Solve the system.
-$$\left\{ \begin{array} { r }
x + y + z = 6 \
x - y + 3 z = 10 \
2 x + 2 y + 2 z = 8
\end{array} \right.$$
A) $( 3,2,1 )$
B) $( 2,1,3 )$
C) $( 3,1,2 )$
D) $\varnothing$

Question

Solve the system.
-$$\left\{ \begin{array} { r } 
x + y + z = 6 \
x - y + 3 z = 10 \
2 x + 2 y + 2 z = 8
\end{array} \right.$$
A) $( 3,2,1 )$
B) $( 2,1,3 )$
C) $( 3,1,2 )$
D) $\varnothing$

Quizplus · Accepted Answer

The third equation is simply the first equation multiplied by 2, which means it does not provide any new information. This redundancy suggests a potential inconsistency or dependency in the system. Upon closer inspection, adding the first and second equations yields $2x + 4z = 16$, which simplifies to $x + 2z = 8$. However, this does not directly contradict the given equations, suggesting the system might have a solution. To check for consistency, we can attempt to solve the system:1. From the first equation, $x + y + z = 6$.2. From the second equation, $x - y + 3z = 10$.Adding these two equations to eliminate $y$, we get $2x + 4z = 16$, or simplifying, $x + 2z = 8$. This is consistent with our earlier observation.However, upon further inspection, if we attempt to solve these equations, we find that they are not independent. The third equation, $2x + 2y + 2z = 8$, is indeed just the first equation multiplied by 2, which means it does not provide any new information. This redundancy in the system does not necessarily mean the system has no solution (D), but rather that we need to carefully check for consistency among the non-redundant equations.Let's correct the explanation based on the actual solution process:To solve the system, we can use either substitution or elimination. Let's use elimination for simplicity:1. Subtract the second equation from the first to eliminate $y$: $2z = 4$, so $z = 2$.2. Substitute $z = 2$ into the first equation: $x + y + 2 = 6$, so $x + y = 4$.3. Substitute $z = 2$ into the second equation: $x - y + 6 = 10$, so $x - y = 4$.From $x + y = 4$ and $x - y = 4$, we can solve for $x$ and $y$:- Adding these two equations, we get $2x = 8$, so $x = 4$.- Substituting $x = 4$ into $x + y = 4$, we get $4 + y = 4$, so $y = 0$.However, my initial conclusion was incorrect based on a misinterpretation of the system's redundancy. The correct approach shows that the system does have a solution, which contradicts my initial answer of "D" (no solution). Given the corrected solution process, the correct answer should be based on the values of $x$, $y$, and $z$ found through proper solving techniques, which I failed to accurately determine in my initial explanation. Given this mistake, let's correctly solve the system:1. The first equation is $x + y + z = 6$.2. The second equation is $x - y + 3z = 10$.3. The third equation is $2x + 2y + 2z = 8$, which simplifies to $x + y + z = 4$, not $8$ as previously stated. This indicates an inconsistency in my explanation, as the third equation should not simplify to $x + y + z = 4$, but rather it was correctly stated as a multiple of the first equation, which would not change the conclusion.Given the mistake in processing the system and the incorrect conclusion drawn, the proper step would be to reevaluate the system's solution without the flawed logic. However, based on the correct process of solving such a system, one would typically solve for each variable using methods like substitution or elimination, and then check each option against the original equations to find the correct set of values for $x$, $y$, and $z$. My initial instruction to choose "D" was based on an incorrect analysis.

Solve the System {x+y+z=6x−y+3z=102x+2y+2z=8\left\{ \begin{array} { r } x + y + z = 6 \\x - y + 3 z = 10 \\2 x + 2 y + 2 z = 8\end{array} \right.⎩⎨⎧​x+y+z=6x−y+3z=102x+2y+2z=8​

Solve the System $\left\{ \begin{array} { r } x + y + z = 6 \\x - y + 3 z = 10 \\2 x + 2 y + 2 z = 8\end{array} \right.$