Second equation:a+3b-4c=-31We multiply it by 2: 2a+6b-8c=-62
Find 2a from the third equation and substitute into the second equation: 11-3c+6b-8c=-62 6b-11c=-73
Find 6b from the previous equation: 6b=11c-73
We multiply it by 2: 12b-22c=-146
First equation: -3a-4b+2c=28
Multiply it by -3: 9a+12b-6c=-84
substitute 6b from previous result into the above mentioned equation: 9a+22c-146-6c=-84
We get systems of 2 equations in 2 variables: 9a+16c=62 2a+3c=11; Subtract from the first equation the second, multiplied by 4: a+4c=18 2a+3c=11 Find a from the first equation: a=18-4c and replace it with the second equation: 2(18-4c)+3c=11 36-5c=11 5c=25 hence c=5 a=18-4c=18-4*5=-2 Substitute solutions a,c found in the second equation of the system: a+3b-4c=-31 -2+3b-20=-31 3b=22-31=-9 hence b=-3 Finally answer is the following: a=-2 b=-3 c=5
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