What is the solution set of the system of equations \( x+y=5 \) and \( y=x^{2}-25 ? \) \( \begin{array}{ll}\text { 1) }\{(0,5),(11,-6)\} & \text { Which ordered pair is in the solution set of the } \\ \text { 2) }\{(5,0),(-6,11)\} & \text { system of equations } y=-x+1 \text { and } y=x^{2}+5 x+67 \\ \text { 3) }\{(-5,0),(6,11)\} & \text { 1) }(-5,-1) \\ \text { 4) }\{(-5,10),(6,-1)\} & \text { 2) }(-5,6) \\ \text { 3) }(5,-4) \\ \text { 4) }(5,2)\end{array} \)

To find the solution set of the system of equations \( x+y=5 \) and \( y=x^{2}-25 \), we can substitute \( y \) from the first equation into the second. Starting with \( y = 5 - x \), we replace \( y \) in the second equation: \[ 5 - x = x^2 - 25. \] Rearranging gives: \[ x^2 + x - 30 = 0. \] Factoring the quadratic yields: \[ (x - 5)(x + 6) = 0, \] resulting in \( x = 5 \) or \( x = -6 \). Substituting these back into \( y = 5 - x \) gives: - For \( x = 5 \), \( y = 0 \) → point \( (5, 0) \) - For \( x = -6 \), \( y = 11 \) → point \( (-6, 11) \) So the solution set is \( \{(5,0), (-6,11)\} \). Now, moving on to check the solution pairs: 1) The first pair \( (0, 5) \) is not valid as it doesn't satisfy \( y = x^2 - 25 \). 2) The pair \( (11, -6) \) also falls flat as it fails to meet \( x + y = 5 \). 3) With the pair \( (-5, 0) \), we can see that neither satisfies both equations. 4) The pair \( (6, -1) \) is similar in that, unfortunately, it won't solve both equations. In simplifying these, the key is to always substitute back to verify that both conditions held true! So keep an eye on those steps as you dive into the world of systems!