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Answer:
The system of equations has no solution.
Step-by-step explanation:
Let the following system of linear equations:
[tex]x +2\cdot y + z = 6[/tex] (1)
[tex]2\cdot x + y + 2\cdot z = 6[/tex] (2)
[tex]x + y + z = 5[/tex] (3)
Since the quantity of equations and variables are the same, there is only one solution to this system. First, we clear [tex]x[/tex] in (3):
[tex]x = 5 - y - z[/tex]
And make substitution both in (1) and (2):
[tex](5-y-z) +2\cdot y + z = 6[/tex]
[tex]2\cdot (5-y-z)+y+2\cdot z = 6[/tex]
[tex]y = 1[/tex] (4)
[tex]y = 4[/tex] (5)
Which lead to a contradiction, since [tex]4\ne 5[/tex]. Therefore, the system of linear equations has no solution.