The obvious method here is to eliminate one of the variables by creating additive inverses and then adding the two equations.
#color(white)(xxxxxxxxx)4acolor(red)(+3b)=22# ....................A
#color(white)(xxxxxxxxx)5acolor(red)(-4b)=43# ....................B
#A xx4color(white)(xxxx)16acolor(red)(+12b)=88# ....................C
#B xx3color(white)(xxxx)15acolor(red)(-12b)=129# ..................D
#C+D color(white)(xx.x)31acolor(white)(xxx.) =217" "larr# No b term!!
#color(white)(xxxxxxxxxx.xxx)color(blue)( a=7)#
Now that we know #a#, subst 7 for #a# in A (or B,C,D)
#color(white)(xxxxxxxx)4(color(blue)(7))+3b=22# ....................A
#color(white)(xxxxxxxxx)28+3b=22#
#color(white)(xxxxxxxxxxxxx)3b=22 -28#
#color(white)(xxxxxxxxxxxxx)3b=-6#
#color(white)(xxxxxxxxxxx.xx)color(magenta)(b=-2)#
Check in B:
#color(white)(xxxxxxxxx)5(7)-4(-2) " should be " 43# ....................B
#color(white)(xxxxxxxxxx)35" + "8=43#
