Solve Systems of Linear Equations in Three Variables
-$$\begin{array} { r }
y + 4 z = 0 \
4 x + y - 2 z = 2 \
- 3 x - 3 z = - 9
\end{array}$$
A) $\{ ( 2 , - 4,1 ) \}$
B) $\{ ( 2 , - 8,2 ) \}$
C) $\{ ( - 1 , - 3,4 ) \}$
D) $\{ ( - 1,14,4 ) \}$

Question

Solve Systems of Linear Equations in Three Variables
-$$\begin{array} { r } 
y + 4 z = 0 \
4 x + y - 2 z = 2 \
- 3 x - 3 z = - 9
\end{array}$$
A) $\{ ( 2 , - 4,1 ) \}$
B) $\{ ( 2 , - 8,2 ) \}$
C) $\{ ( - 1 , - 3,4 ) \}$
D) $\{ ( - 1,14,4 ) \}$

Quizplus · Accepted Answer

To solve the system of equations, we can use substitution or elimination. From the first equation, we can express $y$ in terms of $z$: $y = -4z$. Substituting $y$ in the second and third equations:1. $4x - 4z - 2z = 2$ simplifies to $4x - 6z = 2$.2. The third equation is $-3x - 3z = -9$.From the third equation, we can express $x$ in terms of $z$: $-3x = -9 + 3z$, so $x = 3 - z$.Substituting $x = 3 - z$ into the modified second equation gives us $4(3 - z) - 6z = 2$, which simplifies to $12 - 4z - 6z = 2$, and further to $-10z = -10$, so $z = 1$.Substituting $z = 1$ back into $y = -4z$ gives $y = -4(1) = -4$, and substituting $z = 1$ into $x = 3 - z$ gives $x = 3 - 1 = 2$.Therefore, the solution is $(x, y, z) = (2, -4, 1)$.

Solve Systems of Linear Equations in Three Variables - y+4z=04x+y−2z=2−3x−3z=−9\begin{array} { r } y + 4 z = 0 \\4 x + y - 2 z = 2 \\- 3 x - 3 z = - 9\end{array}y+4z=04x+y−2z=2−3x−3z=−9​

Solve Systems of Linear Equations in Three Variables
- $\begin{array} { r } y + 4 z = 0 \\4 x + y - 2 z = 2 \\- 3 x - 3 z = - 9\end{array}$