$\left\{ \begin{array} { l }
x - y + z = 8 \
x + y + z = 6 \
x + y - z = - 12
\end{array} \right.$
A) x=-2, y=-1, z=-9
B) x=2, y=-1, z=9
C) x=-2, y=-1, z=9
D) x=2, y=-1, z=-9

Question

$\left\{ \begin{array} { l } 
x - y + z = 8 \
x + y + z = 6 \
x + y - z = - 12
\end{array} \right.$
A)  x=-2, y=-1, z=-9 
B)  x=2, y=-1, z=9 
C)  x=-2, y=-1, z=9 
D)  x=2, y=-1, z=-9

Quizplus · Accepted Answer

To solve the system of equations, we can add the second and third equations to eliminate $z$:$$x + y + z = 6$$$$x + y - z = -12$$-------------------$$2x + 2y = -6$$Dividing by 2:$$x + y = -3$$Now, subtract the third equation from the first to eliminate $y$:$$x - y + z = 8$$$$-(x + y - z) = 12$$-------------------$$-2y + 2z = 20$$Dividing by 2:$$-y + z = 10$$Using $x + y = -3$ and $-y + z = 10$, we can solve for $y$ and $z$:From $x + y = -3$, we get $y = -3 - x$.Substituting $y$ in $-y + z = 10$:$$-(-3 - x) + z = 10$$$$3 + x + z = 10$$$$z = 7 - x$$Now, we can substitute $x + y = -3$ into one of the original equations to find $x$:$$x + y + z = 6$$$$x - 3 - x + 7 - x = 6$$$$4 - x = 6$$$$x = -2$$Substituting $x = -2$ into $y = -3 - x$ and $z = 7 - x$:$$y = -3 - (-2) = -1$$$$z = 7 - (-2) = 9$$Therefore, the solution is $x = -2$, $y = -1$, $z = 9$.

{x−y+z=8x+y+z=6x+y−z=−12\left\{ \begin{array} { l } x - y + z = 8 \\x + y + z = 6 \\x + y - z = - 12\end{array} \right.⎩⎨⎧​x−y+z=8x+y+z=6x+y−z=−12​

$\left\{ \begin{array} { l } x - y + z = 8 \\x + y + z = 6 \\x + y - z = - 12\end{array} \right.$