Answer:
We can start by rearranging the second equation as follows:
1/x^2 + 1/y^2 = 5/4
4/x^2 + 4/y^2 = 5
Multiplying both sides by x^2y^2, we get:
4y^2 + 4x^2 = 5x^2y^2
We can rearrange this equation as follows:
5x^2y^2 - 4x^2 - 4y^2 = 0
Using the quadratic formula, we can solve for xy:
xy = (4 ± sqrt(16 + 80))/10
xy = (4 ± 2sqrt(6))/10
xy = (2 ± sqrt(6))/5
Now we can use the fact that xy = (2 ± sqrt(6))/5 to solve for x and y. We can substitute y = (2 ± sqrt(6))/5x into the first equation:
x^2 + ((2 ± sqrt(6))/5x)^2 = 5
Expanding the square and simplifying, we get:
26x^4 - 50x^2 + 4 = 0
This is a quadratic equation in x^2, so we can use the quadratic formula to solve for x^2:
x^2 = (50 ± sqrt(50^2 - 4264))/52
x^2 = (25 ± sqrt(371))/26
Using the positive root for x^2, we can solve for x:
x = sqrt((25 + sqrt(371))/26) or x = -sqrt((25 + sqrt(371))/26)
Similarly, we can solve for y using the equation y = (2 ± sqrt(6))/5x:
y = (2 ± sqrt(6))/5 * sqrt((25 + sqrt(371))/26) or y = (2 ± sqrt(6))/5 * (-sqrt((25 + sqrt(371))/26))
Therefore, the solutions for x and y are:
x = sqrt((25 + sqrt(371))/26), y = (2 + sqrt(6))/5 * sqrt((25 + sqrt(371))/26)
OR
x = -sqrt((25 + sqrt(371))/26), y = (2 - sqrt(6))/5 * sqrt((25 + sqrt(371))/26)
Explanation: