Question

CHECK MY ANSWER WILL MEDAL. Find the solution(s) of the following system using elimination: x^2+y^2=9;x^2-y^2=9
My answer: x^2+y^2=9 ; x^2-y^2=9 (added together to eliminate) 2^2/2 =18/2 , x^2 = 9 , x=3 .. then substitute in, x^2+y^2=9 , 9+y^2=9 , 9y^2=0 , y=0 . The solutions are (3,0) and (-3,0)

Answer 1

Yes (3, 0) and (-3,0) are the correct solutions.

Answer 2

Thank you

Answer 3

No problem :)

Answer 4

I have another question you might be able to help me with @fission-mailed

Answer 5

x^2-2y=8 , x^2+y^2=16 , (add together to elliminate), 3y^2/3 =9/3 , y^2=3 , y=sqrt3 .. substitute, x^2-2y=8, x^2-3=8 , minus 3 from each side, x^2=5, x=5 .. so what would the two solutions be if i did that right? @fission-mailed

Answer 6

I think you will have more success if you try to subtract instead of add in this case. You're trying to eliminate one of the variables so you can solve more easily, so in this case you can get rid of the x's like...

x^2 + y^2 = 16
x^2 - 2y = 8

... if you subtract the 2nd from the 1st equation the x-squareds will cancel out. You're left with y^2 - (-2y) = 16-8 = 8

y^2 + 2y = 8. You can bring the 8 over to make this a quadratic and solve for y.

y^2 + 2y - 8 = 0. Solve for y and you get y = 2, y = -4.

Now you can plug in these y-values to get the x-values.

Answer 7

so by plugging in the values it would look like, x^2+2^2=16 and x^2-2x(-4)=8

Answer 8

@fission-mailed ^^

Answer 9

Yup, you can do that to get the x's (I assume you meant x^2 - 2(-4) = 8 for the second equation :) )

Answer 10

so then how do i get the x values?

Answer 11

Well then you can just solve for x. Using y = 2,

x^2 + 2^2 = 16
x^2 = 16 - 4 = 12
x = root 12, plus or minus
AKA x = 2 root 3, plus or minus

So from this you get solutions (2, 2 root 3) and (2, - 2 root 3).

The same would be done to get the other solutions, you would just use y = - 4 and solve using that instead of 2.

Answer 12

Also here's a tip in case you don't know this yet. By the time you get to this point where you're trying to solve for the other variable you don't have to use both equations anymore. You could if you wanted, but you could also just whichever one is easier. They will give you the same result because they are solutions to both equations.

So you could just solve with y = 2 and y = 4 using only the first equation for example, x^2 + y^2 = 16, for both. It can make things much easier at times. Let me know if there's anything that didn't make sense.

Answer 13

* sorry typo, y = 2 and y = -4

Answer 14

Whoops just realized I wrote the points in the wrong order. So:

(2 root 3, 2) and (- 2 root 3, 2). Sorry for the confusion!

The 3rd solution will be (0, -4), since

x^2 + (-4)^2 = 16
x^2 = 16 - 16
x = 0

Answer 15

SO, there are 3 solutions to the problem, (2 root3, 2) and (-2 root 3, 2) and (0,-4).
x^2+y^2=16
x^2-2y=8
y^2-(-2y)=16-8=8
y^2+2y=8 , y^2+2y-8=0
AND
x^2+(-4)^2=16
x^2=16-16
x=0

@fission-mailed

Answer 16

Yep. Those are the solutions.

Answer 17

As I said earlier, you could have also used just the second equation to solve for x if you'd prefer. It would give you the exact same answers

Example, For y = -4

x^2 - 2(-4) = 8
x^2 = 8 - 8 = 0
x = 0

So you have a lot of freedom in these types of questions :)

Answer 18

alright thank you so much! i was not looking forward to doing these. i really appreciate it!

Answer 19

No problem, glad to help!