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#