Respuesta :

carlosego
For this case we have the following system of equations:
 x + y = 6
 x ^ 3 + y ^ 3 = 144
 Solving the system of equations graphically we have that one of the solutions is:
 x = 3-root (5)
 y = 3 + root (5)
 Then, an ordered pair that satisfies both equations:
 (x, y) = (3-root (5), 3 + root (5))
 Answer:
 (x, y) = (3-root (5), 3 + root (5))

Answer:

No real solution

Step-by-step explanation:

Given that there are two real numbers x and y such that

[tex]x+y =6[/tex] and [tex]x^3+y^3 =144[/tex]

We can substitute for y as

[tex]y=6-x[/tex]

Substitute in II equation as

[tex](6-x)^3+x^3 =144\\216-108x+18x^2-x^3+x^3 =144\\x^2-6x+12=0[/tex]

Solve this using quadratic formula

[tex]x=\frac{-6±\sqrt{36-48} }{2} \\=-3±i\sqrt{3}[/tex]

This shows that there cannot be any two real numbers satisfying the given condition

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