We have to solve for x, y and z using
x + y + z = 6 … (1)
2x –y + 3z = 9 … (2)
-x + 2y + 2z = 9 … (3)
Add (1) and (3)
=> 3y + 3z = 15
=> y + z = 5
Substitute this in (1)
=> x + 5 = 6
=> x = 1
substitute this in (2)
=> 2 – y + 3z = 9
=> 2 – (5 – z) + 3z = 9
=> 2 – 5 + z + 3z = 9
=> -3 + 4z = 9
=> 4z = 12
=> z = 3
y + z = 5
=> y = 5 – 3
=> y = 2
Therefore x = 1, y = 2 and z = 3.
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