One single equation containing three variables can't be solved. Ahmes did not use algebra. He used algorithms or step by step procedures that required a good knowledge of conversions of 1 into a variety of unit fraction series - the more the better. Beginners had to learn those algorithms and conversions. If they excelled - when they could perform the arithmetical operations 'in their sleep' - they were taken to the higher levels of challenging problems and hidden theorems enveloped in the same numbers. We can reach the advanced levels and unveil the secrets promised by Ahmes in the opening lines of the Rhind Mathematical Papyrus (copy of a lost scroll of the Middle Kingdom from around 1850 BC) playing with the numbers, recognizing to our surprise that the same problems could have been used for beginners and 'college' classes and 'graduates' alike. Picture that we could use the same math textbook from primary school to seminary ...
RMP 36 - a pair of granaries
3 '3 '5 times '4 '53 '106 '212 equals 1
Imagine a square whose diagonal measures 2 royal cubits. Its area measures 2 square cubits.
Imagine a circle whose diagonal measures 3 royal cubits. Using the value '135 of 424 for pi, the area of the circle measures 7 '15 square cubits.
6/2 (plus 22/7) 28/9 ... 424/135
Build a granary on the square (inner diagonal 2 royal cubits). Build a granary on the circle (inner diameter 3 royal cubits).
Fill the round granary to a height of 1 royal cubit. You will need 53 quadruple-hekat or 212 hekat of barley. Fill 212 hekat of grain in the square granary. The barley will reach a height of 3 '3 '5 royal cubits.
Fill the square granary to a height of 1 royal cubit. You will need 2 cubic cubits or 3 khar or 15 quadruple-hekat or 60 hekat of barley. Fill 60 hekat of barley in the round granary. It will reach a height of '4 '53 '106 '212 royal cubits.