Step 1) Because the second equation is already solved for #y# we can substitute #(-x + 6)# for #y# in the first equation and solve for #x#:
#4x - 3y = -4# becomes:
#4x - 3(-x + 6) = -4#
#4x - (3 xx -x) - (3 xx 6) = -4#
#4x - (-3x) - 18 = -4#
#4x + 3x - 18 = -4#
#(4 + 3)x - 18 = -4#
#7x - 18 = -4#
#7x - 18 + color(red)(18) = -4 + color(red)(18)#
#7x - 0 = 14#
#7x = 14#
#(7x)/color(red)(7) = 14/color(red)(7)#
#(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = 2#
#x = 2#
Step 2) Substitute #2# for #x# in the second equation and calculate #y#:
#y = -x + 6# becomes:
#y = -2 + 6#
#y = 4#
The solution is: #x = 2# and #y = 4# or #(2, 4)#
