Question

Решите функции 1)x+y=4 X^2-y^2=8 2)4x^2-xy=33 3x^2+xy=30

Answer 1

Вот решение на фото............

Answer 2

1. x + y = 4 x^2 - y^2 = 8

We can rewrite the second equation as (x+y)(x-y) = 8. Since we know that x + y = 4, we can substitute 4 - x for y in the second equation and simplify:

x^2 - (4-x)^2 = 8 x^2 - 16 + 8x - x^2 = 8 8x = 24 x = 3

Substituting x = 3 into the first equation gives us y = 1. Therefore, the solution to the system of equations is x = 3 and y = 1.

2. 4x^2 - xy = 33 3x^2 + xy = 30

Adding the two equations eliminates the xy term:

7x^2 = 63 x^2 = 9 x = 3 or x = -3

Substituting x = 3 into one of the original equations gives us:

4(3^2) - 3y = 33 y = -3

Therefore, one solution to the system of equations is x = 3 and y = -3. Substituting x = -3 gives us:

4(-3^2) - 3y = 33 y = -9

Therefore, the other solution to the system of equations is x = -3 and y = -9.

3. 3x^2 + xy = 30 x^2 + xy = 10

Subtracting the second equation from the first eliminates the xy term:

2x^2 = 20 x^2 = 10 x = sqrt(10) or x = -sqrt(10)

Substituting x = sqrt(10) into one of the original equations gives us:

3(10) + sqrt(10)y = 30 y = (20 - 3sqrt(10))/sqrt(10)

Therefore, one solution to the system of equations is x = sqrt(10) and y = (20 - 3sqrt(10))/sqrt(10). Substituting x = -sqrt(10) gives us:

3(10) - sqrt(10)y = 30 y = (20 + 3sqrt(10))/sqrt(10)

Therefore, the other solution to the system of equations is x = -sqrt(10) and y = (20 + 3sqrt(10))/sqrt(10).