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

a=-2

b=3

c=5

Step-by-step 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

step-by-step explanation: