How do you solve the system of equations #3x + y = 136# and #x - 5y = - 40#?

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Answer 1 · 2017-01-19T06:21:14

#3x+y=136#
#x-5y=-40#

#=> x=40# and #y=16#

This system

#3x+y=136#
#x-5y=-40#

#to# can also be written as

#((3,1),(1,-5))((x),(y))=((136),(-40))# a matrix equation

and if a matrix is invertible it is true that

#Avecx=veccb => vecx=A^(-1)vecb#

To find the inverse #A^(-1)#

#A=((a,b),(c,d)) => A^(-1)=1/(det(A))((d,-b),(-c,a))#

In this case #a=3, b=1, c=1, d=-5#. Therefore

#det(A)=ad-bc => det(A)=-15-1=-16#

following the formula for #A^(-1)# we get

#=> A^(-1)=((5/16,1/16),(1/16,-3/16))#

#=> ((x),(y))=((5/16,1/16),(1/16,-3/16))((136),(-40))#

So we plug in and get

#=> x=((5)(136)-40)/16=640/16=40#

and

#y=(136+120)/16=256/16=16#