#x-3y=20#----------------(1)
# -6x + 5y = 4#-----------------(2)
We can find the value off variable #x# in terms of #y# from equation (1) as the coefficient of #x# is 1 .
#=> x= 20+3y# ----------(3)
Now, substitute this value of #x# in equation (2),
#=> -6(20+3y) +5y = 4#
#=> -120 - 18y +5y = 4#
#=> -18y +5y = 4+120#
#=> -13y =124#
#therefore y= 124/-13 = 9.5384#------(4) --- (truncated value)
Now substitute this value of #y# in equation (3) to get the value of #x#,
#=> x= 20 +3(9.5384) = 20 + 28.615384 = 48.615384#
# therefore x= 48.615384 and y = 9.5384#
Note: The values of #x# and #y# are truncated.
