Answer:
The possible pairs of numbers that satisfy the given conditions are (x, y) = (4, 2) and (x, y) = (8, -2).
Explanation:
To solve this problem, we can use algebra. Let's call the two numbers x and y. We are given that x + y = 6 and x^2 + y^2 = 28. We can square the equation x + y = 6 to get x^2 + 2xy + y^2 = 36. We can then subtract the equation x^2 + y^2 = 28 from the equation x^2 + 2xy + y^2 = 36 to get 2xy = 8. We can then divide both sides by 2 to get xy = 4.
We can then use the equation x + y = 6 to write x = 6 - y, which we can substitute into the equation xy = to get (6 - y)y = 4. Solving this equation for y yields y = 2 or y = -2.
If y = 2, then x = 6 - y = 6 - 2 = 4, so the two numbers are x = 4 and y = 2. If y = -2, then x = 6 - y = 6 + 2 = 8, so the two numbers are x = 8 and y = -2. Therefore, the possible pairs of numbers that satisfy the given conditions are (x, y) = (4, 2) and (x, y) = (8, -2).