1. Two trucks were transporting fertilizers, making the same number of trips. It turned out that 4 tons less could be loaded onto the first truck and 3 tons less onto the second truck than planned, so each truck had to make 10 extra trips. As a result, the first truck transported 60 tons more than the second, as planned. How much fertilizer was loaded onto each truck and how many trips were made?
# Solution:

Let $x, y$ - capacity, $t-$ number of trips as planned.

$$
\left\{\begin{array}{l}
x t=(x-4)(t+10), \\
y t=(y-3)(t+10), \Leftrightarrow\left\{\begin{array} { l } 
{ 1 0 x - 4 t = 4 0 } \\
{ \quad x t - y t = 6 0 }
\end{array} \quad \left\{\begin{array}{l}
10 y-3 t=30, \\
(x-y) t=60
\end{array} \Rightarrow 10(x-y)-t=10 \Rightarrow t^{2}+10 t-600=0\right.\right.
\end{array}\right.
$$

From here $t_{1}=20, t_{2}=-30-$ extraneous root; $x=12, y=9$. The first truck was loaded with 8 tons, the second with 6 tons, and 30 trips were completed.

Answer: 8 tons, 6 tons, 30 trips.