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anonymous

  • 5 years ago

y=x^2+2 y=-x+4 SOMEONE PLEASE HELP ME ON THIS

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  1. anonymous
    • 5 years ago
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    Equate the two equations, and then move everythign to one side so you have x^2+x-2. That factors into (x-1)^2=0, therefore x=1

  2. anonymous
    • 5 years ago
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    Y=y, so x^2+2=-x+4 x^2+x-2=0 (x-1)(x+2)=0 x=1 or -2 Y=-x+4 y=-1+4. Or y=-(-2)+4 y=3 or y=6 so, x=1, y=3 or x=-2, y=6

  3. anonymous
    • 5 years ago
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    @spaceknight ... after you get that 1 , do i plug that into x to get y ?

  4. anonymous
    • 5 years ago
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    &lotsobooks omg thanks . can you like tell me the steps on how you do it

  5. anonymous
    • 5 years ago
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    lol oops i factored wrong sorry just listen to lotsobooks

  6. anonymous
    • 5 years ago
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    i already solved this problem in the one before...

  7. anonymous
    • 5 years ago
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    ok so can i ask you guys another question ? i need help on my entire algebra2 homework

  8. anonymous
    • 5 years ago
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    you should try doing your own homework

  9. anonymous
    • 5 years ago
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    &spaceknight i have been trying . if i knew how to do thiss i wouldntt be on my laptop asking for help . my teacher is the kind that doesnt like to explain too much .... I NEED HELP

  10. anonymous
    • 5 years ago
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    Set both equations equal to one another. Move all of the terms to the left side of the equation by adding or subtracting the terms on both sides of the equation. Combine like terms. Since it is a quadratic equation, you can factor it - use FOIL backwards, sort of. Once you have the 2 possible values of x, then, you can substitute them back inti either of the original equations - I picked -x+4 since it is the easiest. Then, calculate the values of y. Remember to keep the x,y pairs together since they only go with one another.

  11. anonymous
    • 5 years ago
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    see i never got instructions like that. ok how about this . im about to show you my next problem. im going to try and solve it by myself to see if i can do it but can you still help me

  12. anonymous
    • 5 years ago
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    ?

  13. anonymous
    • 5 years ago
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    y=6x^2+5x+3 y==x+3

  14. anonymous
    • 5 years ago
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    \[6x ^{2}+5x+3=-x+3\]

  15. anonymous
    • 5 years ago
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    So far, so good.

  16. anonymous
    • 5 years ago
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    ok so it will go like this i thinkk 6x^2+5x+3=-x+3 +x-3 +x-3 hold on i think i messed up

  17. anonymous
    • 5 years ago
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    i got 6x^2+6x=0 when i subtracted

  18. anonymous
    • 5 years ago
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    That is correct.

  19. anonymous
    • 5 years ago
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    so what do i do now ?

  20. anonymous
    • 5 years ago
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    In this case, it is easiest to subtract 6x from both sides.

  21. anonymous
    • 5 years ago
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    then \[6x ^{2} = -6x ?\]

  22. anonymous
    • 5 years ago
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    Right. Then, divide both sides by 6.

  23. anonymous
    • 5 years ago
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    ok so then that is x^2 = -1x or x^2 = -x ?

  24. anonymous
    • 5 years ago
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    Right. Now, divide both sides by x.

  25. anonymous
    • 5 years ago
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    ok so then after that x = -1 ?

  26. anonymous
    • 5 years ago
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    Right. Plug -1 into one of the original equations and solve for y.

  27. anonymous
    • 5 years ago
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    so whenever i plug it into the second equation how do i do it ? do i ignore the negative there already is ?

  28. anonymous
    • 5 years ago
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    No, keep the negative. -(-1)+3=

  29. anonymous
    • 5 years ago
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    wouldnt that 1 just turn into a positive ?

  30. anonymous
    • 5 years ago
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    Right.

  31. anonymous
    • 5 years ago
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    ok so it would be 1+3 so y = 4 ?

  32. anonymous
    • 5 years ago
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    Yep.

  33. anonymous
    • 5 years ago
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    We missed a second solution when I had you subtract 6x from both sides.

  34. anonymous
    • 5 years ago
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    you are AWESOMEEEE !!!! :p thanks for helping meee

  35. anonymous
    • 5 years ago
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    what step did we miss ?

  36. anonymous
    • 5 years ago
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    Not a step so much as another solution. At 6x^2+6x=0, it would have been better to factor 6x out. So, 6x (x+1)=0 x=-1 or x=0. then, y=4 or y=3.

  37. anonymous
    • 5 years ago
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    You're welcome! Have fun!

  38. anonymous
    • 5 years ago
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    :)

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