## anonymous 4 years ago If xy = 6 and x^2 + y^2 = 16, then what is the value of (x + y)?

1. anonymous

X+y=5

2. anonymous

3. anonymous

im trying to remember how to do this one sec

4. anonymous

$x ^{2}+y ^{2}+2xy =16+6\times2$ $\left( x +y \right)^{2}=28$ $x +y =\pm \sqrt{28}$

5. anonymous

dunno how you came up with that but no way that works

6. anonymous

Thanks

7. 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

8. anonymous

so x+y=5

9. myininaya

$(x+y)^2=x^2+2xy+y^2=x^2+y^2+2(xy)=16+2(6)=16+12=28$ so we have $(x+y)^2=28 => x+y=\pm \sqrt{28}$

10. myininaya

so i agree with lazy :)