To eliminate the x term: We'll multiply the first equation by \(3\) and the second equation by \(-4\). Hence:
\(\left \{ 3\times(4x + 5y) = 3\times7 \atop (-4)\times(3x-2y) = (-4)\times(-12)} \right\iff \left \{ 12x+15y=21 \atop -12x+8y=48 \right.\)
Adding them, we'll obtain:
\(0x+23y=69\iff y=\dfrac{69}{23}\iff \boxed{y=3}\)
Substituting the value obtained in the first equation:
\(4x + 5y = 7\Longrightarrow 4x+5\cdot3=7\iff 4x+15=7\iff 4x=-8\iff\\\\x=\dfrac{-8}{4}\iff \boxed{x=-2}\)
