I forgot the third equation last time so the full question is
a+b+c=28
b=a+3
a+3+b+10c=131
I only got up to the part where I plug in the b but I don't know what to do with the other variables.
c = 28-a-b = 28-a-(a+3)
a+3+(a+3)+10[28-a-(a+3) = 131
Solve for a, then b and c.