Question

(x² - y² = 0,4 + xy = 0;​

Answer 1

$\left \{ {{x^2-y^2=0} \atop {4+xy=0}} \right. =>\left \{ {{(x-y)(x+y)=0} \atop {4+xy=0}} \right. => \\\\x-y=0\\x+y=0\\x_1=y\\x_2=-y\\\\4+y*y=0\\4+y^2=0\\y^2=-4\\y \in \emptyset\\\\4-y*y=0\\4-y^2=0\\y^2=4\\y_{1,2}=\pm2\\\\x_{1,2}=\pm2$

ОТВЕТ: $x_{1,2}=\pm2\\y_{1,2}=\pm2$

Answer 2

To solve the system of equations:

1. $x^2 - y^2 = 0$
2. $4 + xy = 0$

Let's solve the system step by step.

From equation 1, we have:

$x^2 - y^2 = 0$

This equation can be rewritten as:

$x^2 = y^2$

Taking the square root of both sides, we get:

$x = \pm y$

Now let's substitute this result into equation 2:

$4 + xy = 0$

Substituting $x = y$:

$4 + y^2 = 0$

Rearranging the equation:

$y^2 = -4$

Taking the square root of both sides, we get:

$y = \pm \sqrt{-4}$

Since the square root of a negative number is imaginary, there are no real solutions for $y$. Therefore, the system of equations has no real solutions.