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ChrisVBest ResponseYou've already chosen the best response.1
im trying to remember how to do this one sec
 2 years ago

lazypigBest ResponseYou've already chosen the best response.2
\[x ^{2}+y ^{2}+2xy =16+6\times2\] \[\left( x +y \right)^{2}=28\] \[x +y =\pm \sqrt{28}\]
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.1
dunno how you came up with that but no way that works
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.1
its obvious the values of x and y are 2 and 3 but i cannot for the life of me remember how to explain it
 2 years ago

myininayaBest ResponseYou've already chosen the best response.0
\[(x+y)^2=x^2+2xy+y^2=x^2+y^2+2(xy)=16+2(6)=16+12=28\] so we have \[(x+y)^2=28 => x+y=\pm \sqrt{28}\]
 2 years ago

myininayaBest ResponseYou've already chosen the best response.0
so i agree with lazy :)
 2 years ago
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