\( \left. \begin{array} { l }{ ( 5 , - 8 ) ; 9 x + 7 y = - 11 } \\ { - 2 x - 5 y = 30 } \end{array} \right. \)

To solve this system of equations, we can use substitution or elimination. First, let's rearrange the first equation \(9x + 7y = -11\) to express \(y\) in terms of \(x\): 1. Rearranging gives us: \[ 7y = -11 - 9x \implies y = \frac{-11 - 9x}{7} \] Now we can substitute this expression for \(y\) into the second equation: 2. The second equation is: \[ -2x - 5y = 30 \] Substitute \(y\): \[ -2x - 5\left(\frac{-11 - 9x}{7}\right) = 30 \] Simplifying gives: \[ -2x + \frac{55 + 45x}{7} = 30 \] 3. Multiplying through by 7 to eliminate the fraction leads to: \[ -14x + 55 + 45x = 210 \implies 31x + 55 = 210 \] Thus, \[ 31x = 210 - 55 \implies 31x = 155 \implies x = 5 \] 4. Use \(x = 5\) to find \(y\): \[ y = \frac{-11 - 9(5)}{7} = \frac{-11 - 45}{7} = \frac{-56}{7} = -8 \] The solution to the system is \( (5, -8) \). This solution makes sense since it satisfies both equations. To confirm, plug \( (5, -8) \) back into both original equations: - For the first equation: \[ 9(5) + 7(-8) = 45 - 56 = -11 \quad \text{(True)} \] - For the second equation: \[ -2(5) - 5(-8) = -10 + 40 = 30 \quad \text{(True)} \] Thus, the final answer is \( (5, -8) \).