Determine the solution set of the system.
x^2+y^2=9
x+y=3
Very confused on this one! PLease Help!
From the second:
y = 3-x
sub into the first
x^2 + (3-x)^2 = 9
x^2 + 9 - 6x + x^2 = 9
2x^2 - 6x = 0
x^2 - 3x=0
x(x-3) = 0
x = 0 or x=3
if x = 0 , then y = 3-0 = 3
if x=3 , then y = 3-3 = 0
solution set: (0,3) and (3,0)
Thank you!
To determine the solution set of the given system of equations:
1. Start by solving the second equation, x+y=3, for one variable in terms of the other. Let's solve for x:
Subtract y from both sides: x = 3 - y.
2. Now, substitute this expression for x in the first equation: (3-y)^2 + y^2 = 9.
3. Simplify the equation: (9 - 6y + y^2) + y^2 = 9.
Combine the like terms: 2y^2 - 6y = 0.
4. Factor out a common term from the equation: 2y(y - 3) = 0.
5. Apply the Zero Product Property: Set each factor equal to zero and solve for y:
2y = 0 or y - 3 = 0.
Solving the first equation: y = 0.
Solving the second equation: y = 3.
6. Substitute these values of y back into the second equation, x + y = 3, to solve for x:
For y = 0: x + 0 = 3, so x = 3.
For y = 3: x + 3 = 3, so x = 0.
7. Therefore, the solution set of the system is (x, y) = {(3, 0), (0, 3)}.