Step 1) Multiply each side of the first equation by #color(red)(-2)# so the #x# coefficient is #-4#:
#2x = -y + 6#
#color(red)(-2) xx 2x = color(red)(-2)(-y + 6)#
#-4x = (color(red)(-2) xx -y) + (color(red)(-2) xx 6)#
#-4x = 2y - 12#
Step 2) Solve the second equation for #-4x#:
#-4x + 3y = 8#
#-4x + 3y - color(red)(3y) = 8 - color(red)(3y)#
#-4x + 0 = 8 - 3y#
#-4x = 8 - 3y#
Step 3) Because both equations are now solved in terms of #-4x# we can equate the right side of each equation to each other and solve for #y#:
#2y - 12 = 8 - 3y#
#2y - 12 + color(red)(12) + color(blue)(3y) = 8 - 3y + color(red)(12) + color(blue)(3y)#
#2y + color(blue)(3y) - 12 + color(red)(12) = 8 + color(red)(12) - 3y + color(blue)(3y)#
#(2 + color(blue)(3))y - 0 = 20 - 0#
#5y = 20#
#(5y)/color(red)(5) = 20/color(red)(5)#
#(color(red)(cancel(color(black)(5)))y)/cancel(color(red)(5)) = 4#
#y = 4#
Step 4) Substitute #4# for #y# in the solution to either equation at the end of Step 1 or Step 2 and solve for #x#. Either equation can be used. I will use the equation from Step 1:
#-4x = 2y - 12# becomes:
#-4x = (2 * 4) - 12#
#-4x = 8 - 12#
#-4x = -4#
#(-4x)/color(red)(-4) = -4/color(red)(-4)#
#(color(red)(cancel(color(black)(-4)))x)/cancel(color(red)(-4)) = 1#
#x = 1#
The Solution Is: #x = 1# and #y = 4# or #(1, 4)#
