What is the solution set of the following system of equations? \( \begin{array}{l}5 x-3 y-10=0 \\ x+y-7=0 \\ \{(31 / 8,25 / 8)\} \\ \{(31 / 2,45 / 2)\} \\ \{(25 / 8,31 / 8)\} \\ \{(45 / 2,31 / 2)\}\end{array} \)

To solve the system of equations, we can start by rewriting both equations in slope-intercept form: 1. From \(5x - 3y - 10 = 0\), rearranging gives us: \[3y = 5x - 10\] \[y = \frac{5}{3}x - \frac{10}{3}\] 2. From \(x + y - 7 = 0\), we rearrange to get: \[y = 7 - x\] Now, to find the solution set, we can substitute \(y\) from the second equation into the first one: \[\frac{5}{3}x - \frac{10}{3} = 7 - x\] Solving this equation step by step: 1. Multiply by 3 to eliminate the fraction: \[5x - 10 = 21 - 3x\] 2. Combine like terms: \[5x + 3x = 21 + 10\] \[8x = 31\] 3. Hence, \[x = \frac{31}{8}\] Now substitute \(x\) back into the second equation to find \(y\): \[y = 7 - \frac{31}{8}\] \[y = \frac{56}{8} - \frac{31}{8} = \frac{25}{8}\] Thus, the solution set of the system of equations is \(\{(31/8, 25/8)\}\). The answer is: \(\{(31 / 8, 25 / 8)\}\).