Respuesta :
[tex]\begin{gathered} 4x-2y+3z=-3\ldots\ldots\text{.}(1) \\ 2x-4y+2z=1\ldots\ldots\ldots(2) \\ 6x+10z=-2\ldots\ldots\ldots..\ldots(3) \\ \text{Now, from the eqution (1) and (2) } \\ 4(\frac{1-2z+4y}{2})-2y+3z=-3 \\ 2-4z+8y-2y+3z=-3 \\ 2+3-z+6y=0 \\ 6y-z=-5 \\ z=6y+5\ldots\ldots\text{.}(4) \end{gathered}[/tex][tex]\begin{gathered} \text{Now, from equation (4) and (3) } \\ 6x=10(6y+5)-2 \\ 6x=60y+50-2 \\ 6x=60y+48 \\ x=10y+8\ldots\ldots\text{.}(5) \\ \text{Substitute all the value in the equation(1)} \\ 4(10y+8)-2y+3(6y+5)=-3 \\ 40y+32-2y+18y+15=-3 \\ 56y=50 \\ y=\frac{50}{56} \\ y=\frac{25}{28} \end{gathered}[/tex]
Now, from equation (5)
[tex]\begin{gathered} x=10y+8 \\ x=10\times\frac{25}{28}+8 \\ x=16.92 \\ \text{Now, from the eqution(4)} \\ z=6y+5 \\ z=6\times\frac{25}{28}+5 \\ z=10.35 \end{gathered}[/tex]