se Gaussian elimination to find the complete solution to the system of eq \( \begin{array}{r}w+2 x+2 y-z=6 \\ 2 x-2 y+z=4 \\ w-4 x+y\end{array} \) A. There is one solution. The solution set is \( \{(\square \). B. There are infinitely many solutions. The solution set is \( \{\square \). (Type expressions using \( z \) as the variable. Use integers or fraction C. There is no solution. The solution set is \( \varnothing \).

Solve the system of equations \( w+2x+2y-z=6;2x-2y+z=4;w-4x+y=0 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}w+2x+2y-z=6\\2x-2y+z=4\\w-4x+y=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}w=6-2x-2y+z\\2x-2y+z=4\\w-4x+y=0\end{array}\right.\) - step2: Substitute the value of \(w:\) \(\left\{ \begin{array}{l}2x-2y+z=4\\6-2x-2y+z-4x+y=0\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}2x-2y+z=4\\6-6x-y+z=0\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}z=4-2x+2y\\6-6x-y+z=0\end{array}\right.\) - step5: Substitute the value of \(z:\) \(6-6x-y+4-2x+2y=0\) - step6: Simplify: \(10-8x+y=0\) - step7: Move the expression to the right side: \(y=0-\left(10-8x\right)\) - step8: Subtract the terms: \(y=-10+8x\) - step9: Substitute the value of \(y:\) \(z=4-2x+2\left(-10+8x\right)\) - step10: Simplify: \(z=-16+14x\) - step11: Substitute the values of \(y,z:\) \(w=6-2x-2\left(-10+8x\right)-16+14x\) - step12: Simplify: \(w=10-4x\) - step13: Calculate: \(\left(w,x,y,z\right) = \left(10-4x,x,-10+8x,-16+14x\right),x \in \mathbb{R}\) - step14: Alternative Form: \(\textrm{Infinitely many solutions}\) The system of equations has infinitely many solutions. Therefore, the solution set is \( \{(w,x,y,z) \in \mathbb{R}^4 \} \).