Solve the system of linear equations and check any solution algebraically.? $$\left\{\begin{array} { l }
3 x + 3 y = 0 \
6 x + 3 y - z & = 0 \
12 x + 3 y + 3 z & = 0
\end{array}\right.$$ ?
A) $( - 3,6,12 )$ ?
B)?(0,0,0)
C)? $( - 3,6 , - 12 )$
D)? $( 3,6,12 )$
E)? $( 3,6 , - 12 )$

Question

Solve the system of linear equations and check any solution algebraically.? $$\left\{\begin{array} { l } 
3 x + 3 y = 0 \
6 x + 3 y - z & = 0 \
12 x + 3 y + 3 z & = 0
\end{array}\right.$$ ?
A) $( - 3,6,12 )$ ?
B)?(0,0,0)
C)? $( - 3,6 , - 12 )$
D)? $( 3,6,12 )$
E)? $( 3,6 , - 12 )$

Quizplus · Accepted Answer

To solve the system, we can start by simplifying the equations. The first equation is $3x + 3y = 0$, which simplifies to $x + y = 0$ or $y = -x$.Substituting $y = -x$ into the second and third equations:1. $6x + 3(-x) - z = 0$ simplifies to $3x - z = 0$, or $z = 3x$.2. $12x + 3(-x) + 3z = 0$ simplifies to $9x + 3z = 0$. Substituting $z = 3x$ into this equation gives $9x + 9x = 0$, or $18x = 0$, which means $x = 0$.With $x = 0$, $y = -x = 0$, and $z = 3x = 0$, the solution is $(0,0,0)$.Checking the solution algebraically by substituting $x = 0$, $y = 0$, and $z = 0$ into the original equations confirms that all three equations are satisfied, which means the solution is correct.

Solve the System of Linear Equations and Check Any Solution {3x+3y=06x+3y−z=012x+3y+3z=0\left\{\begin{array} { l } 3 x + 3 y = 0 \\6 x + 3 y - z & = 0 \\12 x + 3 y + 3 z & = 0\end{array}\right.⎩⎨⎧​3x+3y=06x+3y−z12x+3y+3z​=0=0​

Solve the System of Linear Equations and Check Any Solution $\left\{\begin{array} { l } 3 x + 3 y = 0 \\6 x + 3 y - z & = 0 \\12 x + 3 y + 3 z & = 0\end{array}\right.$