Step 1) Solve both equations for #x#:
#1/2x + 3y = 5/6#
#color(red)(2)(1/2x + 3y) = color(red)(2) xx 5/6#
#(color(red)(2) xx 1/2x) + (color(red)(2) xx 3y) = cancel(color(red)(2)) xx 5/(color(red)(cancel(color(black)(6)))3)#
#color(red)(2)/2x + 6y = 5/3#
#1x + 6y = 5/3#
#x + 6y = 5/3#
#x + 6y - color(red)(6y) = 5/3 - color(red)(6y)#
#x + 0 = 5/3 - 6y#
#x = 5/3 - 6y#
#1/3x - 5y = 16/9#
#color(red)(3)(1/3x - 5y) = color(red)(3) xx 16/9#
#(color(red)(3) xx 1/3x) - (color(red)(3) xx 5y) = cancel(color(red)(3)) xx 16/(color(red)(cancel(color(black)(9)))3)#
#color(red)(3)/3x - 15y = 16/3#
#1x - 15y = 16/3#
#x - 15y = 16/3#
#x - 15y + color(red)(15y) = 16/3 + color(red)(15y)#
#x - 0 = 16/3 + 15y#
#x = 16/3 + 15y#
Step 2) Because the left side of both equations are equal we can equate the right side of both equations and solve for #y#:
#5/3 - 6y = 16/3 + 15y#
#5/3 - color(blue)(16/3) - 6y + color(red)(6y) = 16/3 - color(blue)(16/3) + 15y + color(red)(6y)#
#(5 - color(blue)(16))/3 - 0 = 0 + (15 + color(red)(6))y#
#-11/3 = 21y#
#1/color(red)(21) xx -11/3 = (21y)/color(red)(21)#
#-(1 xx 11)/(color(red)(21) xx 3) = (color(red)(cancel(color(black)(21)))y)/cancel(color(red)(21))#
#-11/63 = y#
#y = -11/63#
Step 3) Substitute #-11/63# for #y# in the solution to either equation in Step 1 and calculate #x#:
#x = 16/3 + 15y# becomes:
#x = 16/3 + (15 xx -11/63)#
#x = (21/21 xx 16/3) + (15 xx -11)/63#
#x = 336/63 + (-165)/63#
#x = (336- 165)/63#
#x = 171/63#
The Solution Is:
#x = 171/63# and #y = -11/63#
Or
#(171/63, -11/63)#