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