How do you solve the system of equations #x^2+y^2=13# and #2x-y=4#?

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Answer 1 · 2017-05-08T19:53:25

#x=3#

#y=2#

First, think of which two perfect squares would add up to #13#:
#4+9=13# or #9+4=13#

Take the square root of those perfect squares:
#sqrt4=2# and #sqrt9=3#

#x# must equal either #2# or #3#, and #y# must equal #2# or #3#

Let's say #x=2# and #y=3#:
#2(2)-(3)=4# #rarr# #4-3=4# #rarr# #1!=4#

Since that didn't work, let's say #x=3# and #y=2#:
#2(3)-(2)=4# #rarr# #6-2=4# #rarr# #4=4#

Therefore, #x=3# and #y=2#