Answer:
x = 19 and y = 10.
Explanation:
x = 2y - 1
x^2 + y^2 = 461
(2y - 1)^2 + y^2 = 461
4y^2 - 4y + 1 + y^2 = 461 5y^2 - 4y + 1 = 461
5y^2 - 4y - 460 = 0
y = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 5, b = -4, and c = -460. Plugging these values into the quadratic formula, we get:
y = (-(-4) ± √((-4)^2 - 4(5)(-460))) / (2(5))
y = (4 ± √(16 + 9200)) / 10
y = (4 ± √9216) / 10 y = (4 ± 96) / 10
y1 = (4 + 96) / 10 = 100 / 10 = 10
y2 = (4 - 96) / 10 = -92 / 10 = -9.2
Since we are looking for positive integers, we can ignore the negative value for y. Therefore, y = 10.
x = 2(10) - 1 = 20 - 1 = 19 So, the two positive integers that satisfy the given conditions are x = 19 and y = 10.