To solve the system of equations:
- ( x + y = 4 ) (Equation 1)
- ( 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)}
]
