We are first going to find #a# in function of #b# :
#2a-5b=-16#
#2a=-16+5b# ( we add #5b# on each side )
#a=-8+(5b)/2# ( we divide by #2# on each side )
Now that we have #a#, we can find #b# with the first equation :
#5a+3b=-9#
#5(-8+(5b)/2)+3b=-9#
#-40+(25b)/2+3b=-9#
#(25b)/2+3b=-9+40# ( we add 40 on each side )
#3b=(6b)/2#
Thus #(25b+6b)/2 = 31#
#(31b)/2=31#
#b/2=1# ( we divide by 31 on each side )
#b=2# ( we multiply by 2 on each side )
We can thus now find #a# :
#2a-5b=-16#
#2a-5*2=-16#
#2a-10=-16#
#2a=-16+10# ( we add 10 on each side )
#2a=-6#
#a=-3# ( we divide by 2 on each side )
And thus :
#a=-3#
and #b=2#
