Solve the following system of equations. x^2 + y^2 = 9 x^2 - 3y = 9

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Solve the following system of equations. x^2 + y^2 = 9 x^2 - 3y = 9

Mathematics
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These are the options: (-3, 0), (3, 0),(-3, 9),(3, 9) (-3, 9),(-3, 0) (-3, 0), (0, -3), (3, 0) (-3, 9) I am very confused on how to do this problem because there is a 3y and y^2
well since they both = 9 its safe to say they equal each other right @BeccaB003
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well since they both = 9 its safe to say they equal each other right @BeccaB003
I am here. Yes, I suppose they would equal each other.
x^2 + y^2 = x^2 - 3y
so x^2 + y^2 = x^2 - 3y now solve do -x^2 to get y^2+-3y
i mean y^2=-3y
subtract x^2 from both sides y^2 = - 3y
@welshfella @DaBest21 But how do you solve with that problem if there is y and y^2?
can you solve that?
I thought you used substitution or elimination for problems like this.
I don't understand how you would solve for y^2=3y ---they are two different types of variables. you can't combine them or anything.
well i have used elimination - i eliminated x^2
y^2 + 3y = 0 y(y + 3) = 0 y = ?
you factor them as above
y(y+3)=0 <---------- is factored
I don't know how to simplify that more.
when 2 functions are multiplied together and the result is 0 then either one or the other is zero. I other words there are 2 solutions y(y + 3) = 0 y = 0 or y + 3 = 0 y = 0 y= -3 0 and -3 are the 2 solutions
So then you plug them both into the original equation to find x?
yes plug them into x^2 - 3y = 9
so when y = -3 x^2 -3(-3) = 9 x^2 + 9 = 9 x^2 = 9-9 = 0 x = 0 so one solution is (0,-3)
Okay so, x^2 -3(0) = 9 equals: x^2 = 9 = x = 3
So another solution is (3,0)
not quite sqrt x^2 is also -3
right (3,0) is one solution and a third one is (-3,0)
* square root of 9 is also -3
How can the square root of 9 be -3?
so the 3rd option is the correct one.
c
3 * 3 = 9 also -3 * -3 = 9
Okay, Thank you so much for your help! :)
yw

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