Find the complete solution of the system. $\left\{ \begin{array} { l }
x + z = 4 \
y + 2 z = 1 \
- y + z = 0
\end{array} \right.$
A) $x = \frac { 11 } { 3 }$, $y = \frac { 1 } { 3 }$, $z = \frac { 1 } { 3 }$

B) $x = \frac { 11 } { 3 }$, $y = \frac { 4 } { 3 }$, $z = \frac { 1 } { 3 }$

C) $x = \frac { 11 } { 3 }$, $y = \frac { 1 } { 3 }$, $z = - \frac { 2 } { 3 }$

D) $x = 3$, $y = 1$, $z = 11$

E) no solution

Question

Find the complete solution of the system. $\left\{ \begin{array} { l } 
x + z = 4 \
y + 2 z = 1 \
- y + z = 0
\end{array} \right.$
A) $x = \frac { 11 } { 3 }$, $y = \frac { 1 } { 3 }$, $z = \frac { 1 } { 3 }$

B) $x = \frac { 11 } { 3 }$, $y = \frac { 4 } { 3 }$, $z = \frac { 1 } { 3 }$

C) $x = \frac { 11 } { 3 }$, $y = \frac { 1 } { 3 }$, $z = - \frac { 2 } { 3 }$

D) $x = 3$, $y = 1$, $z = 11$

E) no solution

Quizplus · Accepted Answer

We can solve this system by using elimination method.

Adding the first and third equations, we get:

$x+y=4$

Adding the second and third equations, we get:

$y= z$

Substituting $y$ in the first equation, we get:

$x+z=4$

Solving this system, we obtain:

$x = \frac{11}{3}$, $y = \frac{1}{3}$, and $z = \frac{1}{3}$.

Therefore, the answer is A.

Find the Complete Solution of the System {x+z=4y+2z=1−y+z=0\left\{ \begin{array} { l } x + z = 4 \\y + 2 z = 1 \\- y + z = 0\end{array} \right.⎩⎨⎧​x+z=4y+2z=1−y+z=0​

Find the Complete Solution of the System $\left\{ \begin{array} { l } x + z = 4 \\y + 2 z = 1 \\- y + z = 0\end{array} \right.$