11. \( \left\{\begin{array}{l}y^{2}=x^{2}-9 \\ 2 y=x-3\end{array}\right. \)

To solve this system of equations, let's first rewrite the equations for clarity: 1. \( y^2 = x^2 - 9 \) 2. \( 2y = x - 3 \) or rearranged, \( x = 2y + 3 \) Now, substitute \( x \) from the second equation into the first equation: \[ y^2 = (2y + 3)^2 - 9 \] Expanding the right side: \[ y^2 = 4y^2 + 12y + 9 - 9 \] This simplifies to: \[ y^2 = 4y^2 + 12y \] Now, rearranging gives: \[ 0 = 3y^2 + 12y \] Factor out \( 3y \): \[ 0 = 3y(y + 4) \] This gives us two potential solutions for \( y \): 1. \( y = 0 \) 2. \( y = -4 \) For each of these, we can substitute back to find \( x \): 1. For \( y = 0 \): \[ x = 2(0) + 3 = 3 \] Thus, one solution is \( (3, 0) \). 2. For \( y = -4 \): \[ x = 2(-4) + 3 = -8 + 3 = -5 \] Thus, the second solution is \( (-5, -4) \). The complete solution set to the system of equations is \( (3, 0) \) and \( (-5, -4) \).