Question

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

Answer 1

X+y=5

Answer 2

please can you explain

Answer 3

im trying to remember how to do this one sec

Answer 4

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

Answer 5

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

Answer 6

Thanks

Answer 7

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

Answer 8

so x+y=5

Answer 9

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

Answer 10

so i agree with lazy :)