What is the solution to the following system?
1 answer:
From eqn 3,
2z = 10
z = 5
Substitute z = 5 into eqn 1&2
x + 2y + 5 = 9
x + 2y = 4
x - y + 3(5) = 13
x - y = -2
Eqn 1 - eqn 2:
x + 2y - (x - y) = 4 - (-2)
3y = 6
y = 2
Substitute y = 2 into eqn 1
x + 2(2) = 4
x = 0
Thus, x = 0, y = 2, z = 5 (option 3)
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