Answer:
We have two equations:
√x +2y^2 = 15 ----(1)
√4x - 4y^2=6 ----(2)
Let's solve for x:
From (1), we have:
√x = 15 - 2y^2
Squaring both sides, we get:
x = (15 - 2y^2)^2
Expanding, we get:
x = 225 - 60y^2 + 4y^4
From (2), we have:
√4x = 6 + 4y^2
Squaring both sides, we get:
4x = (6 + 4y^2)^2
Expanding, we get:
4x = 36 + 48y^2 + 16y^4
Substituting the expression for x from equation (1), we get:
4(225 - 60y^2 + 4y^4) = 36 + 48y^2 + 16y^4
Simplifying, we get:
900 - 240y^2 + 16y^4 = 9 + 12y^2 + 4y^4
Rearranging, we get:
12y^2 - 12y^4 = 891
Dividing both sides by 12y^2, we get:
1 - y^2 = 74.25/(y^2)
Multiplying both sides by y^2, we get:
y^2 - y^4 = 74.25
Let z = y^2. Substituting, we get:
z - z^2 = 74.25
Rearranging, we get:
z^2 - z + 74.25 = 0
Using the quadratic formula, we get:
z = (1 ± √(1 - 4(1)(74.25))) / 2
z = (1 ± √(-295)) / 2
Since the square root of a negative number is not real, there are no real solutions for z, which means there are no real solutions for y and x.
Therefore, the answer is "no solution".