Solve Systems of Linear Equations in Three Variables
-$$\begin{aligned}
x + 4 y + 5 z & = 26 \
2 y + 3 z & = 11 \
z & = 1
\end{aligned}$$
A) $\{ ( 5,4,1 ) \}$
B) $\{ ( 5,1,4 ) \}$
C) $\{ ( 1,4,5 ) \}$
D) $\{ ( 4,5,1 ) \}$

Question

Quizplus · Accepted Answer

Given $z = 1$, substitute $z$ into the second equation to find $y$: $2y + 3(1) = 11 \Rightarrow 2y = 8 \Rightarrow y = 4$. Then, substitute $y$ and $z$ into the first equation to find $x$: $x + 4(4) + 5(1) = 26 \Rightarrow x + 16 + 5 = 26 \Rightarrow x = 5$. Therefore, the solution is $(x, y, z) = (5, 4, 1)$.

Solve Systems of Linear Equations in Three Variables - x+4y+5z=262y+3z=11z=1\begin{aligned}x + 4 y + 5 z & = 26 \\2 y + 3 z & = 11 \\z & = 1\end{aligned}x+4y+5z2y+3zz​=26=11=1​

Solve Systems of Linear Equations in Three Variables
- $\begin{aligned}x + 4 y + 5 z & = 26 \\2 y + 3 z & = 11 \\z & = 1\end{aligned}$