#-\frac { 1} { 2} x + \frac { 1} { 5} y = 9 #----------(1)
and
#7x - \frac { 1} { 3} y = - \frac { 2} { 3}#------------(2)
Multiplying equation (1) by 10 and equation (2) by 3 to eliminate the fractional part:
#- 5 x + 2 y = 90#--------------(1') and
#21x - 1 y = - 2# -----------(2')
Eliminate any one variable (say #y#):
(1') + (2') x 2 #=>#
#- 5 x + (2xx 21x) + 2 y + (2xx -1y) = 90+ (2xx -2)#
#=> -5x +42x +2y -2y = 90 - 4#
#=> 37x = 86#
#=> x = 86/37 = 2 12/37#
Substitute this value of #x# in (1') or (2') to get the value of #y#
(2')#=> 21 xx(86/37) - 1 y = - 2#
#=> 1806/ 37 -y = -2 #
#=> - y = -2 - 1806/37#
#-y = (-74-1806)/37#
#y = 1880/37#
Cross check the validity of #x and y# by substituting in left hand side (LHS)of any of (1) and (2):
LHS of (1)#=>#
#-\frac { 1} { 2} x + \frac { 1} { 5} y = -1/2 (86/37) + 1/5 xx (1880/37)#
#= -43/37 + 376/ 37 = 333/37 #
#= 9 # = RHS (right hand side ) of equation (1)
#therefore x= 86/37 and y = 1880/37# are the correct values.
