Solve the system. If the system has infinitely many solutions, write the solution set with x arbitrary.
-$$\begin{array} { l }
x + y - 2 z - 8 = 0 \
3 x + z + 6 = 0 \
2 x - y + 3 z + 14 = 0
\end{array}$$
A) $\{ ( x , - 7 x - 4 , - 3 x - 6 ) \}$
B) $\{ ( x , 7 x + 4 , - 3 x - 6 ) \}$
C) $\{ ( x , - 7 x + 4 , - 3 x - 6 ) \}$
D) $\{ ( x , 7 x - 4 , - 3 x - 6 ) \}$

Question

Solve the system. If the system has infinitely many solutions, write the solution set with x arbitrary.
-$$\begin{array} { l } 
x + y - 2 z - 8 = 0 \
3 x + z + 6 = 0 \
2 x - y + 3 z + 14 = 0
\end{array}$$
A) $\{ ( x , - 7 x - 4 , - 3 x - 6 ) \}$
B) $\{ ( x , 7 x + 4 , - 3 x - 6 ) \}$
C) $\{ ( x , - 7 x + 4 , - 3 x - 6 ) \}$
D) $\{ ( x , 7 x - 4 , - 3 x - 6 ) \}$

Quizplus · Accepted Answer

The system of equations is dependent, meaning it has infinitely many solutions. By solving the system, we can express y and z in terms of x, leading to the solution set $\{ ( x , - 7 x - 4 , - 3 x - 6 ) \}$. This involves combining and manipulating the equations to eliminate variables and express y and z as functions of x.

Solve the System x+y−2z−8=03x+z+6=02x−y+3z+14=0\begin{array} { l } x + y - 2 z - 8 = 0 \\3 x + z + 6 = 0 \\2 x - y + 3 z + 14 = 0\end{array}x+y−2z−8=03x+z+6=02x−y+3z+14=0​

Solve the System $\begin{array} { l } x + y - 2 z - 8 = 0 \\3 x + z + 6 = 0 \\2 x - y + 3 z + 14 = 0\end{array}$