2x+y+sqrt(x^2-4y^2)=2, x*sqrt(x^2-4y^2)=0

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вопрос

To solve this system of equations, we will first use the second equation to determine the value of x.

Given: x*sqrt(x^2 - 4y^2) = 0

This implies that either x = 0 or sqrt(x^2 - 4y^2) = 0. Since x*sqrt(x^2 - 4y^2) = 0, we must have that either x = 0 or x^2 - 4y^2 = 0.

Case 1: x = 0

If x = 0, then the first equation becomes:

2(0) + y + sqrt(0^2 - 4y^2) =
y + sqrt(-4y^2) =
y - 2y = 2

So one solution is (0, -2).

Case 2: x^2 - 4y^2 = 0

This implies that x^2 = 4y^2, or x^2 - 4y^2 = 0.

Substitute x^2 = 4y^2 into the first equation:

2x + y + sqrt(4y^2 - 4y^2) =
2x + y =
y = 2 - 2x

Substitute y = 2 - 2x into x^2 = 4y^2:

x^2 = 4(2 - 2x)^
x^2 = 16 - 32x + 16x^
15x^2 - 32x + 16 = 0

Solve this quadratic equation for x:

Using the quadratic formula x = [32±sqrt((-32)^2 - 4(15)(16))]/(2(15)):

x = [32±sqrt(1024 - 960)]/(30
x = [32±sqrt(64)]/(30
x = (32±8)/3
x = 40/30 or 24/3
x = 4/3 or 4/5

Substitute x = 4/3 and x = 4/5 into the equation y = 2 - 2x to find the corresponding y values:

For x = 4/3
y = 2 - 2*(4/3) = 2 - 8/3 = 6/3 - 8/3 = -2/3

So another solution is (4/3, -2/3).

For x = 4/5
y = 2 - 2*(4/5) = 2 - 8/5 = 10/5 - 8/5 = 2/5

Therefore, the solutions to the system of equations are (0, -2), (4/3, -2/3), and (4/5, 2/5).