\( \left. \begin{array} { l l l } { x + 2 = y } & { \rightarrow } & { y = x + 2 } \\ { y - x = - 3 } & { \rightarrow } & { y = \square } \end{array} \right. \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+2=y\\y-x=-3\end{array}\right.\) - step1: Rewrite the expression: \(\left\{ \begin{array}{l}y=x+2\\y-x=-3\end{array}\right.\) - step2: Substitute the value of \(y:\) \(x+2-x=-3\) - step3: Simplify: \(2=-3\) - step4: Calculate: \(\left\{ \begin{array}{l}x \in \varnothing \\y \in \varnothing \end{array}\right.\) - step5: Rewrite: \((x, y) \in \varnothing\) The system of equations given is: 1. \( x + 2 = y \) 2. \( y - x = -3 \) Let's analyze these equations step by step. 1. From the first equation, we can express \( y \) in terms of \( x \): \[ y = x + 2 \] 2. Now, substituting this expression for \( y \) into the second equation: \[ (x + 2) - x = -3 \] 3. Simplifying this gives: \[ 2 = -3 \] This statement is false, indicating that there is no solution to the system of equations. Therefore, the final conclusion is: **No solution**.