What is the value of $a+b+c+d$, if

$$
\begin{gathered}
6 a+2 b=3848 \\
6 c+3 d=4410 \\
a+3 b+2 d=3080
\end{gathered}
$$
I. Solution. At first glance, we have 4 unknowns, and to determine them, the given 3 equations are clearly insufficient. However, the problem only asks for the single value of $(a+b+c+d)$. We can still provide an answer by forming an equation in which the coefficients of $a, b, c$, and $d$ are equal, just like in the expression $a+b+c+d$, similar to one of the known methods of elimination.

Let $\lambda_{1}$, $\lambda_{2}$, and $\lambda_{3}$ denote the unknown numbers by which we multiply (1), (2), and (3) respectively, and then sum their products to achieve this. The equation is:

$$
\left(6 \lambda_{1}+\lambda_{3}\right) a+\left(2 \lambda_{1}+3 \lambda_{3}\right) b+6 \lambda_{2} c+\left(3 \lambda_{2}+2 \lambda_{3}\right) d=3848 \lambda_{1}+4410 \lambda_{2}+3080 \lambda_{3}
$$

We write three equations for our new unknowns from the requirement that 1. the coefficients of $a$ and $b$ are equal, then 2. the coefficients of $b$ and $c$, and finally 3. the coefficients of $c$ and $d$ are also equal, and we immediately rearrange them in the usual way:

$$
\begin{aligned}
6 \lambda_{1}+\lambda_{3} & =2 \lambda_{1}+3 \lambda_{3}, & 2 \lambda_{1}-\lambda_{3} & =0 \\
2 \lambda_{1}+3 \lambda_{3} & =6 \lambda_{2}, & 2 \lambda_{1}-6 \lambda_{2}+3 \lambda_{3} & =0 \\
6 \lambda_{2} & =3 \lambda_{2}+2 \lambda_{3}, & 3 \lambda_{2}-2 \lambda_{3} & =0
\end{aligned}
$$

From (5), we express $\lambda_{1}$, and from (7), $\lambda_{2}$:

$$
\lambda_{1}=\frac{\lambda_{3}}{2}, \quad \lambda_{2}=\frac{2 \lambda_{3}}{3}
$$

Substituting these into (6), we expect a one-variable equation for $\lambda_{3}$:

$$
2 \cdot \frac{\lambda_{3}}{2}-6 \frac{2 \lambda_{3}}{3}+3 \lambda_{3}=0
$$

However, the coefficient of $\lambda_{3}$ here turns out to be 0, so it also disappears. On the other hand, the equation

$$
0 \cdot \lambda_{3}=0
$$

is satisfied by any $\lambda_{3}$, so we can accept a number different from 0, and then from (8) we also get a value for $\lambda_{1}$ and $\lambda_{2}$. Thus, we only get the ratio of the three multipliers, but this is sufficient for our purpose:

$$
\frac{\lambda_{3}}{2}, \frac{2 \lambda_{3}}{3}, \lambda_{3}
$$

Choose $\lambda_{3}$ so that the coefficient of $c$ in (4) is already 1:

$$
6 \lambda_{2}=4 \lambda_{3}=1
$$

that is,

$$
\lambda_{2}=\frac{1}{6}, \quad \lambda_{3}=\frac{1}{4}, \quad \lambda_{1}=\frac{1}{8}
$$

With these, (4) indeed becomes:

$a+b+c+d=481+735+770=1986$, and this answers the question of the problem.

By choosing $\lambda_{3}$ differently, we arrive at the same result.

Remark. 1. Most of the submitted solutions formed new and new equations with more or less "intuition," which is greatly facilitated by the small integer coefficients 1, 2, 3, and 6 in equations (1)-(3). However, there were also some solutions that reached the goal according to the following II. solution.

2. Those who have already learned about systems of equations see that there is a "dependence" in the system (5)-(7), hence it has infinitely many solutions. For example, subtracting (5) from (6) yields $6 \lambda_{2}-4 \lambda_{3}=0$, and (7) is not a new equation relative to this.

However, in our case, this was fortunate. If, for example, $17 \lambda_{3}=0$ had resulted at (8), only $\lambda_{3}=0, \lambda_{1}=0$, and $\lambda_{2}=0$ would have satisfied it, and the degeneration of (4) would have prevented us from reaching our goal. The fact that the right-hand side of each of (5)-(7) is 0 means that such a linear system is called homogeneous.

II. Solution. Since at most 3 unknowns can be determined from 3 equations, consider $d$ as a parameter and calculate the expressions for $a, b, c$ as functions of $d$. From (2) straightforwardly:

$$
c=\frac{4410-3 d}{6}=735-\frac{d}{2}
$$

$a$ and $b$ from the two-variable system (1), (3):

$$
a=336.5+\frac{d}{4}, \quad b=914.5-\frac{3 d}{4}
$$

With these,

$$
a+b+c=1986-d
$$

so $a+b+c+d=1986$.

Remark. The system (1)-(3) is indeterminate for 4 unknowns, but it is determinate for their sum. We could also uniquely determine many other expressions of $a, b, c, d$, but certainly not any expression of them.