Q.6. Complicated simultaneous equations in three unknowns

Question 6:
Solve:
1] 3a= -10a^2 - 64b^2 - 36c^2 + 12ab + 8ac
2] 48b= -4a^2 + 176b^2 - 9c^2 + 96ab + 128bc
3] 27c = -4a^2 - 16b^2 + 63c^2 + 54ac + 108bc  

Answer 6:
Two answers are
(1) a = 0, b = 0 and c = 0 and
(2) a = - 0.3144, b = - 0.0487 and c = - 0.00865
as worked out below.

Rewriting the given equations,
4a^2 + 16b^2 + 9c^2 = (a/4)(6a + 12b + 8c - 3) ... (1)
4a^2 + 16b^2 + 9c^2 = 16b(6a + 12b + 8c - 3) ... (2)
4a^2 + 16b^2 + 9c^2 = 9c(6a + 12b + 8c - 3) ... (3)
=> a/4 = 16b = 9c
=> a = 64b = 36c

Plugging a = 36c and b = 9c/16 in eqn. (3),
4(36c)^2 + 16(9c/16)^2 + 9c^2 = 9c(216c + 27c/4 + 8c - 3)
[Note: c = 0 => a = b = c = 0. If c ≠ 0, then dividing by c leads to the following steps.]
=> (5184 + 81/16 + 9)c = 9(224 + 27/4)c - 27
=> (5193 - 2016 + 81/16 -243/4)c= - 27
=> c = - (27*16)/49941 = - 0.00865
=> b = (9c/16) = - 0.00487 and a = 36c = - 0.3114
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Verification:
3a = (- 0.3114) = - 0.9342
-10a^2 - 64b^2 - 36c^2 + 12ab + 8ac
= -10(0.3114)^2 - 64(0.00487)^2 - 36(0.00865)^2 + 12(-0.3114)(-0.00487) + 8(-0.3114)(-0.00865)
= - 0.9697 - 0.00019 - .00269 + 0.0182 + 0.02155
= -0.9328
[confirms very well].

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