Question #a512b

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Answer 1 · 2017-06-26T19:06:01

Any integer multiple of the lowest solution: #(3,-2)#

Lets label the first equation

#2x+3y=0#

And the second equation

#x+4y=-5#

The first step is to multiply the second equation by #-2#

#-2(x+4y)=-2(-5)#

#-2x-8y=10#

Adding this second equation to the first equation

#2x+3y=0#
#-2x-8y=10#

#=> -5y=10#

#y=-2#

Plugging #y=-2# back into one of the original two equations

#2x+3color(red)(y)=0#

#2x+3color(red)(-2)=0#

#2x-6=0#

#2x=6#

#x=3#

The solution is #(3,-2)#. In fact, any integer multiple of this answer gives another solution:

#2xx(3,-2)=(6,-4)#

#3xx(3,-2)=(9,-6)#

#4xx(3,-2)=(12,-8)#