3. Determine the pair of real numbers $x, y$ which satisfy the system of equations:
$$
\begin{array}{l}
\frac{1}{x}+\frac{1}{y}=1 \\
\frac{2}{x}+\frac{3}{y}=4
\end{array}
$$
Solution 1: Let $a=\frac{1}{x}$ and $b=\frac{1}{y}$. We can rewrite our equations as:
$$
\begin{array}{r}
a+b=1 \\
2 a+3 b=4
\end{array}
$$

Subtracting twice the first equation from the second equation gives $b=2$. Substituting this into one of the equations gives $a=-1$. These give $x=-1$ and $y=\frac{1}{2}$.

Solution 2: Taking 3 times the first equation and subtracting the second equation gives $\frac{1}{x}=-1$, so $x=-1$. Substituting this into the first equation gives $-1+\frac{1}{y}=1$ which yields $y=\frac{1}{2}$. Thus $(x, y)=\left(-1, \frac{1}{2}\right)$