Respuesta :
x = 2
y = -6
z = 3
[tex]\begin{cases}E_1\Rightarrow2x-y-3z=1 \\ E_2\Rightarrow4x+3y+2z=-4 \\ E_3\Rightarrow-3x+2y+5z=-3\end{cases}[/tex]I named the equation to make it more simple.
To solve for elimination, we can multiply E1 by (-2) to eliminate the 4x in E2
[tex]\begin{gathered} (4x+3y+2z=-4)+(-2)(2x-y-3z=1)\Rightarrow0x+5y+8z=-6 \\ \end{gathered}[/tex]Now the system is:
[tex]\begin{cases}E_1\Rightarrow2x-y-3z=1 \\ E_2\Rightarrow5y+8z=-6 \\ E_3\Rightarrow-3x+2y+5z=-3\end{cases}[/tex]We can continue by eliminating the (-3x) in E3. THen multiply E1 by (3/2) and add it to E3:
[tex](-3x+2y+5z=-3)+\frac{3}{2}(2x-y-3z=1)\Rightarrow0x+\frac{1}{2}y+\frac{1}{2}z=-\frac{3}{2}[/tex]Now the system is:
[tex]\begin{cases}E_1\Rightarrow2x-y-3z=1 \\ E_2\Rightarrow5y+8z=-6 \\ E_3\Rightarrow\frac{1}{2}y+\frac{1}{2}z=-\frac{3}{2}\end{cases}[/tex]Let's continue by eliminating y. Then we multiply E2 by -(1/10) and add it to E3:
[tex](\frac{1}{2}y+\frac{1}{2}z=-\frac{3}{2})+(-\frac{1}{10})(5y+8z=-6)\Rightarrow0y-\frac{3}{10}z=-\frac{9}{10}[/tex]The system is now:
[tex]\begin{cases}E_1\Rightarrow2x-y-3z=1 \\ E_2\Rightarrow5y+8z=-6 \\ E_3\Rightarrow-\frac{3}{10}z=-\frac{9}{10}\end{cases}[/tex]Now we can know the value of z:
[tex]-\frac{3}{10}z=-\frac{9}{10}\Rightarrow z=(-\frac{9}{10})\cdot(-\frac{10}{3})=3[/tex][tex]\begin{cases}E_1\Rightarrow2x-y-3z=1 \\ E_2\Rightarrow5y+8z=-6 \\ E_3\Rightarrow z=3\end{cases}[/tex]Now we can replace z in E2:
[tex]\begin{gathered} 5y+8\cdot3=-6 \\ y=-\frac{30}{5} \\ y=-6 \end{gathered}[/tex][tex]\begin{cases}E_1\Rightarrow2x-y-3z=1 \\ E_2\Rightarrow y=-6 \\ E_3\Rightarrow z=3\end{cases}[/tex]Finally, replace y and z in E1:
[tex]\begin{gathered} 2x-y-3z=1\Rightarrow2x-(-6)-3\cdot3=1 \\ x=\frac{4}{2}=2 \end{gathered}[/tex]The solution of the equation is:
[tex]\begin{cases}x=1 \\ y=-6 \\ z=3\end{cases}[/tex]