Trinomial Mystery (Problem 1)

For first pair $x_1,y_1$ ,

$\begin{aligned} x_1+y_1-9&=19\\ \implies x_1+y_1&=28\\ \end{aligned}$

For second pair $x_2,y_2$ ,

$\begin{aligned} x_2+y_2-9&=19\\ \implies x_2+y_2&=28\\ \end{aligned}$

$\implies x_1+y_1+x_2+y_2=28+28=\boxed{56}$

Other equations are just for finding values, which we don't need to.

Let $z = y-9$ . Then we have $x+z=19$ and:

$\begin{aligned} x^2 + z^2 & = 361 \\ (x+z)^2 - 2xz & = 361 \\ 19^2 - 2xz & = 361 \\ 361 - 2xz & = 361 \\ \implies xz & = 0 \\ x(y-9) & = 0 \end{aligned}$

$\implies \begin{cases} x_1 = 0 & \implies 0 + y_1 - 9 = 19 & \implies y_1 = 28 \\ y_2 = 9 & \implies x_2 + 0 = 19 & \implies x_2 = 19 \end{cases}$

Therefore $x_1 + y_1 + x_2 + y_2 = 0+28+19+9 = \boxed{56}$ .

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Note: $x^3 + z^3 = (x+z)(x^2 - xz + z^2) = (x+z)(x^2+z^2) = (x+z)^3 = 19^3 = 6859$

The equations are :

$x+(y-9)=19$

$x^2+(y-9)^2=19^2$

$x^3+(y-9)^3=19^3$ .

Solutions to these equations are

$x=0, y-9=19\implies y=28$

$y-9=0\implies y=9, x=19$ .

So, $x_1=0,x_2=19, y_1=28, y_2=9$

$\implies x_1+x_2+y_1+y_2=\boxed {56}$ .