x + y = 4 and x x + y y = 64 . What are x and y ?

 To solve this system of equations:

1. Start with the first equation: \(x + y = 4\).

2. Rearrange it to solve for one variable, for example, \(x\): \(x = 4 - y\).

3. Substitute this expression for \(x\) into the second equation: \((4 - y)^2 + y^2 = 64\).

4. Expand and simplify: \(16 - 8y + y^2 + y^2 = 64\).

5. Combine like terms: \(2y^2 - 8y + 16 = 64\).

6. Subtract 64 from both sides: \(2y^2 - 8y - 48 = 0\).

7. Divide both sides by 2 to simplify: \(y^2 - 4y - 24 = 0\).

8. Factor the quadratic equation: \((y - 6)(y + 2) = 0\).

9. Solve for \(y\): \(y = 6\) or \(y = -2\).

Now that you have the values for \(y\), you can find \(x\) using the first equation:

If \(y = 6\):

\[x = 4 - y = 4 - 6 = -2\].

If \(y = -2\):

\[x = 4 - y = 4 - (-2) = 6\].

So, the solutions for \(x\) and \(y\) are either \(x = -2\) and \(y = 6\) or \(x = 6\) and \(y = -2\).