To solve the system of equations
\[ \begin{aligned} &-5x + 4y = 3 \quad (1) \ &x = 2y - 15 \quad (2) \end{aligned} \]
we can substitute equation (2) into equation (1).
Substituting \(x\) from (2) into (1):
\[ -5(2y - 15) + 4y = 3 \]
Now, we simplify this equation:
\[ -10y + 75 + 4y = 3 \]
Combine like terms:
\[ -10y + 4y + 75 = 3 \]
This simplifies to:
\[ -6y + 75 = 3 \]
Now, isolate \(y\):
\[ -6y = 3 - 75 \]
\[ -6y = -72 \]
Now divide both sides by -6:
\[ y = \frac{-72}{-6} = 12 \]
Now that we have \(y\), we can find \(x\) by substituting \(y\) back into equation (2):
\[ x = 2(12) - 15 \]
\[ x = 24 - 15 = 9 \]
Thus, the solution to the system of equations is:
\[ \begin{aligned} x &= 9 \ y &= 12 \end{aligned} \]
Final answer:
\[ x = 9, \quad y = 12 \]