Solve the system.
-$\left\{ \begin{aligned}
x + 2 y + 4 z & = 0 \
5 y + 3 z & = 9 \
z & = - 2
\end{aligned} \right.$
A) $( 2,3 , - 2 )$
B) $( - 2,3,2 )$
C) $( 2 , - 2,3 )$
D) $\varnothing$

Question

Quizplus · Accepted Answer

Given $z = -2$, substituting into the second equation gives $5y + 3(-2) = 9$, which simplifies to $5y - 6 = 9$, leading to $5y = 15$, so $y = 3$. Substituting $y = 3$ and $z = -2$ into the first equation gives $x + 2(3) + 4(-2) = 0$, which simplifies to $x + 6 - 8 = 0$, leading to $x = 2$. Therefore, the solution is $(2, 3, -2)$.

Solve the System {x+2y+4z=05y+3z=9z=−2\left\{ \begin{aligned}x + 2 y + 4 z & = 0 \\5 y + 3 z & = 9 \\z & = - 2\end{aligned} \right.⎩⎨⎧​x+2y+4z5y+3zz​=0=9=−2​

Solve the System $\left\{ \begin{aligned}x + 2 y + 4 z & = 0 \\5 y + 3 z & = 9 \\z & = - 2\end{aligned} \right.$