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anonymous

  • one year ago

solve this system of equation using either substitution or elimination method

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  1. anonymous
    • one year ago
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    \[x^{2}+y = 6\] \[x+y=4\]

  2. tkhunny
    • one year ago
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    Well, do it. The isolated 'y' in each is pretty compelling.

  3. anonymous
    • one year ago
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    hey do you still need help? willing to walk you through it

  4. anonymous
    • one year ago
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    yes i neeed a walk through please lol

  5. anonymous
    • one year ago
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    I would consider isolating y in each equation and then setting them equal to each other (not the y's the equation they are equal to) can you do that

  6. anonymous
    • one year ago
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    |dw:1440534476791:dw|

  7. anonymous
    • one year ago
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    like that? :s

  8. anonymous
    • one year ago
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    perfect now set them equal so that y=y

  9. anonymous
    • one year ago
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    what do you mean y=y?

  10. anonymous
    • one year ago
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    set it: 4-x = 6-x^2

  11. anonymous
    • one year ago
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    does that make sense why we would do that?

  12. anonymous
    • one year ago
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    bc if they are both y then they equal each other

  13. anonymous
    • one year ago
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    yes now we have to solve for x, it is going to turn into a quadratic function: 4-x = 6-x^2 -4 on both sides -x = 2-x^2 +x on both sides 2-x^2 + x = 0 Now order it correctly: -x^2 + x + 2 = 0 but -x^2 needs to be positive so switch all the signs: x^2 - x - 2 = 0 Making sense so far? Can you factor this?

  14. anonymous
    • one year ago
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    so i factor it and get the x's and then i plug tose into the equation to get the y points of intersection?

  15. anonymous
    • one year ago
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    yep!!

  16. anonymous
    • one year ago
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    did I help?

  17. anonymous
    • one year ago
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    let me know what you get please

  18. anonymous
    • one year ago
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    hey @sarahg7 what did you end up getting...

  19. anonymous
    • one year ago
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    @twistnflip i got (-1,5) and (2,2)

  20. anonymous
    • one year ago
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    which method was that btw

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