Solving equations involving two sets Out of curiosity, how would one go about solving an equation involving two sets? For example,
$$ \{1, 2, 3\} = \{a + b + c, a + b - c, a - b + c, a - b - c\} $$
An intuitive solution to this is $ \{a = 2, b = 0.5, c = 0.5\} $, but is there a specific process?
 A: hint
In the right set, two elements are equal. The only possibilities are
$$a+b+c=a-b-c$$
or
$$a+b-c=a-b+c$$
which gives
$$b=\pm c$$
The equation  becomes
$$\{a-2b,a,a+2b\}=\{1,2,3\}$$
The sum gives $ 3a=6 $ and $ a=2$.
A: Notice the set on the left has three distinct elements, but the set on the right has four representations so two of those represetations are of the same number.
So we have six options.
$a+b+c = a+b-c$ and $c = 0$ and we have the values $a+b$ and $a-b$.  But that's only two different values (or fewer) and we have exactly $3$ so that's impossible.
$a+b + c = a-b+c$  and $b =0$ and have the values $a+c$ and $a-c$ and that's the same problem.
$a+b+c = a-b-c$ and $b+c= 0$ and $b=-c$ and we have the values $a, a+2b, a-2b$.  We'll get back to this.
$a+b-c = a-b + c$ and $b-c=0$ and $b=c$ and we have the values $a+2b; a; a-2b$.  That will have the same solutions as above.
$a+b-c=a-b-c$ and $b=0$ and we have the same problem we had earlier.
Or $a-b+c = a-b-c$ adn $c=0$ and we have the same problem from the very begining.
So either way we have $a,a+2b, a-2b = 1,2,3$.
$a\pm 2b, a , a\pm 2b$ are in arithmetic progression so we must have either $a-2b < a < a+2b$ and $a-2b = 1, a=2, a+2b=3$ or $a+2b < a < a-2b$ and $a+2b = 1, a=2, a-2b =3$
So we must have $a=2$ and $b =\pm 0.5$ and $c = \pm 0.5$ (so there are four sets of solutions.)
There isn't really any way to do this in general.
