Solve the system.
$\begin{array} { r }
4 x - 2 y - 3 z + 1 = 0 \
12 x - 2 y + 6 z - 4 = 0 \
4 y + 3 z - 3 = 0
\end{array}$
A) $\left\{ \left( - \frac { 1 } { 4 } , \frac { 1 } { 2 } , - \frac { 1 } { 3 } \right) \right\}$
B) $\left\{ \left( \frac { 1 } { 4 } , \frac { 1 } { 2 } , \frac { 1 } { 3 } \right) \right\}$
C) $\varnothing$

D) $\left\{ \left( - \frac { 1 } { 4 } , - \frac { 1 } { 2 } , \frac { 1 } { 3 } \right) \right\}$

Question

Solve the system.
$\begin{array} { r } 
4 x - 2 y - 3 z + 1 = 0 \
12 x - 2 y + 6 z - 4 = 0 \
4 y + 3 z - 3 = 0
\end{array}$
A) $\left\{ \left( - \frac { 1 } { 4 } , \frac { 1 } { 2 } , - \frac { 1 } { 3 } \right) \right\}$
B) $\left\{ \left( \frac { 1 } { 4 } , \frac { 1 } { 2 } , \frac { 1 } { 3 } \right) \right\}$
C) $\varnothing$

D) $\left\{ \left( - \frac { 1 } { 4 } , - \frac { 1 } { 2 } , \frac { 1 } { 3 } \right) \right\}$

Quizplus · Accepted Answer

To solve the system, we can use substitution or elimination methods. By manipulating the equations, we can find that the solution that satisfies all three equations is $\left( \frac { 1 } { 4 } , \frac { 1 } { 2 } , \frac { 1 } { 3 } \right)$, making choice B correct.

Solve the System 4x−2y−3z+1=012x−2y+6z−4=04y+3z−3=0\begin{array} { r } 4 x - 2 y - 3 z + 1 = 0 \\12 x - 2 y + 6 z - 4 = 0 \\4 y + 3 z - 3 = 0\end{array}4x−2y−3z+1=012x−2y+6z−4=04y+3z−3=0​

Solve the System $\begin{array} { r } 4 x - 2 y - 3 z + 1 = 0 \\12 x - 2 y + 6 z - 4 = 0 \\4 y + 3 z - 3 = 0\end{array}$