How do you solve the system of equations #2x + 6y = 3# and #- 2x + 14y = 7#?

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Answer 1 · 2018-04-14T12:34:43

#x=0#
#y=1/2#

#2x+6y=3# --- (1)
#-2x+14y=7# --- (2)

From (1),
#2x+6y=3#
#2x=3-6y#
#x=1/2(3-6y)# --- (3)

Sub (3) into (2)

#-2times1/2(3-6y)+14y=7#
#6y-3+14y=7#
#20y=10#
#y=1/2# --- (4)

Sub (4) into (3)

#x=1/2(3-6times1/2)#
#x=0#

Answer 2 · 2018-04-14T12:35:54

make x in terms of y and then substitute in the second equation

2#x# + #6y# = 3

2#x# = 3 - 6#y#

#x# = #3/2# - 3#y#

substitute in the second equation

-2#x# + 14#y# = 7

-2(#3/2# - 3#y#) + 14 #y# = 7

-3 + 6#y# +14#y# = 7
20#y# = 10

#y# = #1/2#

then substitute y in any of the 2 equations to get x
you get #y# = #1/2# and #x# = 0

Hope this helped