\( \left.\begin{array}{r}\left.\text { 1. } \begin{array}{r}2 x+3 y=8 \\ x+2 y=5\end{array}\right\} \\ 2 a-b=7 \\ 2 a+3 b=-1\end{array}\right\} \)

To solve the system of equations, let’s start with the first set: 1. From \(2x + 3y = 8\) and \(x + 2y = 5\), we can use substitution or elimination. If we multiply the second equation by 2, we get \(2x + 4y = 10\). Subtracting the first equation from this gives \(y = 2\). Plugging \(y\) back into \(x + 2(2) = 5\) results in \(x = 1\). Now, for the second set: 2. For \(2a - b = 7\) and \(2a + 3b = -1\), we can also use elimination. Solving the first equation for \(b\) gives \(b = 2a - 7\). Substituting this into the second equation yields: \(2a + 3(2a - 7) = -1\). Simplifying and solving gives \(a = 4\) and consequently \(b = 1\). So, the solutions are \( (x, y) = (1, 2) \) and \( (a, b) = (4, 1) \).