Solve the system of equations.
-$$\begin{array} { l }
2 x + 3 y + z = - 5 \
3 x - 4 y - z = 22 \
5 x + y + 3 z = - 2
\end{array}$$
A) $\{ ( 3 , - 2 , - 5 ) \}$
B) $\{ ( 3 , - 5 , - 2 ) \}$
C) $\{ ( - 5 , - 2,3 ) \}$
D) $\{ ( - 2,3 , - 5 ) \}$

Question

Solve the system of equations.
-$$\begin{array} { l } 
2 x + 3 y + z = - 5 \
3 x - 4 y - z = 22 \
5 x + y + 3 z = - 2
\end{array}$$
A) $\{ ( 3 , - 2 , - 5 ) \}$
B) $\{ ( 3 , - 5 , - 2 ) \}$
C) $\{ ( - 5 , - 2,3 ) \}$
D) $\{ ( - 2,3 , - 5 ) \}$

Quizplus · Accepted Answer

One way to solve the system of equations is to use elimination. Adding the first and second equations eliminates z: 
$$(2x+3y+z) + (3x-4y-z) = -5+22 \Rightarrow 5x-y = 17$$
Similarly, adding the second and third equations eliminates z:
$$(3x-4y-z) + (5x+y+3z) = 22-2 \Rightarrow 8x-3y = 20$$
Now we have two equations with two variables. Solving for y in the first equation gives $y=5x-17$, and substituting into the second equation gives $8x-3(5x-17) = 20$, which simplifies to $x=3$. Substituting $x=3$ back into $5x-y=17$ gives $y=-2$, and substituting $x=3$ and $y=-2$ into any of the original equations solves for z, giving $z=-5$. Therefore, the solution is $\{(3,-2,-5)\}$, which corresponds to choice letter A.

Solve the System of Equations 2x+3y+z=−53x−4y−z=225x+y+3z=−2\begin{array} { l } 2 x + 3 y + z = - 5 \\3 x - 4 y - z = 22 \\5 x + y + 3 z = - 2\end{array}2x+3y+z=−53x−4y−z=225x+y+3z=−2​

Solve the System of Equations $\begin{array} { l } 2 x + 3 y + z = - 5 \\3 x - 4 y - z = 22 \\5 x + y + 3 z = - 2\end{array}$