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cutiecomittee123

  • one year ago

Solve this system of conic section equations 7y^2+x^2=64 x+y=4

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  1. jim_thompson5910
    • one year ago
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    First solve the equation x+y = 4 for y \[\Large x+y=4\] \[\Large y=4-x\] Now plug this into the other equation so you boil things down to one variable only \[\Large 7y^2+x^2=64\] \[\Large 7(4-x)^2+x^2=64\] \[\Large 7(16-8x+x^2)+x^2=64\] \[\Large 112-56x+7x^2+x^2=64\] \[\Large 112-56x+8x^2=64\] \[\Large 112-56x+8x^2-64=0\] \[\Large 8x^2-56x+48=0\] From this point, you need to solve for x. To do so, I recommend using the quadratic formula. I'll let you do this part. Tell me what x values you get as the solutions.

  2. cutiecomittee123
    • one year ago
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    I get x=1 and x=6

  3. jim_thompson5910
    • one year ago
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    me too

  4. jim_thompson5910
    • one year ago
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    if x = 1, then y = ???

  5. cutiecomittee123
    • one year ago
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    sweet and then just plug that into on of the original equations to solve for y

  6. jim_thompson5910
    • one year ago
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    yes or plug into y = 4-x since that already has y isolated

  7. cutiecomittee123
    • one year ago
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    so like y=4-1 and y=4-6

  8. cutiecomittee123
    • one year ago
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    y=3 and y=-2

  9. jim_thompson5910
    • one year ago
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    correct, so the solutions are these two ordered pairs (1,3) and (6,-2) and this confirms it http://www.wolframalpha.com/input/?i=7y^2%2Bx^2%3D64%2Cx%2By%3D4

  10. cutiecomittee123
    • one year ago
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    wanna help me with another one?

  11. jim_thompson5910
    • one year ago
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    sure

  12. jim_thompson5910
    • one year ago
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    what's your question?

  13. cutiecomittee123
    • one year ago
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    x^2+y^2+2x+2y=0 x^2+y^2+4x+6y+12=0 solve this system of equations

  14. jim_thompson5910
    • one year ago
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    ok at first this looks really complicated, but we can eliminate quite a bit here notice how each equation has x^2+y^2 in it so we can subtract the equations (either equation1 - equation2 or equation2-equation1) to eliminate the x^2+y^2 terms what do you get when you subtract?

  15. cutiecomittee123
    • one year ago
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    2x+4y+12=0

  16. jim_thompson5910
    • one year ago
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    Now let's solve 2x+4y+12=0 for x 2x+4y+12=0 2x+4y+12-12=0-12 2x+4y = -12 2x+4y-4y = -12 - 4y 2x = -4y - 12 2x/2 = (-4y-12)/2 x = -2y - 6

  17. cutiecomittee123
    • one year ago
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    now plug that into the other equation

  18. jim_thompson5910
    • one year ago
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    Next, plug x = -2y - 6 into either original equation. I'll pick the first equation x^2+y^2+2x+2y=0 (-2y-6)^2+y^2+2(-2y-6)+2y=0 ... replace x with -2y-6 (4y^2+24y+36)+y^2+2(-2y-6)+2y=0 4y^2+24y+36+y^2-4y+12+2y=0 5y^2+22y+48=0 solve that for y (use the quadratic formula). Tell me what you get

  19. jim_thompson5910
    • one year ago
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    yes correct

  20. jim_thompson5910
    • one year ago
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    hmm I messed up, let me fix

  21. jim_thompson5910
    • one year ago
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    x^2+y^2+2x+2y=0 (-2y-6)^2+y^2+2(-2y-6)+2y=0 ... replace x with -2y-6 (4y^2+24y+36)+y^2+2(-2y-6)+2y=0 4y^2+24y+36+y^2-4y-12+2y=0 ... it should be -12, not +12 5y^2+22y+24=0

  22. cutiecomittee123
    • one year ago
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    well I tried and I got the b^2-4(a)(c) = -476 you cant take a square root of a negative number

  23. jim_thompson5910
    • one year ago
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    I fixed my mistake and got 5y^2+22y+24=0

  24. cutiecomittee123
    • one year ago
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    oh gotcha let me try that

  25. cutiecomittee123
    • one year ago
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    y=-2 and y=-2.4

  26. jim_thompson5910
    • one year ago
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    -2.4 or -12/5

  27. jim_thompson5910
    • one year ago
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    now use each y value to find the corresponding x value

  28. cutiecomittee123
    • one year ago
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    -12/5? how did you get that?

  29. cutiecomittee123
    • one year ago
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    -22+ or =- the sqrt of 4

  30. cutiecomittee123
    • one year ago
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    over 10

  31. jim_thompson5910
    • one year ago
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    I used the quadratic formula or you can convert -2.4 to fraction form -2.4 = -2.4*(10/10) = -24/10 = -12/5

  32. cutiecomittee123
    • one year ago
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    makes sense

  33. cutiecomittee123
    • one year ago
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    so now we plug in the y values to get x

  34. jim_thompson5910
    • one year ago
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    correct

  35. jim_thompson5910
    • one year ago
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    you can use x = -2y - 6

  36. cutiecomittee123
    • one year ago
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    x=-10.8 and x=-10

  37. jim_thompson5910
    • one year ago
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    incorrect on both

  38. jim_thompson5910
    • one year ago
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    if y = -2, then x = -2y - 6 x = -2(-2) - 6 x = 4 - 6 x = -2

  39. jim_thompson5910
    • one year ago
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    if y = -2.4, then x = -2y - 6 x = -2(-2.4) - 6 x = 4.8 - 6 x = -1.2

  40. cutiecomittee123
    • one year ago
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    so the solutions are (-2,-2) and (-1.2, -12/

  41. jim_thompson5910
    • one year ago
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    correct and the graph confirms it

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  42. jim_thompson5910
    • one year ago
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    I think you meant to say -12/5

  43. jim_thompson5910
    • one year ago
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    if you use -12/5 then convert -1.2 to fraction form as well -1.2 = -1.2*(10/10) = -12/10 = -6/5

  44. cutiecomittee123
    • one year ago
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    Yeah I did mean -12/5 thanks:))

  45. jim_thompson5910
    • one year ago
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    np

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