**Step 1)** Because the second equation is already solved for #x# we can substitute #(5y + 2)# for #x# in the first equation and solve for #y#:

#3x + y = 4# becomes:

#3(5y + 2) + y = 4#

#(3 xx 5y) + (3 xx 2) + y = 4#

#15y + 6 + y = 4#

#15y + 6 - color(red)(6) + y = 4 - color(red)(6)#

#15y + 0 + y = -2#

#15y + y = -2#

#15y + 1y = -2#

#(15 + 1)y = -2#

#16y = -2#

#(16y)/color(red)(16) = -2/color(red)(16)#

#(color(red)(cancel(color(black)(16)))y)/cancel(color(red)(16)) = -2/16#

#y = -1/8#

**Step 2)** Substitute #-1/8# for #y# in the second equation and calculate #x#:

#x = 5y + 2# becomes:

#x = (5 xx -1/8) + 2#

#x = -5/8 + (8/8 xx 2)#

#x = -5/8 + 16/8#

#x = (-5 + 16)/8#

#x = 11/8#

**The Solution Is:**

#x = 11/8# and #y = -1/8#

Or

#(11/8, -1/8)#