Step 1) Solve the first equation for #x#:
#x + 2y = 2#
#x + 2y - color(red)(2y) = 2 - color(red)(2y)#
#x + 0 = 2 - 2y#
#x = 2 - 2y#
Step 2) Substitute #2 - 2y# for #x# in the second equation and solve for #y#:
#5x - 3y = -29# becomes:
#5(2 - 2y) - 3y = -29#
#10 - 10y - 3y = -29#
#10 - 13y = -29#
#-color(red)(10) + 10 - 13y = -color(red)(10) - 29#
#0 - 13y = -39#
#-13y = -39#
#(-13y)/color(red)(-13) = -39/color(red)(-13)#
#(color(red)(cancel(color(black)(-13)))y)/cancel(color(red)(-13)) = 3#
#y = 3#
Step 3) Substitute #3# for #y# in the solution to the first equation at the end of Step 1 and calculate #x#:
#x = 2 - 2y# becomes:
#x = 2 - (2 xx 3)#
#x = 2 - 6#
#x = -4#
The solution is: #x = -4# and #y = 3# or #(-4, 3)#