[tex](2-3i)\cdot x +(4+i)\cdot y = 10 - i \\
2x - 3xi + 4y+yi = 10 - i \\ \\
\text{rezolvam sistemul de ecuatii:} \\ \\
2x + 4y = 10 \\
-3x + y = -1 ~~~\Longrightarrow ~~ \boxed{y=-1+3x~~~~~(Substitutie)}\\
--- \\ \\
2x+4(-1+3x) = 10 \\
2x - 4+12x=10 \\
14x = 10+4 \\
14x = 14 \\ \\
\displaystyle x = \frac{14}{14} =\boxed{1 }\\ \\
y = 1 + 3x = -1 + 3\cdot 1 = -1 + 3= \boxed{2}
[/tex]
