Answer:
x=6,
y=-9,
z=7
Step-by-step explanation:
Rewrite the second equation in the first place
[tex]\left\{\begin{array}{l}x+3y-z=-28\\2x+y-2z=-11\\3x+4y-z=-25\end{array}\right..[/tex]
Multiply the first equation by 2 and subtract the second and then multiply the first equation by 3 and subtract the third:
[tex]\left\{\begin{array}{l}x+3y-z=-28\\2x+6y-2z-(2x+y-2z)=-2\cdot 28-(-11)\\3x+9y-3z-(3x+4y-z)=-3\cdot 28-(-25)\end{array}\right.\Rightarrow \left\{\begin{array}{r}x+3y-z=-28\\5y=-45\\5y-2z=-59\end{array}.[/tex]
Write the variable y into the third column:
[tex]\left\{\begin{array}{r}x-z+3y=-28\\-2z+5y=-59\\5y=-45\end{array}.[/tex]
From the last equation
[tex]y=-\dfrac{45}{5}=-9.[/tex]
Substitute it into the previous equation:
[tex]-2z+5\cdot (-9)=-59,\\ \\-2z=-59+45,\\ \\-2z=-14,\\ \\z=7.[/tex]
Substitute both y and z into the first equation:
[tex]x+3\cdot (-9)-7=-28,\\ \\x=-28+7+27,\\ \\x=6.[/tex]
