solving 3 equations in algebra 2 by elimination
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A: -6a-3b-3c=-27, 2a+b+c=9

B: 6a-b+c=27

C: 3a-b-5c=-18

D: A+B=8a+2c=36; E: A+C=5a-4c=-9

2D+E=16a+5a=72-9=63; 21a=63, a=3, so 15-4c=-9, 4c=24, c=6, so b=9-6-6=-3.

Therefore a=3, b=-3, c=6.
by Top Rated User (719k points)

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