Solving an arithmagon

One of my nephews was working on an “arithmagon” puzzle. Here’s an example puzzle:
The idea is to solve for a, b, and c such that:

a + b = 26
a + c = 33
b + c = 35

and, I guess, the idea at his young age is to try different numbers ad-hoc until you find the ones that work.

Well, of course, that’s not how I think these should be done. 🙂

In general:

we have:

a + b = X
a + c = Y
b + c = Z

Three unknowns (a, b, c) and three linear equations; we can solve for the general case:

b = X - a
c = Y - a

therefore

b + c = Z
using b = X - a and c = Y - a:
X - a + Y - a = Z
X + Y - 2a = Z
X + Y - Z = 2a

a = (1/2) * (X + Y - Z)

So we get a = 12 for the original example shown at the start, and from that (using c = Y – a and b = X – a) c = 21 and b = 14.

Easy! Sorry for ruining all the puzzles for my nephew (once he can understand algebra).

 

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