How do you solve the following system: #11y + 3x = 4, x-2y=3 #?

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Answer 1 · 2017-07-17T08:28:56

#x = 41/17#
#y = -5/17#

Let,
#3x + 11y = 4# be eq 1
#x - 2y = 3# be eq 2

now, multiply eq 2 by -3 so as it will be cancelled out in the next step.
we get,
#-3x + 6y = -9#

#rArr 3x + 11y =4#
# -3x + 6y = -9#

#rArr 17y = -5#
#therefore y =-5/17#
now, substitute value of y in any equation,
u get # x = 41/17 #

ENJOY MATHS !!!!!

Answer 2 · 2017-07-17T08:30:34

By arranging equation, #x=41/17# and #y=-5/17#

#3x + 11y = 4#
#-3x + 6y = -9# (when you enlarge the second equation with -3)
Combine these two equations and get
#17y = -5#

#y = -5/17#

Put this value in any original equation (for instance the first)

#3x - 55/17 = 4#

#3x = (68+55)/17#

#x = 123/51#

#x= 41/17#

This is your answer #x=41/17# and #y=-5/17#