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cutiecomittee123

  • one year ago

How many solutions are there to this system of equations? 2x+y^2=33 x^2+y^2+2x=19 1,2,3, or 4???

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  1. anonymous
    • one year ago
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    mathway.com quickmath.com i promise they help so much

  2. NotTim
    • one year ago
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    are you allowed to substitued 1 into the other?

  3. cutiecomittee123
    • one year ago
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    you can try

  4. NotTim
    • one year ago
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    :/

  5. NotTim
    • one year ago
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    you're suppose to try. i'm suppose to help.

  6. NotTim
    • one year ago
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    yo.

  7. NotTim
    • one year ago
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    why not re-arrange for x in the first equation, and sub it into the next equation.

  8. anonymous
    • one year ago
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    Notice both equations have y^2 +2x in them. The first equation tells us this expression is equal to 33, so you can replace y^2 +2x for 33 in the second equation. Doing this gives x^2 +33 = 19 or x^2 = -14 How many solutions does that have? :)

  9. NotTim
    • one year ago
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    Wait what.

  10. cutiecomittee123
    • one year ago
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    4?

  11. NotTim
    • one year ago
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    this is another question.

  12. NotTim
    • one year ago
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    get that junk into another question.

  13. anonymous
    • one year ago
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    It would be 0. Are you sure 0 isnt one of the answer choices?

  14. cutiecomittee123
    • one year ago
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    its not

  15. anonymous
    • one year ago
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    Unless mobile is screwing with me that badly and the question i see is different than where Im typing :/

  16. anonymous
    • one year ago
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    Although you dont have to be so rude about it either way -_-

  17. NotTim
    • one year ago
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    ah. sorry, i'm working on 0 hrs of sleep, and university work and job applications. forgive me.

  18. anonymous
    • one year ago
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    Alright. So assuming mobile may be messing with me, i see the question as asking for how many solutions there are to the system 2x + y^2 = 33 X^2 + y^2 +2x = 19 Am i seeing the wrong question because of mobile?

  19. NotTim
    • one year ago
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    nah, is good.

  20. NotTim
    • one year ago
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    looking at too many problems at the same time.

  21. anonymous
    • one year ago
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    Okay, then i stand by what i said. Her answer choices dont include 0 solutions. I can check wolfram to back me up, but might be slow on mobile, lol

  22. NotTim
    • one year ago
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    Why is that?

  23. anonymous
    • one year ago
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    Yeah, definitely 0 solutions. Cant copy the wolfram link on mobile. But what i did was perfectly fine and shows there is no solution

  24. NotTim
    • one year ago
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    I worked through the entire problem and got something different.

  25. anonymous
    • one year ago
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    Well, do you see what i did to come across my solution?

  26. NotTim
    • one year ago
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    Yeah, went to wolfram. It made sense. I'm double checking my answer now.

  27. NotTim
    • one year ago
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    http://www.wolframalpha.com/input/?i=2x%2By%5E2%3D33+and+x%5E2%2By%5E2%2B2x%3D19 is what you went thru right.

  28. anonymous
    • one year ago
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    Yes, thats what i saw when i checked. Well, i can either try to better explain what i did or try and see where you had a mistake

  29. NotTim
    • one year ago
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    Alright. I'm going to re-do my thing. If i do the same thing, I'll post it. If my results are in agreement, I'll tell ya. give me 3 minutes.

  30. NotTim
    • one year ago
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    I think I already found my problem

  31. anonymous
    • one year ago
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    Okay, thats good. Either way, were trying to make a substitution that lets us reduce one of the equations to a single variable. Thankfully, we have a common y^2 + 2x in each equation. The first equation explicitly tells us that y^2 + 2x = 33. This will ways be true. Because the 2nd equation contains the same y^2 +2x, we can replace that chunk with 33. x^2 +2x + y^2 = 19 becomes x^2 + 33 = 19. And then from there you can see the no solution result.

  32. NotTim
    • one year ago
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    I dunno. I worked thru it again, and I got 7.81=y.

  33. anonymous
    • one year ago
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    We could also show the same result by an elimination method, which is essentially going to be the same thing. Whichever way makes the result more visible to ya.

  34. anonymous
    • one year ago
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    Can you type out tour steps then, tim?

  35. NotTim
    • one year ago
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    Alright.

  36. NotTim
    • one year ago
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    Rearranged x= (-y^2+33)/2

  37. NotTim
    • one year ago
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    Also, im NOT tim

  38. NotTim
    • one year ago
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    then i subsituted into the other equation

  39. NotTim
    • one year ago
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    ((-y^2+33)/2)^2+y^2+2((-y^2+33)/2)=19

  40. NotTim
    • one year ago
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    (-0.5y^2+16.5)^2+y^2-y^2+33=19

  41. NotTim
    • one year ago
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    -0.5y^2+y^2-y^2+16.5+33-19=0

  42. NotTim
    • one year ago
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    -0.5y^2+30.5=0

  43. NotTim
    • one year ago
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    y=sqrt(-30.5/-0.5)

  44. NotTim
    • one year ago
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    y=7.81

  45. NotTim
    • one year ago
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    Which doesn't make sense in the presence of wolfram alpha. I get that wolfram alpha sometimes get wrong data in context, but my written out answer doesn't collide well with that data.

