Question 80320
three unknowns need three different equations ... the equations for this case are:


A+B+C=12 ... A+B=C/2 ... A+C=3B


subtract second equation from first equation to get C=12-(C/2) ... (3/2)C=12 ... C=8


subtract third equation from first equation to get B=12-3B ... 4B=12 ... B=3


using first equation A+3+8=12 ... A=1