4. The difference of the squares of two numbers is 6, and if each of these numbers is decreased by 2, then the difference of their squares becomes equal to 18. What is the sum of these numbers?
Answer: -2.

## Solution.

Given:

\[
\begin{aligned}
& a^{2}-b^{2}=6 \\
& (a-2)^{2}-(b-2)^{2}=18
\end{aligned}
\]

There are different ways to proceed.

## Method 1.

\((a-2)^{2}-(b-2)^{2}=a^{2}-4a+4-b^{2}+4b-4=a^{2}-b^{2}-4(a-b)\). Since from the first condition \(a^{2}-b^{2}=6\), we get \(6-4(a-b)=18\). Hence, \(a-b=-3\), and from the first equation, we get \(a+b=-2\).

## Method 2.

\[
\begin{aligned}
& a^{2}-b^{2}=(a-b)(a+b)=6 \\
& (a-2)^{2}-(b-2)^{2}=(a-b)(a+b-4)=18
\end{aligned}
\]

Clearly, all factors in the given equations are not zero. Dividing the second equation by the first and denoting the desired sum \(a+b=x\), we get \(\frac{x-4}{x}=3\), from which \(x=-2\).

## Grading Criteria.

- Correct solution - 7 points.
- Obtained the equation \(\frac{a+b-4}{a+b}=3\) but made an arithmetic error in further solution - 4 points.
- Correctly found the difference \(a-b\), but no further progress - 3 points.
- Made a sign error in finding \(a+b\) or \(a-b\) - no more than 3 points.
- Only the correct answer without justification - 2 points.