How do you solve #[(2x), (2x+3y)]=[(y), (12)]#?

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Answer 1 · 2017-08-05T01:36:06

#x=3/2 " and " y=3#

Here we have the system

#[(2x),(2x+3y)]=[(y),(12)] ->_(R2-R1)#

We can subtract the first equation from the second equation to cancel out the #2x# factor

#[(0),(0+3y)]=[(0),(12-y)]#

Which then leaves us with the equation

#3y=12-y#

#<=># Add #y# to both sides

#4y=12#

#<=># Divide both sides by #4#

#y=3#

Then, since we know that #y=2x# from the first equation, we can plug in

#2x=3#

#<=># Divide both sides by #2#

#x=3/2#

We can check out answer by plugging in #(3/2,3)# into the second equation

#2(3/2)+3(3)=12#

#<=>#

#3+9=12#

#<=>#

#12=12#