The following system does not have a unique solution. Solve the system.
-$$\left\{ \begin{array} { r }
x + 3 y + 2 z = 11 \
4 y + 9 z = - 12
\end{array} \right.$$
A) infinitely many solutions of the form $x = 20 + \frac { 19 } { 4 } z , y = - 3 - \frac { 9 } { 4 } z , z = z$
B) infinitely many solutions of the form $x = 20 - \frac { 19 } { 4 } z , y = 3 - \frac { 9 } { 4 } z , z = z ^ { 2 }$
C) infinitely many solutions of the form $x = 20 + \frac { 19 } { 4 } z , y = 3 - \frac { 9 } { 4 } z , z = z$
D) infinitely many solutions of the form $x = 20 + \frac { 19 } { 4 } z , y = 3 + \frac { 9 } { 4 } z , z = z ^ { 2 }$

Question

The following system does not have a unique solution. Solve the system.
-$$\left\{ \begin{array} { r } 
x + 3 y + 2 z = 11 \
4 y + 9 z = - 12
\end{array} \right.$$
A) infinitely many solutions of the form $x = 20 + \frac { 19 } { 4 } z , y = - 3 - \frac { 9 } { 4 } z , z = z$
B) infinitely many solutions of the form $x = 20 - \frac { 19 } { 4 } z , y = 3 - \frac { 9 } { 4 } z , z = z ^ { 2 }$
C) infinitely many solutions of the form $x = 20 + \frac { 19 } { 4 } z , y = 3 - \frac { 9 } { 4 } z , z = z$
D) infinitely many solutions of the form $x = 20 + \frac { 19 } { 4 } z , y = 3 + \frac { 9 } { 4 } z , z = z ^ { 2 }$

Quizplus · Accepted Answer

Given the system of equations, we can solve for $y$ from the second equation: $4y + 9z = -12$, which gives $y = -3 - \frac{9}{4}z$. Substituting $y$ in the first equation, $x + 3(-3 - \frac{9}{4}z) + 2z = 11$, we can solve for $x$ as $x = 20 + \frac{19}{4}z$. Since $z$ is a free variable, the solution is of the form $x = 20 + \frac{19}{4}z$, $y = -3 - \frac{9}{4}z$, and $z = z$, indicating infinitely many solutions.

The Following System Does Not Have a Unique Solution {x+3y+2z=114y+9z=−12\left\{ \begin{array} { r } x + 3 y + 2 z = 11 \\4 y + 9 z = - 12\end{array} \right.{x+3y+2z=114y+9z=−12​

The Following System Does Not Have a Unique Solution $\left\{ \begin{array} { r } x + 3 y + 2 z = 11 \\4 y + 9 z = - 12\end{array} \right.$