anonymous
  • anonymous
If xy = 6 and x^2 + y^2 = 16, then what is the value of (x + y)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
X+y=5
anonymous
  • anonymous
please can you explain
anonymous
  • anonymous
im trying to remember how to do this one sec

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anonymous
  • anonymous
\[x ^{2}+y ^{2}+2xy =16+6\times2\] \[\left( x +y \right)^{2}=28\] \[x +y =\pm \sqrt{28}\]
anonymous
  • anonymous
dunno how you came up with that but no way that works
anonymous
  • anonymous
Thanks
anonymous
  • anonymous
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
anonymous
  • anonymous
so x+y=5
myininaya
  • myininaya
\[(x+y)^2=x^2+2xy+y^2=x^2+y^2+2(xy)=16+2(6)=16+12=28\] so we have \[(x+y)^2=28 => x+y=\pm \sqrt{28}\]
myininaya
  • myininaya
so i agree with lazy :)

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