Which system of equations has a solution of [tex]\((0, 1, 2)\)[/tex]?

A.
[tex]\[
\begin{cases}
-2x + 2y = 2 \\
y + 7z = 13 \\
x - y - z = 0 \\
2x + 4y = 4
\end{cases}
\][/tex]

B.
[tex]\[
\begin{cases}
-2x + 2y = 2 \\
y - 5z = 9 \\
x + y + z = 2
\end{cases}
\][/tex]

C.
[tex]\[
\begin{cases}
-2x + 2y = 2 \\
2y - z = 4 \\
x - y + z = -1 \\
x + 3y - z = 1
\end{cases}
\][/tex]

D.
[tex]\[
\begin{cases}
-2x + 2y = 2 \\
4x + 2y = 2 \\
5y - 2z = 1
\end{cases}
\][/tex]



Answer :

To determine which system of equations is satisfied by the solution [tex]\((0, 1, 2)\)[/tex], we need to test each system with these values for [tex]\( (x, y, z) = (0, 1, 2) \)[/tex].

System A:
1. Equation 1: [tex]\( y + 7z = 13 \)[/tex]
[tex]\[ 1 + 7 \cdot 2 = 1 + 14 = 15 \neq 13 \][/tex]
This equation is not satisfied.

2. Equation 2: [tex]\( x - y - z = 0 \)[/tex]
[tex]\[ 0 - 1 - 2 = -3 \neq 0 \][/tex]
This equation is not satisfied.

3. Equation 3: [tex]\( 2x + 4y = 4 \)[/tex]
[tex]\[ 2 \cdot 0 + 4 \cdot 1 = 0 + 4 = 4 \][/tex]
This equation is satisfied.

The first and second equations of System A are not satisfied, so System A is not the correct system.

System B:
1. Equation 1: [tex]\( y - 5z = 9 \)[/tex]
[tex]\[ 1 - 5 \cdot 2 = 1 - 10 = -9 \neq 9 \][/tex]
This equation is not satisfied.

2. Equation 2: [tex]\( x + y + z = 2 \)[/tex]
[tex]\[ 0 + 1 + 2 = 3 \neq 2 \][/tex]
This equation is not satisfied.

Both equations of System B are not satisfied, so System B is not the correct system.

System C:
1. Equation 1: [tex]\( 2y - z = 4 \)[/tex]
[tex]\[ 2 \cdot 1 - 2 = 2 - 2 = 0 \neq 4 \][/tex]
This equation is not satisfied.

2. Equation 2: [tex]\( x - y + z = -1 \)[/tex]
[tex]\[ 0 - 1 + 2 = 1 \neq -1 \][/tex]
This equation is not satisfied.

3. Equation 3: [tex]\( x + 3y - z = 1 \)[/tex]
[tex]\[ 0 + 3 \cdot 1 - 2 = 3 - 2 = 1 \][/tex]
This equation is satisfied.

The first and second equations of System C are not satisfied, so System C is not the correct system.

System D:
1. Equation 1: [tex]\( 4x + 2y = 2 \)[/tex]
[tex]\[ 4 \cdot 0 + 2 \cdot 1 = 0 + 2 = 2 \][/tex]
This equation is satisfied.

2. Equation 2: [tex]\( 5y - 2z = 1 \)[/tex]
[tex]\[ 5 \cdot 1 - 2 \cdot 2 = 5 - 4 = 1 \][/tex]
This equation is satisfied.

Both equations of System D are satisfied, so System D is the correct system.

Therefore, the system of equations that is satisfied by the solution [tex]\((0, 1, 2)\)[/tex] is:

D.
[tex]\[ \begin{cases} 4x + 2y = 2 \\ 5y - 2z = 1 \end{cases} \][/tex]