Select the correct systems of equations. Which systems of linear equations have no solution?

A.
[tex]\[
\begin{array}{l}
x + y + z = 1{,}100 \\
x - 2y - z = -500 \\
2x + 3y + 2z = 2{,}600 \\
\end{array}
\][/tex]

B.
[tex]\[
\begin{array}{l}
x + y + z = 1{,}400 \\
x - 2y - z = -500 \\
2x + 2y + 2z = 2{,}700 \\
\end{array}
\][/tex]

C.
[tex]\[
\begin{array}{l}
x + y + z = 1{,}900 \\
x - y - 2z = -2{,}000 \\
2x + 2y + z = 1{,}100 \\
\end{array}
\][/tex]

D.
[tex]\[
\begin{array}{l}
x + y + z = 1{,}500 \\
x - y - z = -500 \\
2x + y + z = 2{,}000 \\
\end{array}
\][/tex]

E.
[tex]\[
\begin{array}{l}
x + y + z = 1{,}400 \\
-0.5x - 0.5y - 0.5z = -900 \\
2x + 3y + 2z = 3{,}000 \\
\end{array}
\][/tex]

F.
[tex]\[
\begin{array}{l}
x + y + z = 2{,}400 \\
2x - 2y + 2z = 700 \\
x + 3y + z = 2{,}400 \\
\end{array}
\][/tex]



Answer :

To determine which systems of linear equations have no solution, let's analyze each system one by one. Here’s a step-by-step analysis of each system, summarizing whether they have solutions or not.

### System 1:
[tex]\[ \begin{array}{l} x + y + z = 1100 \\ x - 2y - z = -500 \\ 2x + 3y + 2z = 2600 \\ \end{array} \][/tex]

### System 2:
[tex]\[ \begin{array}{l} x + y + z = 1400 \\ x - 2y - z = -500 \\ 2x + 2y + 2z = 2700 \\ \end{array} \][/tex]

### System 3:
[tex]\[ \begin{array}{l} x + y + z = 1900 \\ x - y - 2z = -2000 \\ 2x + 2y + z = 1100 \\ \end{array} \][/tex]

### System 4:
[tex]\[ \begin{array}{l} x + y + z = 1500 \\ x - y - z = -500 \\ 2x + y + z = 2000 \\ \end{array} \][/tex]

### System 5:
[tex]\[ \begin{array}{l} x + y + z = 1400 \\ -0.5x - 0.5y - 0.5z = -900 \\ 2x + 3y + 2z = 3000 \\ \end{array} \][/tex]

### System 6:
[tex]\[ \begin{array}{l} x + y + z = 2400 \\ 2x - 2y + 2z = 700 \\ x + 3y + z = 2400 \\ \end{array} \][/tex]

After analyzing each system, we conclude that the following systems:

- System 2
- System 5
- System 6

have no solution.

Thus, the systems of linear equations that have no solution are Systems [tex]\( \mathbf{2, 5, \text{ and } 6} \)[/tex].