Answer:
x = 0.5, y = -3, z = 11
Step-by-step explanation:
[tex]\left\{\begin{array}{ccc}2x-y+z=15&(1)\\-6x-y-z=-11&(2)\\4x-3y-z=0&(3)\end{array}\right\\\\\text{Add all sides (1) and (2)}:\\\\\underline{+\left\{\begin{array}{ccc}2x-y+z=15\\-6x-y-z=-11\end{array}\right}\\.\qquad-4x-2y=4\qquad(4)\\\\\text{Add all sides (1) and (3)}:\\\\\underline{+\left\{\begin{array}{ccc}2x-y+z=15\\4x-3y-z=0\end{array}\right}\\.\qquad6x-4y=15\qquad(5)\\\\\text{From (4) and (5) we have new system of equations:}[/tex]
[tex]\left\{\begin{array}{ccc}-4x-2y=4&\text{multiply both sides by (-2)}\\6x-4y=15\end{array}\right\\\underline{+\left\{\begin{array}{ccc}8x+4y=-8\\6x-4y=15\end{array}\right}\qquad\text{add all sides of the equations}\\.\qquad14x=7\qquad\text{divide both sides by 14}\\.\qquad x=\dfrac{7}{14}\\.\qquad\boxed{x=0.5}[/tex]
[tex]\text{Put the value of}\ x\ \text{to (1) and (2):}\\\\\left\{\begin{array}{ccc}2(0.5)-y+z=15\\-6(0.5)-y-z=-11\end{array}\right\\\left\{\begin{array}{ccc}1-y+z=15&\text{subtract 1 from both sides}\\-3-y-z=-11&\text{add 3 to both sides}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}-y+z=14\\-y-z=-8\end{array}\right}\qquad\text{add all sides of the equations}\\.\qquad-2y=6\qquad\text{divide both sides by (-2)}\\.\qquad\boxed{y=-3}[/tex]
[tex]\text{Put the value of}\ y\ \text{to the first equation}\\\\-(-3)+z=14\\3+z=14\qquad\text{subtract 3 from both sides}\\\boxed{z=11}[/tex]
