how to solve a system of equations 2a-3b+5c =58, -5a+b-4c=-51, -6a-8b+c= 22

Question

Solve for the variables a, b, and c.

Answer 1

a-3b+5c =58 -5a+b-4c=-51 -6a-8b+c= 22

from equaton i- a=58+3b-5c------------------(A)

now putting the value of a in eqn ii & iii

-5(58+3b-5c)+b-4c=-51,=>-290-15b+25c+b-4c=-51,=>-14b+21c=239----(iv)

-6(58+3b-5c)-8b+c=22 ,=>-348-18b+30c-8b+c=22 ,=>-26b+31c=370-----(v)

solving (iv) & (v) we get the value of b& c

putting the value of b and c in eqn A we get the value of a

Answer 2

(2a+5c)-(15+9c)

Answer 3

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