How do you solve the following system: #-x+6y=12, 3x+4y=-10 #?

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Answer 1 · 2016-09-08T13:04:04

#x=-54/11#

#y=13/11#

#-x+6y=12# and #3x+4y=-10#

or

#x=6y-12#

or

#x=6(y-2)#

Multiplying #-x+6y=12# by #3#

We get

#-3x+18y=36#

By adding #-3x+18y=36# with #3x+4y=-10#

We get

#-3x+18y+3x+4y=36-10#

or

#22y=26#

or

#y=26/22#

or

#y=13/11#----------------------Ans #1#

By plugging the value #y=13/11# in the equation #x=6(y-2)#

We get

#x=6(13/11-2)#

or

#x=6 times (13-22)/11#

or

#x=6times (-9/11)#

or

#x=-54/11#---------------------Ans #2#