How do you solve the system of equations #2x-5y=3# and #x-3y=1#?

1 Answer
Taner Müftüoğlu
Sep 30, 2016

Try to get rid of one of the unknown...

Explanation:

#2x-5y=3#
#x-3y=1# multiply this one by - 2 and add to first one so #x# term will disappear and we will get an equation with only one unknown parameter which is #y#

#2x-5y=3#
#-2x+6y=-2#
#0+y=1 => y=1#
Put 1 instead of #y# in one of the equations to obtain #x#
#x-3*(1)=1#
#x-3=1 =>x=4#

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