Answer:
a=-2
b=3
c=5
Explanation:
I assume you want us to solve for te values of a, b, and c
since the last equation has eliminated b, we will eliminate b in the other 2 equations to solve
-3a-4b+2c=28
a+3b-4c=-31
eliminate b
lcm of 3 and -4 is -12 because that's math
multiply first equation by 3 and 2nd by 4 to get
-9a-12b+6c=84
4a+12b-16c=-124
add the equations together
-9a-12b+6c=84
4a+12b-16c=-124 +
-5a+0b-10c=-40
-5a-10c=-40
divide both sides by -5
a+2c=8
now compare this to the 3rd equation
a+2c=8 and 2a+3c=11
eliminate a from both equations
multiply first equation by -2
-2a-4c=-16
add this to other equation
-2a-4c=-16
2a+3c=11 +
0a-c=-5
-c=-5
c=5
subsitute back
a+2c=8
a+2(5)=8
a+10=8
a=-2
subsitute way back to find b
a+3b-4c=-31
-2+3b-4(5)=-31
-2+3b-20=-31
3b-22=-31
3b=9
b=3
a=-2
b=3
c=5