Solve the system.
-$$\left\{ \begin{array} { r }
x + 5 y + 2 z = 35 \
5 y + 3 z = 37 \
z = 4
\end{array} \right.$$
A) $( 2,4,5 )$
B) $( 2,5,4 )$
C) $( 4,5,2 )$
D) $\varnothing$

Question

Solve the system.
-$$\left\{ \begin{array} { r } 
x + 5 y + 2 z = 35 \
5 y + 3 z = 37 \
z = 4
\end{array} \right.$$
A) $( 2,4,5 )$
B) $( 2,5,4 )$
C) $( 4,5,2 )$
D) $\varnothing$

Quizplus · Accepted Answer

Given $z = 4$, we substitute this value into the second equation to find $y$: $5y + 3(4) = 37 \Rightarrow 5y + 12 = 37 \Rightarrow 5y = 25 \Rightarrow y = 5$. Then, substituting $y = 5$ and $z = 4$ into the first equation to find $x$: $x + 5(5) + 2(4) = 35 \Rightarrow x + 25 + 8 = 35 \Rightarrow x = 35 - 33 \Rightarrow x = 2$. Therefore, the solution is $(x, y, z) = (2, 5, 4)$.

Solve the System {x+5y+2z=355y+3z=37z=4\left\{ \begin{array} { r } x + 5 y + 2 z = 35 \\5 y + 3 z = 37 \\z = 4\end{array} \right.⎩⎨⎧​x+5y+2z=355y+3z=37z=4​

Solve the System $\left\{ \begin{array} { r } x + 5 y + 2 z = 35 \\5 y + 3 z = 37 \\z = 4\end{array} \right.$