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A system in $3$ unknowns

Linear algebraMedium

Solve the linear system $x + y + z = 6$ , $2 x - y + z = 3$ , $x + 2 y - z = 2$ . What is $z$ ?

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Hint

Use the first equation to remove $z$ from the other $2$ equations.

Solution

The first equation gives $z = 6 - x - y$ .
Substituting into the second gives $2 x - y + 6 - x - y = 3$ , so $x - 2 y = - 3$ .
Substituting into the third gives $x + 2 y - 6 + x + y = 2$ , so $2 x + 3 y = 8$ .
From $x = 2 y - 3$ we get $4 y - 6 + 3 y = 8$ , so $y = 2$ and $x = 1$ .
Then $z = 6 - 1 - 2 = 3$ .

Answer: $3$