[Daily puzzle] Tricky system of equations \ stacker news ~science

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Here is a tricky one to keep you busy while waiting for the AP to announce the election results...

Can you solve the following system of equations for real variables $x$ , $y$ and $z$

⎩ ⎨ ⎧ \frac{x}{1 + a} + \frac{y}{2 + a} + \frac{z}{3 + a} = 1 \frac{x}{1 + b} + \frac{y}{2 + b} + \frac{z}{3 + b} = 1 \frac{x}{1 + c} + \frac{y}{2 + c} + \frac{z}{3 + c} = 1

where $a$ , $b$ and $c$ are distinct real numbers?

Previous iteration: #753021 (visual and written answers in #753251 and #753272, respectively).

$x = A (1 + a) (1 + b) (1 + c)$

$y = B (2 + a) (2 + b) (2 + c)$

$z = C (3 + a) (3 + b) (3 + c)$

$\frac{x}{1 + a} + \frac{y}{2 + a} + \frac{z}{3 + a}$

$= A (1 + b) (1 + c) + B (2 + b) (2 + c) + C (3 + b) (3 + c)$

$= A (1 + b + c + b c) + B (4 + 2 b + 2 c + b c) + C (9 + 3 b + 3 c + b c)$

All that has to equal to $1$ , so we get a system of equations:

$A + 4 B + 9 C = 1$

$A + 2 B + 3 C = 0$

$A + B + C = 0$

Which if we solve gives us: $A = 0.5$ , $B = - 1$ , $C = 0.5$ .

By symmetry it's easy to show that this solutions works for all three of the above equations.

Correct!

I don't want to spoil anyone's fun, but the answer is pretty obvious on this one....

I'm pretty sure you know that I know. 😏

I'm only 42% sure.

I'm not a mathematician, but if I studied it maybe I could understand it a little.

Never too late~~

You can click through to earlier iterations, I've posted easier ones that rely on some logical reasoning rather than equations.

That's right, I also want to be good at mathematics, I have to try and not be late in learning mathematics.