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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