Question

Система. х+у=7 у²-2х=10 заранее спасибо

Answer 1

Решение
х+у=7
у²-2х=10

x = 7 - y
y² - 2*(7 - y) - 10 = 0

x = 7 - y
y² - 14 + 2y - 10 = 0

x = 7 - y
y² + 2y - 24 = 0

y² + 2y - 24 = 0
y₁ = - 6
y₂ = 4

y₁ = - 6
x₁ = 7 - (-6)
x₁ = 13
(13; - 6)

y₂ = 4
x₂ = 7 - 4
x₂ = 3
(3;4)

Ответ: (13; - 6); (3;4)

Answer 2

Solving the Equation System

To solve the given system of equations x - y = 7 and y^2 - 2x = 10, we can use the substitution method or the elimination method. Let's proceed with the substitution method.

First, we can solve the first equation for x in terms of y: x = y + 7.

Now, we can substitute this expression for x into the second equation: y^2 - 2(y + 7) = 10.

Expanding and simplifying the equation gives us a quadratic equation in terms of y: y^2 - 2y - 14 = 10.

Subtracting 10 from both sides yields: y^2 - 2y - 24 = 0.

We can now solve this quadratic equation for y using the quadratic formula: y = (-b ± √(b^2 - 4ac)) / (2a), where a = 1, b = -2, and c = -24.

Calculating the discriminant (b^2 - 4ac), we get: (-2)^2 - 4*1*(-24) = 4 + 96 = 100.

The solutions for y are then: y = (2 ± √100) / 2, which simplifies to: y = (2 ± 10) / 2, giving us: y = 6 or y = -4.

Now that we have the values for y, we can find the corresponding values for x using the equation x = y + 7: For y = 6, x = 6 + 7 = 13. For y = -4, x = -4 + 7 = 3.

Therefore, the solutions to the system of equations are: x = 13, y = 6 and x = 3, y = -4.