1976 AHSME Problems/Problem 30

Problem 30

How many distinct ordered triples $(x,y,z)$ satisfy the equations \[x+2y+4z=12\] \[xy+4yz+2xz=22\] \[xyz=6\]


$\textbf{(A) }\text{none}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }6$

Solution

The first equation suggests the substitution $a = x$, $b = 2y$, and $c = 4z$. Then $x = a$, $y = b/2$, and $z = c/4$. Substituting into the given equations, we get


a + b + c = 12

ab + ac + bc = 44

abc = 48.

Then by Vieta's formulas, $a$, $b$, and $c$ are the roots of the equation \[x^3 - 12x^2 + 44x - 48 = 0,\] which factors as \[(x - 2)(x - 4)(x - 6) = 0.\] Hence, $a$, $b$, and $c$ are equal to 2, 4, and 6 in some order.

Since our substitution was not symmetric, each possible solution $(a,b,c)$ leads to a different solution $(x,y,z)$, as follows:


a | b | c | x | y | z


2 | 4 | 6 | 2 | 2 | 3/2

2 | 6 | 4 | 2 | 3 | 1

4 | 2 | 6 | 4 | 1 | 3/2

4 | 6 | 2 | 4 | 3 | 1/2

6 | 2 | 4 | 6 | 1 | 1

6 | 4 | 2 | 6 | 2 | 1/2


Hence, there are $\boxed{6}$ solutions in $(x,y,z)$. The answer is (E).

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