2001 AMC 10 Problem 10

1. Compare the first two equations: $xy = 24$ and $xz = 48$. Both contain $x$. The second is twice the first, and the only difference is $y$ versus $z$. So $z$ must be twice $y$: $z = 2y$.
2. (In symbols: divide the second equation by the first. The $x$ cancels and $\frac{z}{y} = \frac{48}{24} = 2$.)
3. Put $z = 2y$ into the third equation $yz = 72$:
$$y \cdot 2y = 72, \quad\text{so}\quad y^2 = 36.$$
4. The numbers are positive, so $y = 6$.
5. Then $z = 2 \times 6 = 12$, and from $xy = 24$ we get $x = 24 \div 6 = 4$.
6. Check all three: $4 \times 6 = 24$, $4 \times 12 = 48$, $6 \times 12 = 72$. All correct.
7. The sum is $4 + 6 + 12 = 22$.

The answer is $\boxed{\textbf{(D)}\ 22}$.

Each equation multiplies two of the three unknowns. Dividing two such equations removes the letter they share, much like subtracting two equations removes a letter in a linear system. The word "positive" matters: it tells you to use $y = 6$ and not $y = -6$.

Multiply all three equations together. The left side is $xy \cdot xz \cdot yz = (xyz)^2$. The right side is $24 \times 48 \times 72 = 82944$, which is $288^2$. So $xyz = 288$. Now divide by each given product: $z = 288 \div 24 = 12$, $y = 288 \div 48 = 6$, $x = 288 \div 72 = 4$. The sum is again $22$.

Divide the second equation by the first: $\dfrac{xz}{xy} = \dfrac{48}{24}$, so $\dfrac{z}{y} = 2$ and $z = 2y$.

Substitute into the third equation: $y \cdot 2y = 72$, so $y^2 = 36$ and $y = 6$ (positive). Then $z = 12$, and from $xy = 24$ we get $x = 4$.

Therefore $x + y + z = 4 + 6 + 12 = 22$.

The answer is $\boxed{\textbf{(D)}\ 22}$.