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📅 May 4, 2026 👤 scraface

May 04, 2026

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27a + 9b + 3c + d = 19 \quad \text{(4)}
Substitute \( d = 1 \) into (1), (3), and (4):
Now solve the system:
From (1): \( -a + b - c = 3 \)
From (3): \( 8a + 4b + 2c = 9 \) → divide by 2: \( 4a + 2b + c = 4.5 \...
From (4): \( 27a + 9b + 3c = 18 \) → divide by 3: \( 9a + 3b + c = 6 \...
Now add (1) and \( 4a + 2b + c = 4.5 \):
Add \( (-a + b - c) + (4a + 2b + c) = 3 + 4.5 \Rightarrow 3a + 3b = 7....
Now subtract \( (4a + 2b + c) \) from \( 9a + 3b + c \):
Now solve (A) and (B):

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