\( \left\{ \begin{array} { l } { x + y \quad 1200 } \\ { ( x + 8 ) ( x + 6 ) = 1200 + 528 } \end{array} \right. \)

To solve the system, let's first rewrite the equations for clarity: 1. \( x + y = 1200 \) 2. \( (x + 8)(x + 6) = 1728 \) (since \( 1200 + 528 = 1728 \)) From the second equation, we can expand and rearrange it: \[ x^2 + 14x + 48 = 1728 \] Which simplifies to: \[ x^2 + 14x - 1680 = 0 \] Next, we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1, b = 14, c = -1680 \). Calculating the discriminant: \[ b^2 - 4ac = 14^2 - 4 \times 1 \times (-1680) = 196 + 6720 = 6916 \] Now, calculate the square root of 6916: \[ \sqrt{6916} = 83.1 \] (approximately). Substituting back into the quadratic formula: \[ x = \frac{-14 \pm 83.1}{2} \] Calculating both possible values for \( x \): 1. \( x = \frac{-14 + 83.1}{2} \approx 34.55 \) 2. \( x = \frac{-14 - 83.1}{2} \approx -48.55 \) (not valid since \( x \) should be positive). Now we can use the valid \( x \) to find \( y \): \[ y = 1200 - x \approx 1200 - 34.55 \approx 1165.45 \] Thus, the solutions are (approximately): \( x \approx 34.55 \) and \( y \approx 1165.45 \).