Respuesta :
The first step to solve the system of equation is to solve one of the equations for one variable:
[tex]\begin{gathered} -3x+6y-3z=5 \\ 6y-3z-5=3x \\ \frac{6y-3z-5}{3}=x \\ 2y-z-\frac{5}{3}=x \\ x=2y-z-\frac{5}{3} \end{gathered}[/tex]Replace this variable for the obtained expression in the other two equations:
[tex]\begin{gathered} 5(2y-z-\frac{5}{3})-5y-2z=4 \\ 4(2y-z-\frac{5}{3})-3y-3z=-3 \end{gathered}[/tex]Now, solve the system just as if it were a 2x2 system:
[tex]\begin{gathered} 10y-5z-\frac{25}{3}-5y-2z=4 \\ 5y-7z-\frac{25}{3}=4 \\ 5y-7z=4+\frac{25}{3} \\ 5y-7z=\frac{37}{3} \\ y=\frac{37}{15}+\frac{7}{5}z \end{gathered}[/tex][tex]\begin{gathered} 8y-4z-\frac{20}{3}-3y-3z=-3 \\ 5y-7z=\frac{20}{3}-3 \\ 5y-7z=\frac{11}{3} \\ y=\frac{11}{15}+\frac{7}{5}z \end{gathered}[/tex]Make both expressions for y equal and solve for z:
[tex]\begin{gathered} \frac{37}{15}+\frac{7}{5}z=\frac{11}{15}+\frac{7}{5}z \\ \frac{7}{5}z-\frac{7}{5}z=\frac{11}{15}-\frac{37}{15} \\ 0=\frac{26}{15} \end{gathered}[/tex]Since 0 is not equal to 26/15, we can determine that this system does not have any solution.
The correct answer is No solution.
