Name. Q.2 Exit Thoket Solve. Show your work. \begin{tabular}{|c|c|}\hline\( x-6=23 \) & \( x+14=52 \) \\ \hline & \\ \hline\end{tabular}

Solve the system of equations by following steps: - step0: Solution: \(\left\{ \begin{array}{l}x-6=23\\x+14=52\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}x=29\\x=38\end{array}\right.\) - step2: Find the intersection: \(x \in \varnothing \) Let's solve the system of equations step by step. We have the following equations: 1. \( x - 6 = 23 \) 2. \( x + 14 = 52 \) ### Step 1: Solve the first equation Starting with the first equation: \[ x - 6 = 23 \] To isolate \( x \), we add 6 to both sides: \[ x = 23 + 6 \] Calculating the right side: \[ x = 29 \] ### Step 2: Solve the second equation Now, let's solve the second equation: \[ x + 14 = 52 \] To isolate \( x \), we subtract 14 from both sides: \[ x = 52 - 14 \] Calculating the right side: \[ x = 38 \] ### Conclusion We found two different values for \( x \): - From the first equation, \( x = 29 \) - From the second equation, \( x = 38 \) Since these values are not equal, the system of equations has no solution. Thus, the solution set is empty: \[ x \in \varnothing \]