total mass and sugar mass are conserved:
0.10 ⋅ 20 + 0.40x = 0.20(20 + x).
The left side is sugar mass in the initial portions; the right side is sugar mass in the final mixture. Thus 2 + 0.4x = 4 + 0.2x, so x = 10 kg. Check: 2 + 4 = 6 kg of sugar in 30 kg of mixture; 6/30 = 20%. We do not add litres as if they were kilograms or assume liquid volumes always add directly.
Branch · The percentage baseA link to our earlier discussion of percentages.
Why not average the two percentages?
(10% + 40%)/2 = 25% would describe equal masses. Here the first syrup weighs 20 kg and the second 10 kg: the weaker syrup has twice the weight in the average. Multiplying each fraction by its mass recovers the sugar quantities, which can then be added.
An equation comes from a relationship: completed shares adding up, a difference of times, or conservation of mass. Name the unknown and its unit, write that relationship, solve, then put the answer back into the original conditions.