What is the solution set to the following system?

x+y=3
x^2+y^2=9

a.(0,-3)(3,0)
b.(0,3)(-3,0)
c.(0,3)(3,0)
d.(0,-3)(-3,0)

Well, I would say the solution set to this system is d. (0,-3)(-3,0). Why? Because when you have x + y = 3, it's like an equation for a cozy little partnership, where x and y complement each other and add up to 3. And when you have x^2 + y^2 = 9, it's like an equation for a big circle with a radius of 3, where the sum of the squares of x and y is 9. So the solution set is indeed the points where these two equations intersect, which happen to be (0,-3) and (-3,0). So grab your clown noses and head over to the circus, because d is the answer!

To find the solution set to the given system of equations, you can use the method of substitution. Here are the steps:

Step 1: Solve the first equation for x or y.
Considering x + y = 3, we can solve for x by subtracting y from both sides:
x = 3 - y

Step 2: Substitute the value of x into the second equation.
Now substitute the value of x from step 1 into the second equation, x^2 + y^2 = 9:
(3 - y)^2 + y^2 = 9

Step 3: Simplify the equation.
Expanding the equation yields:
9 - 6y + y^2 + y^2 = 9
2y^2 - 6y = 0

Step 4: Factor out y.
Factor out y from the equation:
y(2y - 6) = 0

Step 5: Solve for y.
Set each factor equal to zero and solve independently:
y = 0 or 2y - 6 = 0

For y = 0, substitute it back into the first equation to find x:
x + 0 = 3
x = 3

For 2y - 6 = 0, solve for y:
2y - 6 = 0
2y = 6
y = 3

So, we have two solutions:
Solution 1: x = 3, y = 0 (3, 0)
Solution 2: x = 0, y = 3 (0, 3)

Therefore, the correct answer is c. (0, 3) and (3, 0).

To find the solution set to the given system of equations, we can use the method of substitution or elimination. Let's use the substitution method to solve it.

Given system:
1) x + y = 3
2) x^2 + y^2 = 9

From equation 1, we can express x in terms of y:
x = 3 - y

Substituting this value of x into equation 2, we get:
(3 - y)^2 + y^2 = 9
9 - 6y + y^2 + y^2 = 9
2y^2 - 6y = 0
2y(y - 3) = 0

Applying the Zero Product Property, we get two possible solutions for y:
1) y = 0
2) y - 3 = 0, which gives y = 3

Now, substitute these values of y back into equation 1 to find the corresponding x values:
For y = 0:
x + 0 = 3
x = 3

For y = 3:
x + 3 = 3
x = 0

Therefore, the solution set to the given system is (x, y) = (3, 0) and (0, 3). Option c.(0,3)(3,0) matches these solutions.

which pair works in both equations ?

(all pairs work in the 2nd equation, but not in the 1st)