To solve this system of equations, we can use substitution or elimination.
Let's first solve the first equation:
x + y = 5y = 5 - x
Now we can substitute this expression for y into the second equation:
x^3 + (5 - x)^3 = 35x^3 + 125 - 75x + x^3 = 352x^3 - 75x + 90 = 0
Now we can solve this cubic equation to find the values of x. One possible solution is x = 3.
Using this value of x, we can find y:
y = 5 - xy = 5 - 3y = 2
Therefore, the solution to the system of equations is x = 3 and y = 2.
To solve this system of equations, we can use substitution or elimination.
Let's first solve the first equation:
x + y = 5
y = 5 - x
Now we can substitute this expression for y into the second equation:
x^3 + (5 - x)^3 = 35
x^3 + 125 - 75x + x^3 = 35
2x^3 - 75x + 90 = 0
Now we can solve this cubic equation to find the values of x. One possible solution is x = 3.
Using this value of x, we can find y:
y = 5 - x
y = 5 - 3
y = 2
Therefore, the solution to the system of equations is x = 3 and y = 2.