solve the system of equations algebraically for real values of x and y
{ x+u=16
x^2 +y^2 =128
I will assume your first equation has a typo, then
y = 16-x
plug into the 2nd
x^2 + (16-x)^2 = 128
2x^2 - 32x +128 = 0
x^2 - 16x + 64 = 0
(x-8)^2 = 0
take over, compare your answer to
http://www.wolframalpha.com/input/?i=plot+x%2By%3D16,++x%5E2+%2By%5E2+%3D128
To solve the system of equations algebraically for real values of x and y, we'll use the method of substitution.
Step 1: Solve the first equation for x in terms of u.
x + u = 16
x = 16 - u
Step 2: Substitute the expression for x in the second equation.
(16 - u)^2 + y^2 = 128
Step 3: Expand the equation.
(16 - u)(16 - u) + y^2 = 128
(256 - 32u + u^2) + y^2 = 128
256 - 32u + u^2 + y^2 = 128
Step 4: Rearrange the equation and combine like terms.
u^2 - 32u + y^2 = -128
Step 5: Simplify the equation further.
u^2 - 32u + y^2 + 128 = 0
The resulting equation is a quadratic equation in terms of u and y. To determine the values of u and y that satisfy this equation, you can use the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / (2a)
Since the equation does not include x, we are solely interested in the values of u and y. The quadratic formula will give you the real values of u and y that satisfy the equation in terms of x and y.