4. When multiplying two positive numbers, one of which is 9 greater than the other, the student made a mistake by placing a number that is three less than the correct one in the tens place. When dividing that number by the smaller multiplier, the student got a quotient of 42 and a remainder of 4. What numbers did the student multiply?
Solution. Let the smaller factor be $x$, and the number obtained by the student be $y$. Then the conditions of the problem give:

$$
\begin{aligned}
& x(x+9)=y+30 \\
& y=42 x+4
\end{aligned}
$$

Solving the system for $x$ yields $x_{1}=34$ and $x_{2}=-1$. Therefore, the numbers are 34 and 43.

## III year