Step 1) Because the second equation is already solved in terms of #y# we can substitute #-2x + 6# for #y# in the first equation and solve for #x#:
#5x - 10y = 20# becomes:
#5x - 10(-2x + 6) = 20#
#5x + (10 xx 2x) - (10 xx 6) = 20#
#5x + 20x - 60 = 20#
#25x - 60 = 20#
#25x - 60 + color(red)(60) = 20 + color(red)(60)#
#25x - 0 = 80#
#25x = 80#
#(25x)/color(red)(25) = 80/color(red)(25)#
#(color(red)(cancel(color(black)(25)))x)/cancel(color(red)(25)) = (5 xx 16)/color(red)(5 xx 5)#
#x = (color(red)(cancel(color(black)(5))) xx 16)/color(red)(cancel(5) xx 5)#
#x = 16/5#
Step 2) Substitute #16/5# for #x# in the second equation and calculate #y#:
#y = -2x + 6# becomes:
#y = (-2 xx 16/5) + 6#
#y = -32/5 + (5/5 xx 6)#
#y = -32/5 + 30/5#
#y = -2/5#
The solution is: #x = 16/5# and #y = -2/5# or #(16/5, -2/5)#
