Question

Решить систему X^2=4y+1 X^2+3=4y+y^2 Помогите,пожалуйстаааааа

Answer 1

$\left \{ {{x^{2} =4y+1} \atop {x^{2} =4y+y^{2}-3} } \right.$

4y+1=4y+y²-3

y²=4 ⇒ y=±2

x²=4*(-2)+1=-7 (не имеет решений)

x²=4*2+1=9 ⇒ x=±3

Ответ: (-3;2) и (3;2)

Answer 2

Solving the System of Equations

To solve the system of equations X^2 = 4y + 1 and X^2 + 3 = 4y + y^2, we can use the given equations to eliminate one variable and solve for the other.

Let's start by subtracting the first equation from the second equation:

(X^2 + 3) - (X^2) = (4y + y^2) - (4y + 1)

Simplifying this equation, we get:

3 = y^2 - 1

Rearranging the equation, we have:

y^2 = 4

Taking the square root of both sides, we find:

y = ±2

Now that we have the values of y, we can substitute them back into one of the original equations to solve for X.

For y = 2:

X^2 = 4(2) + 1

Simplifying, we get:

X^2 = 9

Taking the square root of both sides, we find:

X = ±3

For y = -2:

X^2 = 4(-2) + 1

Simplifying, we get:

X^2 = -7

Since the square root of a negative number is not a real number, there are no real solutions for X when y = -2.

Therefore, the solutions to the system of equations are:

X = ±3, y = ±2.

Please let me know if you need any further assistance!