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

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y=5\\y=x^{2}-25\end{array}\right.\) - step1: Substitute the value of \(y:\) \(x+x^{2}-25=5\) - step2: Move the expression to the left side: \(x+x^{2}-25-5=0\) - step3: Subtract the numbers: \(x+x^{2}-30=0\) - step4: Factor the expression: \(\left(x-5\right)\left(x+6\right)=0\) - step5: Separate into possible cases: \(\begin{align}&x-5=0\\&x+6=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=5\\&x=-6\end{align}\) - step7: Calculate: \(x=5\cup x=-6\) - step8: Rearrange the terms: \(\left\{ \begin{array}{l}x=5\\y=x^{2}-25\end{array}\right.\cup \left\{ \begin{array}{l}x=-6\\y=x^{2}-25\end{array}\right.\) - step9: Calculate: \(\left\{ \begin{array}{l}x=5\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=-6\\y=11\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=-6\\y=11\end{array}\right.\cup \left\{ \begin{array}{l}x=5\\y=0\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=-6\\y=11\end{array}\right.\cup \left\{ \begin{array}{l}x=5\\y=0\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(-6,11\right)\cup \left(x,y\right) = \left(5,0\right)\) Solve the system of equations \( y=-x+1;y=x^{2}+5 x+67 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=-x+1\\y=x^{2}+5x+67\end{array}\right.\) - step1: Substitute the value of \(y:\) \(-x+1=x^{2}+5x+67\) - step2: Move the expression to the left side: \(-x+1-\left(x^{2}+5x+67\right)=0\) - step3: Calculate: \(-6x-66-x^{2}=0\) - step4: Rewrite in standard form: \(-x^{2}-6x-66=0\) - step5: Multiply both sides: \(x^{2}+6x+66=0\) - step6: Solve using the quadratic formula: \(x=\frac{-6\pm \sqrt{6^{2}-4\times 66}}{2}\) - step7: Simplify the expression: \(x=\frac{-6\pm \sqrt{-228}}{2}\) - step8: The expression is undefined: \(x \notin \mathbb{R}\) - step9: The system of equations has no solution in the set of real numbers: \(\left(x,y\right) \notin \mathbb{R}^{2}\) - step10: Alternative Form: \(\textrm{No real solution}\) Let's break down the solution to the given systems of equations step by step. ### System 1: The equations are: 1. \( x + y = 5 \) 2. \( y = x^2 - 25 \) To find the solution set, we can substitute the expression for \( y \) from the first equation into the second equation. 1. From \( x + y = 5 \), we can express \( y \) as: \[ y = 5 - x \] 2. Substitute \( y \) into the second equation: \[ 5 - x = x^2 - 25 \] 3. Rearranging gives: \[ x^2 + x - 30 = 0 \] 4. Now, we can factor the quadratic: \[ (x - 5)(x + 6) = 0 \] 5. Setting each factor to zero gives: \[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \] \[ x + 6 = 0 \quad \Rightarrow \quad x = -6 \] 6. Now, substituting back to find \( y \): - For \( x = 5 \): \[ y = 5 - 5 = 0 \quad \Rightarrow \quad (5, 0) \] - For \( x = -6 \): \[ y = 5 - (-6) = 11 \quad \Rightarrow \quad (-6, 11) \] Thus, the solution set for the first system is: \[ \{(5, 0), (-6, 11)\} \] ### System 2: The equations are: 1. \( y = -x + 1 \) 2. \( y = x^2 + 5x + 67 \) To find the solution set, we can set the two expressions for \( y \) equal to each other: 1. Set the equations equal: \[ -x + 1 = x^2 + 5x + 67 \] 2. Rearranging gives: \[ x^2 + 6x + 66 = 0 \] 3. Now, we can calculate the discriminant to check for real solutions: \[ D = b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 66 = 36 - 264 = -228 \] Since the discriminant is negative, there are no real solutions for this system. ### Summary of Results: 1. The solution set for the first system \( x + y = 5 \) and \( y = x^2 - 25 \) is: \[ \{(5, 0), (-6, 11)\} \] 2. The second system \( y = -x + 1 \) and \( y = x^2 + 5x + 67 \) has: \[ \text{No real solution} \] ### Answering the Options: 1. The ordered pairs in the solution set of the first system are \( (5, 0) \) and \( (-6, 11) \). 2. The second system has no real solutions, so none of the provided pairs can be solutions.