$\left\{ \begin{array} { l }
x + y + z = 6 \
x - z = - 2 \
y + 3 z = 11
\end{array} \right.$
A) x=-1, y=2, z=-3
B) x=1, y=2, z=3
C) x=3, y=-2, z=5
D) x=0, y=4, z=2

Question

$\left\{ \begin{array} { l } 
x + y + z = 6 \
x - z = - 2 \
y + 3 z = 11
\end{array} \right.$
A)  x=-1, y=2, z=-3 
B)  x=1, y=2, z=3 
C)  x=3, y=-2, z=5 
D)  x=0, y=4, z=2

Quizplus · Accepted Answer

Solving the system of equations, we can substitute the second equation into the first to find y, and then use the third equation to solve for z. From the second equation, $x = z - 2$. Substituting this into the first equation gives $z - 2 + y + z = 6$, which simplifies to $2z + y = 8$. Using the third equation, $y + 3z = 11$, we can solve these two equations simultaneously to find $y = 2$ and $z = 3$. Substituting $z = 3$ into $x = z - 2$ gives $x = 1$. Therefore, the solution is $x = 1$, $y = 2$, and $z = 3$.

{x+y+z=6x−z=−2y+3z=11\left\{ \begin{array} { l } x + y + z = 6 \\x - z = - 2 \\y + 3 z = 11\end{array} \right.⎩⎨⎧​x+y+z=6x−z=−2y+3z=11​

$\left\{ \begin{array} { l } x + y + z = 6 \\x - z = - 2 \\y + 3 z = 11\end{array} \right.$