1 Solve for \( x \) and \( y \) in the f a \( 3 y+x=2 \) \( y^{2}+x=x y+y \) d \( y+13-6 x=0 \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3y+x=2\\y^{2}+x=xy+y\\y+13-6x=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=2-3y\\y^{2}+x=xy+y\\y+13-6x=0\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left\{ \begin{array}{l}y^{2}+2-3y=\left(2-3y\right)y+y\\y+13-6\left(2-3y\right)=0\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}y^{2}+2-3y=3y-3y^{2}\\19y+1=0\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}y=1\cup y=\frac{1}{2}\\19y+1=0\end{array}\right.\) - step5: Evaluate: \(\left\{ \begin{array}{l}y=1\\19y+1=0\end{array}\right.\cup \left\{ \begin{array}{l}y=\frac{1}{2}\\19y+1=0\end{array}\right.\) - step6: Calculate: \(y \in \varnothing \cup \left\{ \begin{array}{l}y=\frac{1}{2}\\19y+1=0\end{array}\right.\) - step7: Calculate: \(y \in \varnothing \cup y \in \varnothing \) - step8: Rearrange the terms: \(y \in \varnothing \) - step9: Calculate: \(\left\{ \begin{array}{l}x \in \varnothing \\y \in \varnothing \end{array}\right.\) - step10: Rewrite: \((x, y) \in \varnothing\) The system of equations provided is: 1. \( 3y + x = 2 \) 2. \( y^2 + x = xy + y \) 3. \( y + 13 - 6x = 0 \) After solving this system, we find that there is **no solution**. This means that the equations are inconsistent and do not intersect at any point in the coordinate plane.