Question

{-х-у=3 {х^+у^=5 помогите пожалуйста это все под одной скобкой

Answer 1

-x-y=3
x^2+y^2=5
x=-y-3
y^2+6y+9+y^2-5=0
2y^2+6Y+4=0
y&2+3y+2=0
y1=-1
x=-2
y2=-2
x=-1

Answer 2

Solving the Equation

To solve the equation -x - y = 3 and x^2 + y^2 = 5 within the same set of brackets, we can use substitution to solve for one variable and then substitute that value into the other equation.

First, let's solve the equation -x - y = 3 for one of the variables. We can rearrange this equation to express y in terms of x:

-y = 3 + x y = -3 - x

Now, we can substitute this value of y into the second equation x^2 + y^2 = 5:

x^2 + (-3 - x)^2 = 5 x^2 + 9 + 6x + x^2 = 5 2x^2 + 6x + 4 = 0

We can then solve this quadratic equation to find the values of x.

Quadratic Equation Solution

Using the quadratic formula, where ax^2 + bx + c = 0:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation 2x^2 + 6x + 4 = 0, the coefficients are: - a = 2 - b = 6 - c = 4

Plugging these values into the quadratic formula gives us: x = (-6 ± √(6^2 - 4*2*4)) / (2*2) x = (-6 ± √(36 - 32)) / 4 x = (-6 ± √4) / 4 x = (-6 ± 2) / 4

So, we have two possible solutions for x: 1. x = (-6 + 2) / 4 = -1 2. x = (-6 - 2) / 4 = -2

Finding the Corresponding y Values

Now that we have the values of x, we can substitute them back into the equation y = -3 - x to find the corresponding y values: 1. For x = -1, y = -3 - (-1) = -2 2. For x = -2, y = -3 - (-2) = -1

Therefore, the solutions to the system of equations are: - x = -1, y = -2 - x = -2, y = -1