Solve the system. $$\left( \begin{array} { r l }
3 x - 2 y + 3 z & = 23 \
3 y - 5 z & = - 18 \
2 z & = 6
\end{array} \right)$$
A) x = 6, y = -2, z = 0
B) x = 5, y = 1, z = 0
C) x = 4, y = -1, z = 3
D) x = 6, y = -4, z = 3

Question

Solve the system. $$\left( \begin{array} { r l } 
3 x - 2 y + 3 z & = 23 \
3 y - 5 z & = - 18 \
2 z & = 6
\end{array} \right)$$
A) x = 6, y = -2, z = 0
B) x = 5, y = 1, z = 0
C) x = 4, y = -1, z = 3
D) x = 6, y = -4, z = 3

Quizplus · Accepted Answer

From the third equation, we have $z=3$. Substituting this into the second equation, we have $3y-5(3)=-18$, which gives $y=-1$. Finally, substituting $z=3$ and $y=-1$ into the first equation, we have $3x-2(-1)+3(3)=23$, which gives $x=4$. Therefore, the solution is $(x,y,z)=(4,-1,3)$, which corresponds to choice C.

Solve the System (3x−2y+3z=233y−5z=−182z=6)\left( \begin{array} { r l } 3 x - 2 y + 3 z & = 23 \\3 y - 5 z & = - 18 \\2 z & = 6\end{array} \right)​3x−2y+3z3y−5z2z​=23=−18=6​​

Solve the System $\left( \begin{array} { r l } 3 x - 2 y + 3 z & = 23 \\3 y - 5 z & = - 18 \\2 z & = 6\end{array} \right)$