Answer to Question #35172 in Algebra for Sam

Question

-3a-4b+2c=28 a+3b-4c=-31 2a+3c=11

Answer 1

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