8. \( \left\{\begin{array}{l}-5 x+15 y=-35 \\ -3 y-2 x=-14\end{array}\right. \)

To solve this system of equations, let’s first simplify each equation. The first equation can be rearranged: 1. Starting with \(-5x + 15y = -35\), we can simplify it by dividing everything by -5, giving us: \(x - 3y = 7\). 2. For the second equation, \(-3y - 2x = -14\), let's rearrange it to match the form \(Ax + By = C\): \(2x + 3y = 14\) (by multiplying through by -1). Now we have the new system: \(\begin{cases} x - 3y = 7 \\ 2x + 3y = 14 \end{cases}\). Next, we can use the method of substitution or elimination to find the values of \(x\) and \(y\). Let’s add both equations to eliminate \(y\): Adding gives: \((x - 3y) + (2x + 3y) = 7 + 14\) \(\Rightarrow 3x = 21 \Rightarrow x = 7\). Now substitute \(x = 7\) back into one of the original equations; using \(x - 3y = 7\): \(7 - 3y = 7 \Rightarrow -3y = 0 \Rightarrow y = 0\). Thus, the solution to the system is \(x = 7\) and \(y = 0\), or as an ordered pair: \((7, 0)\).