Step 1) Solve each equation for #x#:
#x + 2y = -2#
#x + 2y - color(red)(2y) = -2 - color(red)(2y)#
#x + 0 = -2 - 2y#
#x = -2 - 2y#
#x - 3y = -12#
#x - 3y + color(red)(3y) = -12 + color(red)(3y)#
#x - 0 = -12 + 3y#
#x = -12 + 3y#
Step 2) Because the left side of both equations are the same we can equate the right sides and solve for #y#:
#-2 - 2y = -12 + 3y#
#-2 + color(blue)(12) - 2y + color(red)(2y) = -12 + color(blue)(12) + 3y + color(red)(2y)#
#10 - 0 = 0 + (3 + color(red)(2))y#
#10 = 5y#
#10/color(red)(5) = (5y)/color(red)(5)#
#2 = (color(red)(cancel(color(black)(5)))y)/cancel(color(red)(5))#
#2 = y#
#y = 2#
Step 3) Substitute #2# for #y# in the solutions to either of the equations in Step 1 and calculate #x#:
#x = -12 + 3y# becomes:
#x = -12 + (3 xx 2)#
#x = -12 + 6#
#x = -6#
The Solution Is:
#x = -6# and #y = 2#
Or
#(-6, 2)#
