#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.