x+y=4 x^2-y^2=8

To solve the system of equations:

1. \( x + y = 4 \) (Equation 1)
2. \( x^2 - y^2 = 8 \) (Equation 2)

We can start by using Equation 1 to express \( y \) in terms of \( x \):

\[ y = 4 - x \]

Now we can substitute this expression for \( y \) into Equation 2. Recall that \( x^2 - y^2 \) can be factored using the difference of squares:

\[ x^2 - y^2 = (x - y)(x + y) \]

Substituting \( y = 4 - x \) into \( x + y \):

\[ x + y = x + (4 - x) = 4 \]

Now substituting into \( x - y \):

\[ x - y = x - (4 - x) = x - 4 + x = 2x - 4 \]

Now replace \( x^2 - y^2 \) in Equation 2:

\[ (x - y)(x + y) = (2x - 4)(4) = 8 \]

So we have:

\[ 4(2x - 4) = 8 \]

Divide both sides by 4:

\[ 2x - 4 = 2 \]

Add 4 to both sides:

\[ 2x = 6 \]

Divide by 2:

\[ x = 3 \]

Now substitute \( x = 3 \) back into Equation 1 to find \( y \):

\[ 3 + y = 4 \implies y = 4 - 3 = 1 \]

Thus, the solution to the system of equations is:

\[ (x, y) = (3, 1) \]

To verify, substitute \( x = 3 \) and \( y = 1 \) back into the second equation:

\[ x^2 - y^2 = 3^2 - 1^2 = 9 - 1 = 8 \]

Both equations are satisfied, confirming that the solution is correct.

Thus, the final solution is:

\[ \boxed{(3, 1)} \]