How do you solve #[(x+3y),(3x+y)]=[(-13),(1)]#?

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Answer 1 · 2018-01-11T19:08:31

#x=2# and #y=-5#

We can get two equations from the given:

(1) #x+3y=-13#
(2) #3x+y=1#

If we multiply the second equation from #-3# we get:
#-9x-3y=-3#. Adding this new equation to the first equation gives:

#-8x=-16 rarr x =2#.

Taking this value of #x# and substituting into (2) gives

#3(2)+y=1 rarr6+y=1 rarry=-5#,

so #x=2# and #y=-5#