How do you solve the following system: #3x+2y =0, y + 4x = 16 #?

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Answer 1 · 2018-06-22T16:05:36

#x=32/5,y=-48/5#

By the first equation we get

#y=-3/2x#
plugging this in the second one:

#-3/2x+4x=16#
#x=32/5#

#y=-48/5#

Answer 2 · 2018-06-22T16:09:42

#x = 6(2/5), y = -9(3/5)#

#3x + 2y = 0# (1)

#2y = -3x#

#y + 4x = 16# (2)

Substituting y in terms of x in (2),

#-(3/2)x + 4x = 16#

#-3x + 8x = 32#

#5x = 32# or #x = 6(2/5)#

#y = 16 - 4x = 16 - (128/5) = -48/5 = -9(3/5)#