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

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Answer 1 · 2017-06-14T14:22:04

By arranging equations. #x=32/13# and #y=-4/13#

#11y + 3x = 4#
#3y + 2x = 4#

Enlarge the first equation by 2 and the second by -3:

#22y + 6x = 8#
#-9y - 6x = -12#

Now combine these:

#22y - 9y + 6x - 6x =-4#

#13 y = -4#

#y=-4/13#

From here, you can find x using any original equation

#(11times-4/13) + (3x) = 4#

#3x = 4+(44/13)#

#3x = 96/13#

#x = 96/39#

or

#x=32/13#

Your answer is #x=32/13# and #y=-4/13#