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(A) Given equations are:$(i) x = x^2 + y^2$$(ii) y = 2xy$From equation $(ii)$:$y - 2xy = 0$$y(1 - 2x) = 0$This implies $y = 0$ or $x = \frac{1}{2}$.Ca

Vedclass · Accepted Answer

$4$

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(A) Given equations are:$(i) x = x^2 + y^2$$(ii) y = 2xy$From equation $(ii)$:$y - 2xy = 0$$y(1 - 2x) = 0$This implies $y = 0$ or $x = \frac{1}{2}$.Case $1$: If $y = 0$,substitute into $(i)$:$x = x^2 + 0^2$$x^2 - x = 0$$x(x - 1) = 0$So,$x = 0$ or $x = 1$.This gives pairs $(0, 0)$ and $(1, 0)$.Case $2$: If $x = \frac{1}{2}$,substitute into $(i)$:$\frac{1}{2} = (\frac{1}{2})^2 + y^2$$\frac{1}{2} = \frac{1}{4} + y^2$$y^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$y = \pm \frac{1}{2}$.This gives pairs $(\frac{1}{2}, \frac{1}{2})$ and $(\frac{1}{2}, -\frac{1}{2})$.Thus,the total number of pairs $(x, y)$ is $4$.

The number of pairs of real numbers $(x, y)$ such that $x = x^2 + y^2$ and $y = 2xy$ is:

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