A awesome mathematics problem | Olympiad Question | can you solve this problem | x=?,y=?

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[ManojkantSamal]

*= read as square root
^=read as to the power
According to the question

X^2+y^2+xy=84....eqn2
Now simplify eqn1
X+y-*(xy)=6
X+y=6+*(xy)
Squaring the equation
X^2+y^2+2xy=36+xy+12.*(xy)
X^2+y^2+xy=36+12.*(xy)
84=36+12.*(xy)
12.*(xy)=84-36=48
*(xy)=48/12=4
Xy=4^2=16
Put the value of *(xy) in eqn1
X+y-*(xy)=6
X+y-4=6

We know
(X-y)^2=(x+y)^2-4xy
=(10)^2-(4×16)
=100-64=36

Eqn3 +eqn4
X+y+x-y=10+6=16
2x=16
X=16/2=8
Again
X+y=10
Y=10-x
=10-8=2
Hence
X=8, y=2

[souzasilva]

Escreva o radical corretamente professor, assim não dá. (Write the radical correctly, teacher, it won't work.)

[michaeldoerr]

The answers are (8, 2) and (2, 8). Also I actually took that fairly easy route of squaring the RHS and adding 2xy to 36 substituting in what xy means which is 16