\begin{tabular}{|c|l|}\hline & \( \begin{array}{l}\bigcirc \text { The system has no solution. } \\ \text { System A } \\ \text { The system has a unique solution: } \\ 5 x+3 y=2 \\ 4 x+3 y=1\end{array} \) \\ \( (x, y)=(\square, \square) \) \\ & The system has infinitely many solutions. \end{tabular}

The given system of equations is: 1. \( 5x + 3y = 2 \) 2. \( 4x + 3y = 1 \) To find the unique solution, we can use the elimination method. First, we can subtract the second equation from the first: \[ (5x + 3y) - (4x + 3y) = 2 - 1 \] This simplifies to: \[ x = 1 \] Now, we can substitute \( x \) back into one of the original equations to find \( y \). Using the first equation: \[ 5(1) + 3y = 2 \] leads to: \[ 5 + 3y = 2 \quad \Rightarrow \quad 3y = -3 \quad \Rightarrow \quad y = -1 \] So, the solution to the system is: \[ (x, y) = (1, -1) \] As for the other descriptions, when you see a system with no solution, it often indicates parallel lines (e.g., \( 5x + 3y = 2 \) and \( 5x + 3y = 3 \)). In contrast, infinitely many solutions arise when two equations are actually the same line, leading to an overlap. So, here we have our unique solution boxed neatly as \( (1, -1) \)!