solve by elimination ; -6a+6b+5c=-12 and -3a-5b+c=-21 and -2a+5b+c=9

Question

Need solution

Answer 1

Call the equations A, B and C.

B+C: -5a+2c=-12, so c=(5a-12)/2. Call this D.

5A+6B: -48a+31c=-186, so c=(48a-186)/31. Call this E.

Equate D and E: (5a-12)/2=(48a-186)/31=48a/31-6; 5a-12=96a/31-12; a=0.

Therefore, substituting a=0 in each of A, B and C: 6b+5c=-12; -5b+c=-21 and 5b+c=9. Add the last two equations together: 2c=-12, so c=-6, and 5b=15 so b=3.

Check each equation for consistency: A: 18-30=-12; -15-6=-21; 15-6=9. All OK. So a=0, b=3, c=-6 is the unique solution.