# Help with math ## Recommended Posts  Can someone explain how to solve this?

x2 + y2 = 34

x + y + xy = 23

ty. c:

##### Share on other sites  ##### Share on other sites  Do a systems of equations, I believe

##### Share on other sites  Do a systems of equations, I believe

I don't know how to solve the system.

##### Share on other sites  ty. Much appreciated

##### Share on other sites  Just looking to find x and y ?

simultaneous equations?

##### Share on other sites  Just looking to find x and y ?

simultaneous equations?

Yes.

##### Share on other sites  solve the top function for y

then plug the result in for y into the second one

after that you will have what x should be

then plug x into the first one and solve again for y

##### Share on other sites  I looked at it for a few seconds, figured out that it has to be five and three. 25+9 = 34, 5+3+15 = 23.

However, I don't know which one is 5 and which one is three.

##### Share on other sites  I looked at it for a few seconds, figured out that it has to be five and three. 25+9 = 34, 5+3+15 = 23.

However, I don't know which one is 5 and which one is three.

I know the answer, but I need the solution.

##### Share on other sites  Can someone explain how to solve this?

x2 + y2 = 34

x + y + xy = 23

ty. c:

x^2 + y^2 = 34

y^2 = 34 - x^2

y = sqrt(34-x^2)

x + sqrt(34-x^2) + x(sqrt(34-x^2) = 23

(square everything) x^2 + 34-x^2 + x^2 * 34-x^2 = 23*23

x^4 + 34x^2 + 34 = 529

x^4 + 34x^2 = 495

Dunno how. Hardest system of equations i've seen.

##### Share on other sites  x = 3, y = 5 and then the reverse would hold true as well since they are identical y = 3, x= 5

##### Share on other sites  x = 3, y = 5 and then the reverse would hold true as well since they are identical y = 3, x= 5

I know the answer, but I need the solution.

##### Share on other sites  x^2 + y^2 = 34

y^2 = 34 - x^2

y = sqrt(34-x^2)

x + sqrt(34-x^2) + x(sqrt(34-x^2) = 23

(square everything) x^2 + 34-x^2 + x^2 * 34-x^2 = 23*23

x^4 + 34x^2 + 34 = 529

x^4 + 34x^2 = 495

Dunno how. Hardest system of equations i've seen.

I did the same shiz, then I get a biquadratic equation, but the discriminant is negative. Maybe there's a mistake in one of the equations :c

##### Share on other sites  What type of math are you in? Can you use derivatives or have a graphing calculator to use?

##### Share on other sites  Alright, this is how you do it.

x2 + y2 = 34 . . . (1)

x + y + xy = 23 . . . (2)

Factorizing both equations:

(x+y)2-2xy=34 . . . (3)

(x+y)2-xy=23 . . . (4)

-xy=11

xy=-11

Substituting xy=-11 into (2):

x+y-11=23

x+y=34 . . . (5)

x=34-y

x=-y+34

Substituting x=-10-y into (1)

(-y+34)2+y2=34

y2-68y+1156+y2=34

2y2+20y=-1122

2y2+20y+1122=0

y2+10y+561=0

And shit, I'm stuck here.

However, I'm sure this is how it should be done. I'll probably try again later.

##### Share on other sites  sample text

##### Share on other sites  x + y + xy = 23 . . . (2)

(x+y)2-xy=23 . . . (4)

How are these the same?

##### Share on other sites  This shows a step by step solution without needing to graph or use derivatives, I would think you can ignore the imaginary solutions unless you are asked for those too.

https://www.symbolab.com/solver/system-of-equations-calculator/x%5E%7B2%7D%20%2B%20y%5E%7B2%7D%20%3D%2034%2C%20x%2By%2Bxy%3D23/?origin=button

Let know if you need any explanations.

##### Share on other sites  How are these the same?

Oh no :c

##### Share on other sites  This shows a step by step solution without needing to graph or use derivatives, I would think you can ignore the imaginary solutions unless you are asked for those too.

https://www.symbolab.com/solver/system-of-equations-calculator/x%5E%7B2%7D%20%2B%20y%5E%7B2%7D%20%3D%2034%2C%20x%2By%2Bxy%3D23/?origin=button

Let know if you need any explanations.

Thank you this helped/explained a lot. <3