2a-3b+5c=58 ··· Eq.1, -5a+b-4c=-51 ··· Eq.2, -6a-8b+c=22 ··· Eq.3 Multiply both sides of Eq.3 by 5, then by 4, getting 2 equqtions shown below:
-30a-40b+5c=110 ··· Eq.4, -24a-32b+4c=88 ··· Eq.5 Subtract Eq.4 from Eq.1, and add Eq.5 to Eq.2, getting 2 equations shown below:
32a+37b=-52 ··· Eq.6, -29a-31b=37 ··· Eq.7 Multiply Eq.6 by 29, and Eq.7 by 32, getting
928a+1073b=-1508 ··· Eq.8, -928a-992b=1184 ··· Eq.9 Add Eq.8 to Eq.9, getting
81b=-324 We have: b=-4 Substitute b=-4 into Eq.6, getting 32a+37(-4)=-52, so 32a=96
We have: a=3 Substitute a=3, and b=-4 into Eq.3, getting -6(3)-8(-4)+c=22 We have: c=8
CK: Substitute a=3, b=-4, and c=8 into LHS of Eq.1. 2(3)-3(-4)+5(8)=6+12+40=58 CKD.
The answer is: a=3, b=-4, and c=8