Because the second equation has already be solved for #y# we can substitute #color(red)(-3x - 3)# for #y# in the first equation and solve for #x#:

#color(red)(-3x - 3) = 4x - 3#

#color(red)(-3x - 3) + color(blue)(3x + 3) = 4x - 3 + color(blue)(3x + 3)#

#color(red)(-3x) + color(blue)(3x) - color(red)(3) + color(blue)(3) = 4x + color(blue)(3x) - 3 + color(blue)(3)#

#0 - 0 = 4x + color(blue)(3x) - 0#

#0 = 4x + color(blue)(3x)#

#0 = (4 + 3)x#

#0 = 7x#

#0/color(red)(7) = (7x)/color(red)(7)#

#0 = (color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7))#

#0 = x# or #x = 0#

Now that we have #x# we can substitute #0# for #x# in the second equation and calculate #y#:

#y = (-3 * 0) - 3#

#y = 0 - 3#

#y = -3#