Solve the system. If the system has infinitely many solutions, write the solution set with x arbitrary.
-$$\begin{aligned}
4 x + y + z & = 0 \
x - y + 2 z & = - 8 \
4 x + y + z & = - 3
\end{aligned}$$
A) $\{ ( - 3 , - 1,4 ) \}$
B) $\{ ( - 1,4 , - 3 ) \}$
C) $\varnothing$
D) $\{ ( - 3,4 , - 1 ) \}$

Question

Quizplus · Accepted Answer

The first and third equations are contradictory (they have the same left-hand side but different right-hand sides), indicating that the system has no solution.

Solve the System 4x+y+z=0x−y+2z=−84x+y+z=−3\begin{aligned}4 x + y + z & = 0 \\x - y + 2 z & = - 8 \\4 x + y + z & = - 3\end{aligned}4x+y+zx−y+2z4x+y+z​=0=−8=−3​

Solve the System $\begin{aligned}4 x + y + z & = 0 \\x - y + 2 z & = - 8 \\4 x + y + z & = - 3\end{aligned}$