  46. NotTim
    • one year ago
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    @cutiecomittee123 In my conclusion, I believe it is 1. But don't trust me. both of us got different answers. But what you can take out of this conversation, is to rearrange 2x+y^2=33 into x=____ and then sub it into the other equation for x.

  47. anonymous
    • one year ago
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    Sorry, i kost wifi. Ill explain where in your steps you can get no solutiom nottim

  48. anonymous
    • one year ago
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    At the step youwhere you have (-0.5y^2 +16.5)^2 +y^2 -y^2 +33 = 19. So let me cancel the y^2's and move the 33 to the other side. That would give (-0.5y^2 +16.5)^2 = -14 Notice we have something squares equal to a negative value. This isnt possible and from here we can also conclude no solution

  49. anonymous
    • one year ago
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    @NotTim

  50. anonymous
    • one year ago
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    Wow, it deleted my wxplanatiom

  51. anonymous
    • one year ago
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    Let me retype

  52. anonymous
    • one year ago
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    So, the step where you have (-0.5y^2 +16.5)^2 +y^2 -y^2 +33 = 19. Let me cancel the y^2's and move the 33 to the other side. This gives (-0.5y^2 +16.5)^2 = -14. Notice how we have a squared term equal to a negative value. From here we can see the no solution result @NotTim

  53. anonymous
    • one year ago
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    Well, it got deleted again. This time ill copy it in case that happens again, haha

  54. anonymous
    • one year ago
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    At the step where you have (-0.y^2 + 16.5)^2 +y^2 - y^2 + 33 = 19 Ill cancel the y^2's and move the 33 to the other side. This gives (-0.5y^2+16.5)^2 = -14. Thus we have something squared equal to a negative value. This shows that we would have no solution since a value squared cannot equal something negative

  55. NotTim
    • one year ago
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    You forgot to subtract 16.5^2 to both sides.

  56. NotTim
    • one year ago
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    But the order is 3 I think.

  57. NotTim
    • one year ago
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    wait wrong place what is that

  58. anonymous
    • one year ago
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    Im just looking at your steps and i dont see where that would be. There wouldnt be a 16.5^2 to subtract. You have the entire quantity of (-0.5y^2 +16.5)^2, i cannot subtract 16.5^2 from here.

  59. NotTim
    • one year ago
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    (-0.5y^2+16.5)^2 = -14.

  60. NotTim
    • one year ago
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    Unless you typed that wrong.

  61. anonymous
    • one year ago
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    Thats after i cancelled y^2's and subtracted 33.

  62. NotTim
    • one year ago
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    that would result in -0.25y^4+272.25= -14

  63. NotTim
    • one year ago
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    then -0.25y^4+272.25=-14

  64. anonymous
    • one year ago
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    No, it wouldnt. (-0.5y^2 +16.5)^2 = .25y^4 -16.5y^2 +272.25

  65. NotTim
    • one year ago
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    why does 16.5^2 become -16.5y^2?

  66. anonymous
    • one year ago
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    Its a foiling problem if you were to expand it. That and squaring -0.5y^2 includes squaring the negative, it would disappear

  67. NotTim
    • one year ago
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    that's true.

  68. anonymous
    • one year ago
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    Bleh, sick of it deleting my messages. Stupid mobile, lol

  69. NotTim
    • one year ago
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    Alright, you reasoning sounds perfect. Of course, I'm going to check once more, but everything you stated is sound.

  70. anonymous
    • one year ago
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    But okay, lets write out each individual step of the foil -0.5y^2 * -0.5y^2 = 0.25y^4 -0.5y^2 * 16.5 = -8.25y^2 16.5 * -0.5y^2 = -8.25y^2 16.5 * 16.5 = 272.25 put it together and you have 0.25y^4 -16.5y^2 + 272.25

  71. NotTim
    • one year ago
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    Darnit i'm confused again.

  72. anonymous
    • one year ago
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    Which step/part?

  73. NotTim
    • one year ago
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    (-0.5y^2+16.5)(-0.5y^2+16.5)+33= 19 was a step?

  74. anonymous
    • one year ago
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    Correct

  75. NotTim
    • one year ago
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    Alright, figured it out. Finally. Thanks mate.

  76. anonymous
    • one year ago
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    Haha, okay. As long as everything makes sense. Do you come to no solution now?

  77. NotTim
    • one year ago
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    Yeah, finally.

  78. NotTim
    • one year ago
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    Same problem I had in high school. Didn't check over my work.

  79. anonymous
    • one year ago
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    Lol, alright. Well, cutie closed the question, so not sure what conclusion she came to or if she made another question

  80. anonymous
    • one year ago
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    Alright, sick of my wifi. Ill log back on once im in a position to set up my laptop. Laterz :)

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