Question

2) 22 + xy - 652 = 0, x2 - 5xy + 252 + 4 = 0;​

Answer 1

Ответ:

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Объяснение:

Answer 2

It seems like you have a system of two equations with two variables, x and y. Let's solve this system of equations:

1. 22 + xy - 652 = 0
2. x^2 - 5xy + 252 + 4 = 0

Let's start with equation (1):

1. 22 + xy - 652 = 0

First, let's simplify it:

xy = 652 - 22 xy = 630

Now, let's move on to equation (2):

2. x^2 - 5xy + 252 + 4 = 0

Substitute the value of xy from equation (1):

x^2 - 5(630) + 252 + 4 = 0

Now, simplify this equation:

x^2 - 3150 + 252 + 4 = 0

x^2 - 2894 = 0

Now, let's solve for x:

x^2 = 2894

Take the square root of both sides:

x = ±√2894

x can be both positive and negative, so we have two possible values for x:

x = √2894 x ≈ 53.79 (rounded to two decimal places)

x = -√2894 x ≈ -53.79 (rounded to two decimal places)

So, the solutions for x are approximately 53.79 and -53.79.

Now, let's find the corresponding values of y using the value of xy from equation (1):

xy = 630

For x = 53.79: 53.79y = 630 y ≈ 630 / 53.79 y ≈ 11.70 (rounded to two decimal places)

For x = -53.79: -53.79y = 630 y ≈ 630 / (-53.79) y ≈ -11.70 (rounded to two decimal places)

So, the solutions for y are approximately 11.70 and -11.70.

Therefore, the solutions to the system of equations are:

1. (x ≈ 53.79, y ≈ 11.70)
2. (x ≈ -53.79, y ≈ -11.70)