**Step 1)** Because the second equation is already solve for #y# we can substitute #(2x - 8)# for #y# in the first equation and solve for #x#:

#x + 4y = -14# becomes:

#x + 4(2x - 8) = -14#

#x + (4 xx 2x) - (4 xx 8) = -14#

#x + 8x - 32 = -14#

#1x + 8x - 32 = -14#

#(1 + 8)x - 32 = -14#

#9x - 32 = -14#

#9x - 32 + color(red)(32) = -14 + color(red)(32)#

#9x - 0 = 18#

#9x = 18#

#(9x)/color(red)(9) = 18/color(red)(9)#

#(color(red)(cancel(color(black)(9)))x)/cancel(color(red)(9)) = 2#

#x = 2#

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

#y = 2x - 8# becomes:

#y = (2 xx 2) - 8#

#y = 4 - 8#

#y = -4#

**The Solution Is:**

#x = 2# and #y = -4#

Or

#(2, -4)#