1) \( \left\{\begin{array}{l}x+y+z=1 \\ 2 u+3 y+6 z=10 \\ 2 n-y y+12 z=-10\end{array}\right. \)

Solve the system of equations \( x+y+z=1; 2u+3y+6z=10; 2n-y^2+12z=-10 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y+z=1\\2u+3y+6z=10\\2n-y^{2}+12z=-10\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=1-y-z\\2u+3y+6z=10\\2n-y^{2}+12z=-10\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left\{ \begin{array}{l}2u+3y+6z=10\\2n-y^{2}+12z=-10\end{array}\right.\) - step3: Solve the equation: \(\left\{ \begin{array}{l}2u+3y+6z=10\\n=\frac{-10+y^{2}-12z}{2}\end{array}\right.\) - step4: Substitute the value of \(n:\) \(2u+3y+6z=10\) - step5: Move the expression to the right side: \(2u=10-\left(3y+6z\right)\) - step6: Remove the parentheses: \(2u=10-3y-6z\) - step7: Divide both sides: \(\frac{2u}{2}=\frac{10-3y-6z}{2}\) - step8: Divide the numbers: \(u=\frac{10-3y-6z}{2}\) - step9: Calculate: \(\left(n,u,x,y,z\right) = \left(\frac{-10+y^{2}-12z}{2},\frac{10-3y-6z}{2},1-y-z,y,z\right),\left(y,z\right) \in \mathbb{R}^{2}\) - step10: Alternative Form: \(\textrm{Infinitely many solutions}\) Le système d'équations donné a une infinité de solutions.