dave13 wrote:
pushpitkc wrote:
Hey
dave13We are given in the question stem that \(x+y = 2\) and \(x^2 + y^2 = 2\)
After you have figured that \((x+y)^2 = x^2 + y^2 + 2xy\)
Substituting these values, we ge t\(2^2 = 2 + 2xy\) -> \(4 - 2 = 2xy\) -> \(xy = \frac{2}{2} = 1\)(Option D)
Hope this helps!
pushpitkc many thanks for explanation. yes it helps just partially
ok from this \(x^2 + y^2 = 2\) we got this \(x^2 + y^2 + 2xy =2\) ok I understand this part so far
now you say we substitute values - which values are you talking about? and in which equation do you substitute ? how do you get this \(2^2 = 2 + 2xy\) also how do you get in it \(2^2\)
I would appreciate your detailed explanation for dummies
thanks!
Hey
dave13So we have an equation \((x+y)^2 = x^2 + y^2 + 2xy\) -> (1)
The question stem has given us the following details:
\(x+y = 2\)
\(x^2 + y^2 = 2\) -> (2)
Because \(x+y = 2\), value of
\((x+y)^2 = 2^2 = 4\) -> (3)
Substituting values from (2) and (3) in equation (1)
\((x+y)^2\) =
\(x^2 + y^2\)\(+ 2xy\)
4 =
2 + 2xy
2xy = 4-2 = 2
Therefore, xy = \(\frac{2}{2} = 1\)
Hope it is clear now.
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