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[Daily puzzle] Tricky system of equations

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

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

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

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

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

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

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

$$A + 4B + 9C = 1$$

$$A + 2B + 3C = 0$$

$$A + B + C = 0$$

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.





