How to solve for #x,y# this system #{(x = y-2),(4y+x^2=16):}# ?

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Answer 1 · 2017-05-26T16:55:33

See below.

Calling #x, y# the two numbers such that #x < y# we have:

#{(x = y-2),(4y+x^2=16):}#

Substituting the first equation into the second

#4(x+2)+x^2=16# now solving for #x# we obtain

#x = -2(1pm sqrt3)# so

#y = 2-2(1pm sqrt3)#

If we discard negative solutions, then

#x = 2(sqrt3-1)# and

#y = 2 sqrt3#