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gjhfdfg Group Title

Having trouble with the substitution method,

  • one year ago
  • one year ago

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  1. gjhfdfg Group Title
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    I've attempted it several times but answers come out different each time. & I'm clueless to why.

    • one year ago
  2. alrightatmaths Group Title
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    What are the equations?

    • one year ago
  3. Kweku Group Title
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    can you give us the equation for us to try it and see where the problem is?

    • one year ago
  4. gjhfdfg Group Title
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    Its, |dw:1360705833711:dw|

    • one year ago
  5. gjhfdfg Group Title
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    My answers came out from 4 to somehow I got .14446 etc..

    • one year ago
  6. gjhfdfg Group Title
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    So I know Im doing it wrong somehow

    • one year ago
  7. alrightatmaths Group Title
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    Get the first equation equal to x. So x = 2y-5. Then sustitute that in.

    • one year ago
  8. alrightatmaths Group Title
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    (2y-5)^2 + y^2 - 25 = 0

    • one year ago
  9. alrightatmaths Group Title
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    4y^2 - 20y +25 - 25 = 0

    • one year ago
  10. alrightatmaths Group Title
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    5y^2 - 20y = 0

    • one year ago
  11. alrightatmaths Group Title
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    5y(y-4)=0

    • one year ago
  12. alrightatmaths Group Title
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    y=4 or y=0

    • one year ago
  13. Kweku Group Title
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    I think alrightatmaths is right.

    • one year ago
  14. alrightatmaths Group Title
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    Thank you Kweku :)

    • one year ago
  15. gjhfdfg Group Title
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    Im lost

    • one year ago
  16. Kweku Group Title
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    WHERE?

    • one year ago
  17. gjhfdfg Group Title
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    Would the 0 & 4 have a fraction beside?

    • one year ago
  18. ArTFis Group Title
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    Just make sure you have the variables on one side and the the whole numbers or fractions on the other side. {2y - x = 5 {2x + 2y - 25 = 0 In this case, subtract 2y on both sides and you should get -x = 5 - 2y. Then divide -1 due to the -x on both sides. You'll then have x = -5 + 2y. Now plug in the x = -5 + 2y into the second equation. This should help you solve for y. It'll end up being 2(-5 + 2y) + 2y - 25 = 0 Next it'll be substituted to -10 + 4y +2y -25 = 0. Put variables on one side and the whole numbers on the other side. The whole numbers are "negative," so you should now "add" 10 and 25 on both sides. You should get in the end, 6y = 35. Divide 6 on both sides. It should be fine if you get fractions or not. This is how I learned it. Not every set of system of equations are going to be easy. It may deal with whole numbers and/or fractions. Once you find one of the values such as y or x, then plug in the y or x value into the opposite equation that you didn't use to find that value. That's how it should be I believe.

    • one year ago
  19. gjhfdfg Group Title
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    Alrighty, I got it thanks.!

    • one year ago
  20. ArTFis Group Title
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    ^__^

    • one year ago
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