**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)#