Solve the system of equations algebraically for real values of x and y
{ x^2 + y^2 =26 x+y=6
a. (1, 5), (1, 4)
b. (4, 5), (5, 4)
c. (5, 1), (5, 1)
d. (1, 5), (5, 1)
e. no solution
+23245 Solve: x2 + y2 = 26 and x + y = 6
Solve the linear equation for either x or y: x + y = 6 ---> y = 6 - x
Replace this value into the other equation: x2 + y2 = 26
x2 + (6 - x)2 = 26
Simplify: x2 + 36 - 12x + x2 = 26
2x2 - 12x + 36 = 26
2x2 - 12x + 10 = 0
Divide by 2: x2 - 6x + 5 = 0
Factor: (x - 5)(x - 1) = 0
Solve: x = 5 or x = 1
Substitute these values into the other equation:
x = 5 ---> x + y = 6 x = 1 ---> x + y = 6
5 + y = 6 1 + y = 6
y = 1 ---> (5, 1) y = 5 ---> (1, 5)
