Solve simultaneously..? #x = 3y# and #x = 1/2 (3 + 9y)#

2 Answers
Nov 22, 2017

#x = - 3#
#y = - 1#

Explanation:

The two given equations are both equal to #x#.
Therefore they are equal to each other.

#3y = x# and #x=##1/2# #(3 + 9y)#

#3y# #=##1/2# #(3 + 9y)#

First solve for y

1) Clear the fraction by multiplying both sides by 2 and letting the denominator cancel.
After you have multiplied and canceled, you will have this:
#6y = 3 + 9y#

2) Subtract #6y# from both sides to get all the #y# terms together
#0 = 3 + 3y#

3) Subtract 3 from both sides to isolate the #3y# term
#- 3 = 3y#

4) Divide both sides by 3 to isolate #y#
#-1 = y# #larr# answer for #y#

Next solve for #x#

Sub in #-1# in the place of #y# in one of the given equations.

#x = 3y#

Replace #y# with #- 1#
#x = (3)(-1)#

Clear the parentheses
#x = - 3# #larr# answer for #x#
........................

Check
Sub in #-1# and #-3# for #y# and #x# in one of the given equations

#x=##1/2# #(3 + 9y)#

#-3=##1/2# #(3 + 9(-1))#

Clear the inner parentheses
#-3=##1/2# #(3 -9)#

Solve inside the parentheses
#-3=##1/2# #(- 6)#

Clear the parentheses by distributing the #1/2#
# - 3 = - 3#
Check!

Nov 22, 2017

#y=-1,x=-3#

Explanation:

#x=3y#-----------(1)
#x=9/2y+3/2#---(2)

#(1)-(2)# #0=-3/2y-3/2#

#:.-3/2y-3/2=0#

#:.-3/2y=3/2#

#:.y=(3/2)/(-3/2)#

#:.y=-1#

substitute #y=-1 # in (1)

#:.x=3(-1)#

#:.x=-3#

~~~~~~~~~~~~

check:-

substitute #y=-1 # and #x=-3# in (2)

#:.-3=9/2(-1)+3/2#

#:.-3=-4.5+1.5#

#:.-3=-3#