If x 2 -1 y 2 -1 -9-6x-y where x-y- R- then the value of x 2 - y 2 is-

Given, (x^2+1)(y^2+1)+9=6(x+y) where x, y ∈ R x^2y^2+x^2+y^2+1+9=6x+6y ⇒ x^2y^2+1+x^2+y^2+9 6x 6y=0 ⇒ x^2y^2+1 2xy+2xy+x^2+y^2+9 6x 6y ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

If x 2 +1 ...

Question

If $(x^{2} + 1) (y^{2} + 1) + 9 = 6 (x + y)$ (where $x, y \in R$ ), then the value of $(x^{2} + y^{2})$ is,

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Solution

Given, $(x^{2} + 1) (y^{2} + 1) + 9 = 6 (x + y)$ where $x, y \in R$
$x^{2} y^{2} + x^{2} + y^{2} + 1 + 9 = 6 x + 6 y$
$\Rightarrow x^{2} y^{2} + 1 + x^{2} + y^{2} + 9 - 6 x - 6 y = 0$
$\Rightarrow x^{2} y^{2} + 1 - 2 x y + 2 x y + x^{2} + y^{2} + 9 - 6 x - 6 y = 0$
$\Rightarrow (x y - 1)^{2} + (x + y)^{2} + 9 - 6 (x + y) = 0$
$\Rightarrow (x y - 1)^{2} + (x + y)^{2} + 3^{2} - 6 (x + y) = 0$
$\Rightarrow (x y - 1)^{2} + ((x + y) - 3)^{2} = 0$ ............(1)
from equation (1) we can say that
$x y - 1 = 0$ and $(x + y) - 3 = 0$
$\Rightarrow x y = 1$ $x + y = 3$
$∴ x^{2} + y^{2} = (x + y)^{2} - 2 x y$
$= (3)^{2} - 2 (1)$
$= 9 - 2$
$= 7$

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If (x2+1)(y2+1)+9=6(x+y) (where x,y∈R), then the value of (x2+y2) is,

If $(x^{2} + 1) (y^{2} + 1) + 9 = 6 (x + y)$ (where $x, y \in R$ ), then the value of $(x^{2} + y^{2})$ is,