For natural numbers $x$, $y$, and $z$, if $xy + yz = 19$ and $yz + xz = 51$, then the minimum possible value of $xyz$ is

Given Information:

• $xy + yz = 19$
• $yz + xz = 51$
• $x$, $y$, and $z$ are natural numbers (positive integers)

Our Goal: Find the minimum possible value of $xyz$.

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From $xy + yz = 19$, factoring out the common term $y$:

$y(x + z) = 19$

Since $x$, $y$, and $z$ are natural numbers, both $y$ and $(x + z)$ must be positive integers whose product equals 19.

Since 19 is a prime number, its only positive factor pairs are:

$(1, 19)$: meaning $y = 1$ and $x + z = 19$

$(19, 1)$: meaning $y = 19$ and $x + z = 1$

The second case $(y = 19, x + z = 1)$ is impossible because $x$ and $z$ are natural numbers (at least 1 each), so $x + z \geq 2$.

Therefore: $y = 1$ and $x + z = 19$ ... (Equation 1)

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From $yz + xz = 51$, factoring out the common term $z$:

$z(y + x) = 51$

Finding factor pairs of 51:

$51 = 1 \times 51$

$51 = 3 \times 17$

$51 = 17 \times 3$

$51 = 51 \times 1$

This gives us these possibilities:

Case A: $z = 1$ and $y + x = 51$

Case B: $z = 3$ and $y + x = 17$

Case C: $z = 17$ and $y + x = 3$

Case D: $z = 51$ and $y + x = 1$

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We know from the first constraint that $y = 1$ and $x + z = 19$.

Checking which cases are compatible:

Case A: $z = 1$ and $y + x = 51$

Since $y = 1$, we get $1 + x = 51$, so $x = 50$

Check: $x + z = 50 + 1 = 51$

$\neq 19$ This case is incompatible.

Case B: $z = 3$ and $y + x = 17$

Since $y = 1$, we get $1 + x = 17$, so $x = 16$

Check: $x + z = 16 + 3 = 19$ This case is compatible.

Solution: $x = 16, y = 1, z = 3$

Case C: $z = 17$ and $y + x = 3$

Since $y = 1$, we get $1 + x = 3$, so $x = 2$

Check: $x + z = 2 + 17 = 19$ This case is compatible.

Solution: $x = 2, y = 1, z = 17$

Case D: $z = 51$ and $y + x = 1$

Since $y = 1$, we get $1 + x = 1$, so $x = 0$

But $x$ must be a natural number (positive), so this is impossible.

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Solution 1: $x = 16, y = 1, z = 3$

$xyz = 16 \times 1 \times 3 = 48$

Solution 2: $x = 2, y = 1, z = 17$

$xyz = 2 \times 1 \times 17 = 34$

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Comparing our two valid solutions:

Solution 1: $xyz = 48$

Solution 2: $xyz = 34$

Therefore, the minimum possible value of $xyz$ is $34$.