Minimum Value of xyz

Given:

1. $y(x + z) = 19$
2. $z(y + x) = 51$

Since $x, y, z$ are natural numbers: From (1), $y(x + z) = 19$. Since 19 is a prime number, either: Case I: $y = 1$ and $x + z = 19$ Case II: $y = 19$ and $x + z = 1$ Since $x, z \ge 1$, $x + z \ge 2$, so Case II is impossible. Therefore, $y = 1$ and $x + z = 19$.

Substitute $y = 1$ into equation (2): $z(1 + x) = 51$

We have $x + z = 19 \implies x = 19 - z$. Substitute $x = 19 - z$ into $z(1 + x) = 51$: $z(1 + 19 - z) = 51$ $z(20 - z) = 51$ $20z - z^2 = 51$ $z^2 - 20z + 51 = 0$ $(z - 3)(z - 17) = 0$

So $z = 3$ or $z = 17$.

• If $z = 3$: $x = 19 - 3 = 16$. Then $xyz = 16 \times 1 \times 3 = 48$.
• If $z = 17$: $x = 19 - 17 = 2$. Then $xyz = 2 \times 1 \times 17 = 34$.

The minimum possible value of $xyz$ is $34$.