Assignment 2 1. please use Gauss Elimination to solve \( \left\{\begin{array}{l}x+y+z=3 \\ x-y-z=-1 \\ y+2 z=3\end{array}\right. \)

Solve the system of equations \( x+y+z=3;x-y-z=-1;y+2z=3 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y+z=3\\x-y-z=-1\\y+2z=3\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=3-y-z\\x-y-z=-1\\y+2z=3\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left\{ \begin{array}{l}3-y-z-y-z=-1\\y+2z=3\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}3-2y-2z=-1\\y+2z=3\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}3-2y-2z=-1\\y=3-2z\end{array}\right.\) - step5: Substitute the value of \(y:\) \(3-2\left(3-2z\right)-2z=-1\) - step6: Simplify: \(-3+2z=-1\) - step7: Move the constant to the right side: \(2z=-1+3\) - step8: Add the numbers: \(2z=2\) - step9: Divide both sides: \(\frac{2z}{2}=\frac{2}{2}\) - step10: Divide the numbers: \(z=1\) - step11: Substitute the value of \(z:\) \(y=3-2\times 1\) - step12: Simplify: \(y=1\) - step13: Substitute the values of \(y,z:\) \(x=3-1-1\) - step14: Simplify: \(x=1\) - step15: Calculate: \(\left\{ \begin{array}{l}x=1\\y=1\\z=1\end{array}\right.\) - step16: Check the solution: \(\left\{ \begin{array}{l}x=1\\y=1\\z=1\end{array}\right.\) - step17: Rewrite: \(\left(x,y,z\right) = \left(1,1,1\right)\) L'ensemble des solutions du système d'équations est \( (x, y, z) = (1, 1, 1) \).