How do you solve the system of equations #9y + 4x - 11= 0# and #3y + 10x + 31= 0#?

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Answer 1 · 2017-05-13T00:37:13

Using elimination:
#x=-4#
#y=3#

We have the equations:

#9y+4x-11=0#
#3y+10x+31=0#

Multiply the second equation by #3#. Our two equations are now:

#9y+4x-11=0#
#9y+30x+93=0#

Now we can subtract the second equation from the first equation. This leaves:

#-26x-104=0#

Solve for #x#

#-26x=104#

#x=-4#

Now plug this into one of the original equations. I like small numbers so I'll plug into the second equation:

#3y+10(-4)+31=0#

Solve for #y#:

#3y-40+31=0#

#3y-9=0#

#3y=9#

#y=3#

My answers agree with Donny's.