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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