Solve the system of linear equations.
$$\left\{ \begin{array} { c }
x + 4 y + 2 z = - 2 \
- 4 x + y - 4 z = 9 \
3 x - y + 2 z = - 3
\end{array} \right.$$
A) $( 3,6,3 )$
B) $( 2,1 , - 4 )$
C) $( 1 , - 4,2 )$
D) $( 6,3,3 )$
E) inconsistent

Question

Solve the system of linear equations.
$$\left\{ \begin{array} { c } 
x + 4 y + 2 z = - 2 \
- 4 x + y - 4 z = 9 \
3 x - y + 2 z = - 3
\end{array} \right.$$
A) $( 3,6,3 )$
B) $( 2,1 , - 4 )$
C) $( 1 , - 4,2 )$
D) $( 6,3,3 )$
E) inconsistent

Quizplus · Accepted Answer

One way to solve this system is to use elimination method. We eliminate $y$ first by adding the second equation to three times the third equation:
\begin{align*}
-4x+y-4z+3(3x-y+2z)&=9-9\
-4x+4x+8z&=0\
z&=0
\end{align*}
Now we can substitute $z=0$ into the first and third equation to get a system of two equations in two variables $x$ and $y$:
$$\left\{\begin{aligned} x+4y&=-2\ 3x-y&=-3 \end{aligned}\right.$$
We can multiply the second equation by 4 and add it to the first equation to eliminate $y$:
\begin{align*}
x+4y+12x-4y&=-2-12\cdot 3\
13x&=-38\
x&=-2
\end{align*}
Finally, we can substitute $x=-2$ into the second equation to get $-3y=-9$, or $y=3$. Therefore, the solution to the system is $(x,y,z)=\boxed{(2,1,-4)}$.

Solve the System of Linear Equations {x+4y+2z=−2−4x+y−4z=93x−y+2z=−3\left\{ \begin{array} { c } x + 4 y + 2 z = - 2 \\- 4 x + y - 4 z = 9 \\3 x - y + 2 z = - 3\end{array} \right.⎩⎨⎧​x+4y+2z=−2−4x+y−4z=93x−y+2z=−3​

Solve the System of Linear Equations $\left\{ \begin{array} { c } x + 4 y + 2 z = - 2 \\- 4 x + y - 4 z = 9 \\3 x - y + 2 z = - 3\end{array} \right.$