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cutiecomittee123

  • one year ago

How do I solve this system of equations 2x^2+y^2=33 x^2+y^2+2y=19

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  1. freckles
    • one year ago
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    I would attempt to eliminate the x^2 term and solve for y first then go back and solve for x

  2. freckles
    • one year ago
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    notice if you multiply the second equation by -2 you will be able to then add the equations together to eliminate the x^2 term

  3. cutiecomittee123
    • one year ago
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    I did that but I got y=sqrt5

  4. cutiecomittee123
    • one year ago
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    then I plugged it back into the first equation to solve for x and it came out as x=15.4

  5. freckles
    • one year ago
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    I'm not seeing how you obtained that

  6. freckles
    • one year ago
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    can you should your work so I can see how you first obtained y

  7. freckles
    • one year ago
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    If you wouldn't mind what do you get right after you add the two equations (you know after the second one has been multiplied by -2)

  8. cutiecomittee123
    • one year ago
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    2x^2+sqrt5=33 -sqrt5 -sqrt5 2x^2=30.76 divide by 2 on both sides and take the square root! (I forgot about that) So it would be x=sqrt 15.38

  9. freckles
    • one year ago
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    how did you get sqrt(5)?

  10. cutiecomittee123
    • one year ago
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    After adding the equations I get -y^2=-5

  11. freckles
    • one year ago
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    2x^2+y^2=33 x^2+y^2+2y=19 first step multiply second equation by -2 2x^2+y^2=33 -2x^2-2y^2-4y=-38 now add the equations y^2-2y^2-4y=33-38 simplify -y^2-4y=-5 multiply both sides by -1 to make it prettier y^2+4y=5 now subtract 5 on both sides y^2+4y-5=0

  12. cutiecomittee123
    • one year ago
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    I got square root 5 because I multiplied both sides by -1 and then took the square root to get y=sqrt5

  13. cutiecomittee123
    • one year ago
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    So y^2 +4y=5

  14. freckles
    • one year ago
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    you can factor y^2+4y-5

  15. cutiecomittee123
    • one year ago
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    Help me with that?

  16. freckles
    • one year ago
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    what two numbers multiply to be -5 and add up to be 4?

  17. freckles
    • one year ago
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    think -5 can be written as the following: -5(1) or 5(-1) -5+1=? 5+(-1)=?

  18. cutiecomittee123
    • one year ago
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    -4 or 4

  19. freckles
    • one year ago
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    so which pair of numbers satisfy my question?

  20. freckles
    • one year ago
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    "what two numbers multiply to be -5 and add up to be 4?"

  21. cutiecomittee123
    • one year ago
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    oh 5 and -1

  22. freckles
    • one year ago
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    y^2+4y-5=0 (y+5)(y-1)=0

  23. cutiecomittee123
    • one year ago
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    (x+5) (y-1) = 0

  24. freckles
    • one year ago
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    so now you just set both factor equal to 0 y+5=0 or y-1=0

  25. cutiecomittee123
    • one year ago
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    y=-5 and or y=1

  26. freckles
    • one year ago
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    sounds great now you can go back and find x so use either equation Let's just use the first one 2x^2+y^2=33 replace y with -5 then solve for x then after you have completed that step replace y with 1 then solve for x

  27. cutiecomittee123
    • one year ago
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    x=2

  28. freckles
    • one year ago
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    you will have two different solutions two different ordered pairs (x,-5) and (x,1) where you have to solve for both of those x's

  29. freckles
    • one year ago
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    actually you might end up with 4 solutions since x has a square on it

  30. freckles
    • one year ago
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    yep there is one more solution to x^2=4 I assume that is where you got x=2 from

  31. cutiecomittee123
    • one year ago
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    x also =-2

  32. freckles
    • one year ago
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    so we have (2,-5) and (-2,-5) now we need to solve for x when y is 1

  33. freckles
    • one year ago
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    you can still use 2x^2+y^2=33

  34. cutiecomittee123
    • one year ago
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    uh y=sqrt31

  35. freckles
    • one year ago
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    did you replace x with 1?

  36. cutiecomittee123
    • one year ago
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    yeah

  37. freckles
    • one year ago
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    remember we had y=-5 or y=1 this is why we are trying to find when x when y=-5 and also why we are trying to find x when y=1

  38. freckles
    • one year ago
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    in other words we never said anything about x being 1

  39. freckles
    • one year ago
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    \[2x^2+y^2=33 \\ 2x^2+1^2=33 \\ 2x^2+1=33\] solve for x

  40. cutiecomittee123
    • one year ago
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    2(1)^2+Y^2=33 2(1)^2=2 RIGHT? then 2+y^2=33 minus 2 from both sides then take the square root so y=sqrt31

  41. freckles
    • one year ago
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    why are you replacing x with 1? we never said anything about x being 1

  42. cutiecomittee123
    • one year ago
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    Oh lol, I get x=4 after setting y=1

  43. freckles
    • one year ago
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    or also?

  44. freckles
    • one year ago
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    I assume you got that from x^2=16

  45. freckles
    • one year ago
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    but there is one more solution to that

  46. cutiecomittee123
    • one year ago
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    x=-4

  47. freckles
    • one year ago
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    \[(2,-5); (-2,-5) ; (-4,1);(4,1)\]

  48. cutiecomittee123
    • one year ago
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    sweet

  49. cutiecomittee123
    • one year ago
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    okay now the question is which one of these is in quadrant 4 of the graph?

  50. freckles
    • one year ago
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    the first two pairs come from replacing the y with -5 and solving for x (got x=2 or -2) and the last two pairs come from replacing the y with 1 and solving for x (got x=4 or -4)

  51. freckles
    • one year ago
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    |dw:1433531008630:dw|

  52. cutiecomittee123
    • one year ago
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    (2,-5)

  53. freckles
    • one year ago
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    |dw:1433531053701:dw|

  54. freckles
    • one year ago
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    yep sounds great

  55. cutiecomittee123
    • one year ago
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    Yes:)

  56. NotTim
    • one year ago
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    lol this again

  57. freckles
    • one year ago
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    |dw:1433531133632:dw|

  58. cutiecomittee123
    • one year ago
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    Can you help me with something else?:)

  59. freckles
    • one year ago
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    I can try

  60. cutiecomittee123
    • one year ago
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    Okay its pretty simple what is the maximum number of solutions to this system x^2+4y=64 and x+y=5

  61. freckles
    • one year ago
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    well one is a parabola and one is a line

  62. freckles
    • one year ago
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    what is the max number of times a line can intersect a parabola?

  63. cutiecomittee123
    • one year ago
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    2 times

  64. freckles
    • one year ago
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    yep so without solving the system we could say there is a max number of solutions to the system above of 2 and there is minimum number of 0 solutions so you can you have 2 solutions,1 solution, or 0 solutions to a system containing a parabola and a line.

  65. cutiecomittee123
    • one year ago
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    Awesome thanks

  66. cutiecomittee123
    • one year ago
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    I have more but I think I am going to ask them in other questions so things dont get confusing

  67. freckles
    • one year ago
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    k

  68. NotTim
    • one year ago
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    Wait so i was right in the other question? Dude convinced me otherwise. goddarnit.

  69. NotTim
    • one year ago
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    wait no, here it was 2. i only got 1.

  70. freckles
    • one year ago
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    what other question?>

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