Question

Find one pair (x,y) of real numbers such that x+y=4 and x³+y³=100.

a)x=2,y=2
b)x=1,y=3
c)x=3,y=1
d)x=−2,y=−2

Answer 1

To find a pair (x, y) of real numbers that satisfy the given equations x + y = 4 and x^3 + y^3 = 100, we can solve for xy and then use it to find the values of x and y. The pair (x, y) that satisfies the given conditions is (3, 1) or (1, 3).

Step-by-step explanation:

To find a pair (x, y) of real numbers that satisfy the given equations, we can use the fact that (x + y)^3 = x^3 + y^3 + 3xy(x + y). From the equation x^3 + y^3 = 100 and x + y = 4, we can rewrite (x + y)^3 = 100 + 3xy(4). Simplifying further, we get 64 = 100 + 12xy. Solving this equation for xy, we find xy = -9. Using xy = -9 and x + y = 4, we can solve for x and y. Let's assume x = 4 - y. Substituting this value into the equation xy = -9 gives (4 - y)y = -9. Expanding and rearranging this equation, we get y^2 - 4y - 9 = 0. Solving this quadratic equation, we find y = 1 or y = 3. Substituting these values back into x + y = 4, we find x = 3 or x = 1. Therefore, the pair (x, y) that satisfies the given conditions is (3, 1) or (1, 3).