A \( -13 x+12=13 x+13 \) \( -13 x+12=13 x-13 \) B \( 12 x+12=13 x+12 \) \( 12 x+12=13 x-12 \)

Solve the system of equations by following steps: - step0: Solution: \(\left\{ \begin{array}{l}12x+12=13x+12\\12x+12=13x-12\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}12x=13x\\12x+12=13x-12\end{array}\right.\) - step2: Calculate: \(\left\{ \begin{array}{l}x=0\\x=24\end{array}\right.\) - step3: Find the intersection: \(x \in \varnothing \) Solve the system of equations \( -13 x + 12 = 13 x + 13; -13 x + 12 = 13 x - 13 \). Solve the system of equations by following steps: - step0: Solution: \(\left\{ \begin{array}{l}-13x+12=13x+13\\-13x+12=13x-13\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}x=-\frac{1}{26}\\x=\frac{25}{26}\end{array}\right.\) - step2: Find the intersection: \(x \in \varnothing \) Let's analyze the two systems of equations provided: ### System A: 1. \( -13x + 12 = 13x + 13 \) 2. \( -13x + 12 = 13x - 13 \) **Step 1: Solve the first equation.** Rearranging the first equation: \[ -13x - 13x = 13 - 12 \implies -26x = 1 \implies x = -\frac{1}{26} \] **Step 2: Solve the second equation.** Rearranging the second equation: \[ -13x - 13x = -13 - 12 \implies -26x = -25 \implies x = \frac{25}{26} \] Since the two equations yield different values for \(x\), they do not intersect, leading to no solution. ### System B: 1. \( 12x + 12 = 13x + 12 \) 2. \( 12x + 12 = 13x - 12 \) **Step 1: Solve the first equation.** Rearranging the first equation: \[ 12x - 13x = 12 - 12 \implies -x = 0 \implies x = 0 \] **Step 2: Solve the second equation.** Rearranging the second equation: \[ 12x - 13x = -12 - 12 \implies -x = -24 \implies x = 24 \] Again, since the two equations yield different values for \(x\), they do not intersect, leading to no solution. ### Conclusion: Both systems A and B have no solutions, denoted as \( x \in \varnothing \).