Step 1) Because the second equation is already solved for #y# we can substitute #(2x + 2)# for #y# in the first equation and solve for #x#:
#2y = -2x - 10# becomes:
#2(2x + 2) = -2x - 10#
#(2 * 2x) + (2 * 2) = -2x - 10#
#4x + 4 = -2x - 10#
#color(red)(2x) + 4x + 4 - color(blue)(4) = color(red)(2x) - 2x - 10 - color(blue)(4)#
#(color(red)(2) + 4)x + 0 = 0 - 14#
#6x = -14#
#(6x)/color(red)(6) = -14/color(red)(6)#
#(color(red)(cancel(color(black)(6)))x)/cancel(color(red)(6)) = -7/3#
#x = -7/3#
Step 2) Substitute #-7/3# for #x# in the first equation and calculate #y#:
#y = 2x + 2# becomes:
#y = (2 xx -7/3) + 2#
#y = -14/3 + 2#
#y = -14/3 + (3/3 xx 2)#
#y = -14/3 + 6/3#
#y = -8/3#
The Solution Is: #x = -7/3# and #y = -8/3# or #(-7/3, -8/3)#
