Question

[x² + y² = 34,[xy = 15; решите систему упавнений​

Answer 1

Ответ:(3;5),(5;3),(-3;-5),(-5;-3).

Объяснение:{x²+y²=34, 2xy=30;║(1)+(2)

{x²+2xy+y²=64,xy=15;

{(x+y)²=64;xy=15;

{x+y=8,xy=15           или    {x+y=-8,xy=15;  ( по т.Виета)  ↓

{x=3,y=5 или  {x=5,y=3;       {x=-3,  y=-5   или   {x=-5,y=-3.

Answer 2

To solve the given system of equations:

Equation 1: x² + y² = 34 Equation 2: xy = 15

We can use the method of substitution to find the solutions.

From Equation 2, we can express one variable in terms of the other. Let's solve Equation 2 for x:

xy = 15 x = 15 / y

Now substitute this value of x into Equation 1:

(15 / y)² + y² = 34

Simplifying the equation:

225 / y² + y² = 34 225 + y⁴ = 34y² y⁴ - 34y² + 225 = 0

Let's introduce a new variable z = y²:

z² - 34z + 225 = 0

Now we have a quadratic equation in z. Factoring it:

(z - 25)(z - 9) = 0

Setting each factor equal to zero and solving for z:

z - 25 = 0 or z - 9 = 0

z = 25 or z = 9

Substituting back for z = y²:

Case 1: z = 25 y² = 25 y = ±5

Case 2: z = 9 y² = 9 y = ±3

Now, substitute these values of y back into Equation 2 to find the corresponding values of x:

For y = 5: xy = 15 x * 5 = 15 x = 3

For y = -5: xy = 15 x * -5 = 15 x = -3

For y = 3: xy = 15 x * 3 = 15 x = 5

For y = -3: xy = 15 x * -3 = 15 x = -5

Therefore, the solutions to the system of equations are: (x, y) = (3, 5), (-3, -5), (5, 3), (-5, -3)