The number of ordered pairs (x, y) of real nu | vedclass

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Question

(d)

We have,

$x \quad x+y^2=x^2+y =12$

$x+y^2 =12$

$x^2+y =12 \ldots 	ext {...(ii) }$

On subtracting Eq.$(i)$ from Eq. $(ii)$, we get

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(d)

We have,

$x \quad x+y^2=x^2+y =12$

$x+y^2 =12$

$x^2+y =12 \ldots 	ext {...(ii) }$

On subtracting Eq.$(i)$ from Eq. $(ii)$, we get

$\left(x^2-xight)+\left(y-y^2ight)=0$

$\Rightarrow \quad\left(x^2-y^2ight)-(x-y)=0$

$\Rightarrow \quad(x-y)(x+y-1)=0$

$\Rightarrow \quad x=y, x+y=1$

When $x=y$

$	herefore \quad x^2+x=12 \Rightarrow x^2+x-12=0$

$\Rightarrow(x+4)(x-3)=0$

$x=-4,3, y=-4,3$

When $x+y=1$

$	herefore \quad x^2+1-x=12$

$\Rightarrow x^2-x-11=0$

$\Rightarrow \quad x=\frac{1 \pm \sqrt{45}}{2}$

$	herefore$ Total number of ordered pair of $(x, y)$ is $4 .$

The number of ordered pairs $(x, y)$ of real numbers that satisfy the simultaneous equations $x+y^2=x^2+y=12$ is

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