Solve the system. $\left\{ \begin{array} { l }
x + \frac { 1 } { 3 } y + z = 12 \
\frac { 1 } { 3 } x - y + \frac { 1 } { 2 } z = - 2 \
x + \frac { 1 } { 2 } y - \frac { 1 } { 2 } z = 7
\end{array} \right.$
A) $( 6 , - 6,4 )$
B) $( 6,6,4 )$
C) $( - 6,6 , - 4 )$
D) The system is inconsistent.
E) The equations are dependent.

Question

Solve the system. $\left\{ \begin{array} { l } 
x + \frac { 1 } { 3 } y + z = 12 \
\frac { 1 } { 3 } x - y + \frac { 1 } { 2 } z = - 2 \
x + \frac { 1 } { 2 } y - \frac { 1 } { 2 } z = 7
\end{array} \right.$
A) $( 6 , - 6,4 )$
B) $( 6,6,4 )$
C) $( - 6,6 , - 4 )$
D) The system is inconsistent.
E) The equations are dependent.

Quizplus · Accepted Answer

We can solve this system by either substitution or elimination method. Let's use the elimination method.

Multiplying the first equation by 3, we get:
$$3x+y+3z=36$$
Multiplying the second equation by 6, we get:
$$2x-6y+3z=-12$$
Adding these two equations gives us:
$$5x-5y+6z=24$$
Multiplying the third equation by 2, we get:
$$2x+y-z=14$$
Multiplying the first equation by 2, we get:
$$2x+\frac{2}{3}y+2z=24$$
Adding these two equations gives us:
$$2x+y-z=14$$
Notice that this is the same as the third equation. Therefore, the system is dependent and has infinitely many solutions. We can solve for two of the variables in terms of the third variable. From the third equation, we have:
$$x = 7 - \frac{1}{2}y + \frac{1}{2}z$$
Substituting this into the first equation, we get:
$$7 - \frac{1}{2}y + \frac{5}{6}z + \frac{1}{3}y + z = 12$$
Simplifying gives us:
$$\frac{1}{6}y + \frac{11}{6}z = 5$$
Solving for y in terms of z, we get:
$$y = 30 - 11z$$
Substituting this into the third equation, we get:
$$x + \frac{1}{2}(30-11z) - \frac{1}{2}z = 7$$
Simplifying gives us:
$$x = 6 + 6z$$
Therefore, the solutions are of the form:
$$(6+6z, 30-11z, z)$$
Substituting z=4 gives us the solution:
$$(6,6,4)$$

Solve the System A) (6,−6,4)( 6 , - 6,4 )(6,−6,4) B) (6,6,4)( 6,6,4 )(6,6,4)

Solve the System A) $( 6 , - 6,4 )$ B) $( 6,6,4 )$