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Some chickens and rabbits have a total of 100 feet. If each chicken is exchanged for a rabbit and each rabbit is exchanged for a chicken there would be a total of 86 feet. How many chickens are there? How many rabbits? I know that for a problem like: The Mathematics Stack Exchange camp is holding a party at their HQ in Las Vegas. They want to save energy by carpooling. They are able to fit 450 people into 80 vehicles that are either 5-seat sedans or 7-seat minivans. How many sedans and how many minivans are they allocating? It would be $$s+m=80$$ $$5s+7m=450$$ $$s=80-m$$ $$7m-5m+400=450$$ $$2m=50$$ $$m=25$$ $$80-25=55$$ $$s=55$$

But this one is exchanging. I need the equation to solve this problem. If you give the answer also, I don't mind, but I only need the equation.

thunderbolt
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suomynonA
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1 Answers1

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Let $x$ be the number of chickens, and $y$ the number of rabbits.

Then (if we assume that the members of each species have the regulation number of feet) we have $$2x+4y=100.$$ After the hypothetical species change operations, we have $$4x+2y=86.$$ Solve.

Remark: We can do it without algebra. We lost $14$ feet in the interchanges, so there must have been initially $7$ more rabbits than chickens. Let us "guess" that there were initially $0$ chickens, and therefore $7$ rabbits. Let us check, maybe we got lucky.

Unfortunately, that's $28$ feet, well short of the $100$. So add chicken/rabbit couples ($6$ feet each) until we have $100$ feet. Thus we need to add is $12$ couples. So initially there were $7$ chickens and $19$ rabbits.

Systematic versions of the idea we used were taught for many centuries, starting in China. They persisted in Western European school curricula until the early nineteenth century.

André Nicolas
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