2026 AMC 10A Problem 7

There are three unknowns but only two equations, so we will not get a single answer; we will get a short list. Notice that $y$ and $z$ always appear together with the same number in front ($1$ in the first equation, $3$ in the second). That suggests getting rid of $x$ and treating $y + z$ as one quantity.

1. Subtract the first equation from the second. The $x$'s cancel:
$$(x + 3y + 3z) - (x + y + z) = 41 - 31,$$
$$2y + 2z = 10, \qquad\text{so}\qquad y + z = 5.$$

1. Find $x$. Put $y + z = 5$ into the first equation: $x + 5 = 31$, so $x = 26$. This is the same for every solution.

1. List the choices for $y$ and $z$. Both must be positive whole numbers (at least $1$) with $y + z = 5$:

The triples are ordered, so $(26, 1, 4)$ and $(26, 4, 1)$ count as different. That makes $4$ triples.

1. Check one. $(26, 2, 3)$: $26 + 2 + 3 = 31$ and $26 + 6 + 9 = 41$. Both equations hold.

The answer is $\boxed{\textbf{(A)}\ 4}$.

When two variables always come together in the same combination, treat that combination as a single unknown. Here the system really had only two unknowns, $x$ and $y + z$, so it could be solved exactly. The leftover freedom was just how to split $5$ into two positive parts, and a two-row table counts those without missing any.

Subtracting the equations gives $2(y+z) = 10$, so $y + z = 5$ and $x = 31 - 5 = 26$. The positive solutions of $y + z = 5$ are $(y, z) = (1,4), (2,3), (3,2), (4,1)$, giving $4$ ordered triples.

The answer is $\boxed{\textbf{(A)}\ 4}$.