solve the system of equations algebraically for real values of x and y

{ x+y=16 x^2+y^2=128

a. (6,6)
b. (8,6)
c. (9,8)
d. (8,8)
e. no solution

hint:

from x+y = 16 ----> y = 16-x
sub that into the 2nd equation and solve for x
using your method of solving quadratics

let me know what you get

To solve the system of equations algebraically, we can use the method of substitution or elimination. Let's use the method of substitution.

We have the first equation: x + y = 16.

Solving for x, we get: x = 16 - y.

Substituting this into the second equation, we have: (16 - y)^2 + y^2 = 128.

Expanding and simplifying this equation, we get: 256 - 32y + y^2 + y^2 = 128.

Combining like terms, we have: 2y^2 - 32y + 128 = 0.

Dividing by 2, we get: y^2 - 16y + 64 = 0.

This is a quadratic equation. We can solve it by factoring or using the quadratic formula.

Factoring, we have: (y - 8)(y - 8) = 0.

From this, we can see that y = 8.

Substituting this back into the equation x = 16 - y, we get: x = 16 - 8 = 8.

Therefore, the solution to the system of equations is (x, y) = (8, 8).

So, the correct answer is d. (8, 8).