Question

X^2+y^2=10 x^4+x^2y^2=90

Answer 1

Ответ:

Я так понимаю: это система

x^2 + y^2 = 10

x^4 + x^2y^2= 90

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Выносим за скобки:

x^2 + y^2 = 10

x^2 * (x^2 + y^2)= 90

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Видно, что верхнее уравнение присутствует в нижнем, поэтому просто подставляем:

x^2 + y^2 = 10

x^2 * 10 = 90

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x^2 + y^2 = 10

x^2 = 90 / 10

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x^2 + y^2 = 10

x^2 = 9

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x^2 + y^2 = 10 x^2 + y^2 = 10

x = -3 x = 3

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9 + y^2 = 10 9 + y^2 = 10

x = -3 x = 3

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y^2 = 1 y^2 = 1

x = -3 x = 3

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y = 1 y = 1

y = -1 y = - 1

x = -3 x = 3

Ответ: (-3; 1); (-3;-1); (3; 1); (3;-1)

Answer 2

To solve the system of equations:

1. x^2 + y^2 = 10
2. x^4 + x^2y^2 = 90

You can use substitution to eliminate one of the variables. In this case, I'll solve for y in terms of x from equation 1 and then substitute it into equation 2.

From equation 1:

x^2 + y^2 = 10 y^2 = 10 - x^2 y = ±√(10 - x^2)

Now, substitute this expression for y into equation 2:

x^4 + x^2y^2 = 90 x^4 + x^2(10 - x^2) = 90

Simplify the equation:

x^4 + 10x^2 - x^4 = 90

The x^4 terms cancel out:

10x^2 = 90

Now, solve for x:

x^2 = 90 / 10 x^2 = 9

Taking the square root of both sides:

x = ±3

Now that we have the values of x, we can find the corresponding values of y using the expression we derived earlier:

For x = 3: y = ±√(10 - 3^2) = ±√(10 - 9) = ±√1 = ±1

So, for x = 3, we have two pairs of solutions: (x, y) = (3, 1) and (3, -1)

For x = -3: y = ±√(10 - (-3)^2) = ±√(10 - 9) = ±√1 = ±1

So, for x = -3, we also have two pairs of solutions: (x, y) = (-3, 1) and (-3, -1)

Therefore, the solutions to the system of equations are:

1. (x, y) = (3, 1)
2. (x, y) = (3, -1)
3. (x, y) = (-3, 1)
4. (x, y) = (-3, -1)