How do you solve the system #1/3x-y=3# and #2x+y=25# using substitution?

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Answer 1 · 2017-10-02T05:43:38

#x = 12# and # y=1#

Given equations are:

(1) ------ #1/3x-y=3#

(2)------#2x+y=25#

Multiply (1) by 3 to eliminate the fractional part,

(1) ------- # x- 3y = 9#

# => x= 9+ 3y# -------- let this be equation (3)

Now substitute this value of #x# from (3) in equation (2),

(2) --------- # 2(9+3y) + y = 25#

# => 18 + 6y +y = 25#

#=> 7y = 25- 18#

#=> y = 7/7#

Therefore,

#y=1#

Substituting this value of #y# in equation (3),

#x= 9 + 3\times 1#

#x = 9+3 #

Therefore,
#x = 12#
So we have the values of #x # and #y# as,

#x = 12# and # y=1#