What is the solution to the system of equations \[ \left\{\begin{array}{l}2 x+3 y=1 \\ 4 x-3 y=-25\end{array}\right. \]

To solve the system of equations, we can use the method of elimination or substitution. Let's solve it step by step. First, we have the system: 1. \(2x + 3y = 1\) (Equation 1) 2. \(4x - 3y = -25\) (Equation 2) We can add both equations together to eliminate \(y\): \[ (2x + 3y) + (4x - 3y) = 1 - 25 \] This simplifies to: \[ 6x = -24 \] Now, divide both sides by 6: \[ x = -4 \] Next, we substitute \(x = -4\) back into Equation 1 to find \(y\): \[ 2(-4) + 3y = 1 \] This simplifies to: \[ -8 + 3y = 1 \] Adding 8 to both sides: \[ 3y = 9 \] Now, divide both sides by 3: \[ y = 3 \] Thus, the solution to the system of equations is: \[ \boxed{(-4, 3)} \]