We are given the following system of equations:
[tex]\begin{gathered} 3x+2y-z=7,(1) \\ -3x+y+2z=-14,(2) \\ 3x+y-z=10,(3) \end{gathered}[/tex]1. We will add equation (1) and equation (2), we get:
[tex]3x+2y-z-3x+y+2z=7-14[/tex]Now we add like terms:
[tex]3y+z=-7,(4)[/tex]We get a new equation which we called equation (4).
2. Now, we will add equation (2) and equation (3), we get:
[tex]-3x+y+2z+3x+y-z=-14+10[/tex]Now, we add like terms:
[tex]2y+z=-4,(5)[/tex]We get a new equation that we called equation (5).
3. equation (4) and equation (5) form a new system of equations:
[tex]\begin{gathered} 3y+z=-7,(4) \\ 2y+z=-4,(5) \end{gathered}[/tex]To solve the system we will subtract equation (5) from equation (4), this way we will be able to cancel out the variable "z". Like this:
[tex]3y+z-2y-z=-7+4[/tex]Now, we add like terms:
[tex]y=-3[/tex]Therefore, the value of "y" is -3.
Now, we substitute the value of "y" in equation (4), we get:
[tex]3(-3)+z=-7[/tex]Solving the product:
[tex]-9+z=-7[/tex]Now, we add 9 to both sides:
[tex]\begin{gathered} -9+9+z=-7+9 \\ z=2 \end{gathered}[/tex]Therefore, we have the following values from the new system:
[tex]\begin{gathered} y=-3 \\ z=2 \end{gathered}[/tex]4. To determine the values of the original system we will substitute the values of "y" and "z" in equation (1):
[tex]3x+2(-3)-(2)=7[/tex]Solving the operations we get:
[tex]\begin{gathered} 3x-6-2=7 \\ 3x-8=7 \end{gathered}[/tex]Now, we will add 8 to both sides:
[tex]\begin{gathered} 3x-8+8=7+8 \\ 3x=15 \end{gathered}[/tex]Now, we will divide both sides by 3:
[tex]x=\frac{15}{3}=5[/tex]Therefore, the solution of the system is:
[tex](x,y,z)=(5,-3,2)[/tex]