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Math Help - Who knoes the answer of this:

  1. #16
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    Ok. Thank you verrry much! I understand it now.
    Is every thing here difficult? Because I'd like to stay here for a moment and discover more.
    But if every thing is very very difficult, I can't understand it.
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  2. #17
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    Quote Originally Posted by Archie Meade View Post

    Once you know what y is, then you find x.
    Seems morre complex and you may not be that far yet.
    Too difficult ^^ But thanks!
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  3. #18
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    Quote Originally Posted by Wissem View Post
    wowh ^^ Very difficult English. I must translate it to Dutch so I'll understand it :P
    It's not for school or something, I found it on the web.

    I'm understanding it now.
    x(x)=81 and y(y)= 24
    Is that good?
    Not quite!

    (x-y)^2=9^2=81

    You must multiply out (x-y)(x-y)

    When you do that, part of it is x^2+y^2

    and the other part is -2xy.

    You are given xy=12

    therefore there is enough information to figure out x^2+y^2

    from those clues, which gives the combined area of the 2 rooms.
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  4. #19
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    Quote Originally Posted by Wissem View Post
    Ok. Thank you verrry much! I understand it now.
    Is every thing here difficult? Because I'd like to stay here for a moment and discover more.
    But if every thing is very very difficult, I can't understand it.
    Exactly,
    it's a matter of learning the foundations and building up your skills from there.

    When you don't understand something, there may be something which precedes it
    that you need to go back and master first.

    Something is "easy" to the extent that you know the component parts,
    the "building blocks", the earlier material that should have been covered.
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  5. #20
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    Is everyting here difficult? Or are there some stuff I'll understand?
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  6. #21
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    Quote Originally Posted by Wissem View Post
    Is everyting here difficult? Or are there some stuff I'll understand?
    That depends on the work you do!
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  7. #22
    MHF Contributor Unknown008's Avatar
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    Ok, I'll try to explain some other way to solve the problem...

    You got the equations?

    xy = 12

    y - x = 9

    From the second equation, you know that we can put it in this form: y = 9 + x ?

    From this, you 'know' what is y. Replace This, by the red 'y' in the first equation, like this:

    x(9 + x) = 12

    This done, you can expand, to give:

    9x + x^2 = 12

    Then, you put everything on the same side, changing the order to make it easier, like this:

    x^2 + 9x - 12 = 0

    Now, you'll need to factorise. However, using the ordinary methods, it is impossible to solve this unless you know how to use the quadratic formula. So, we'll change some things a little.

    We are looking for the sum of the rooms' areas. This means, we need to add x^2 and y^2

    We're looking for x^2 + y^2

    But we know that
    y - x = 9

    So... why not square everything to see what happens?

    (y-x)^2 = 9^2

    From your knowledge of expansion, you know that: (y-x)^2 = (y-x)(y-x) = x^2 - xy - xy + y^2 = x^2 + y^2 - 2xy

    and 9^2 = 81

    Hm... we have only part of our solution... there is this '-2xy' which we don't have! But... do you remember the first equation (the one I underlined)?

    xy = 12

    What we need is -2xy... so, let's multiply by -2.

    xy (-2) = 12 (-2)

    -2xy = -24

    Now! We got -2xy. Let's replace this in the equation above, that is: x^2 + y^2 - 2xy = 81

    x^2 + y^2 (- 2xy) = 81

    x^2 + y^2 (-24) = 81

    x^2 + y^2 -24 = 81

    x^2 + y^2 = 105

    Let me know if it's better, or if not, where you get stuck
    Last edited by Unknown008; August 1st 2010 at 10:43 AM.
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  8. #23
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    Dear Wissem,

    You have received an infraction at Math Help Forum.

    Reason: General violations
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    The title of a thread should give some indication of the nature of the question

    CB
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    This infraction is worth 5 point(s) and may result in restricted access until it expires. Serious infractions will never expire.
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    What is this? something good or bad?
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  9. #24
    MHF Contributor Unknown008's Avatar
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    It's actually... bad. Your question needed a more appropriate title. Try not repeating the same mistake again.

    I know it's difficult since you're more used to Dutch... but if you are in this forum, you'll have to learn to be more at ease with English...
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  10. #25
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    I Understand it! But just 1 thing:

    (x-y)^2= (x-y)(x-y)= x^2-xy-xy+y^2 = x^2+y^2-2xy

    (x-y)(x-y)=x^2

    from where did we got that x^2??

    I'll write everyting on paper
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  11. #26
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    Quote Originally Posted by Unknown008 View Post
    It's actually... bad. Your question needed a more appropriate title. Try not repeating the same mistake again.

    I know it's difficult since you're more used to Dutch... but if you are in this forum, you'll have to learn to be more at ease with English...
    Excerly, I'm half Dutch, half Tunisian. So even in the Dutch language, I can't understand math very well.
    I understand it only in Arabic :P
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  12. #27
    MHF Contributor Unknown008's Avatar
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    Hm... I don't know how you learned it, that's the problem...

    The way I learned it, is:
    Multiply x in the first bracket by the first x in the second bracket. This gives x^2
    Then, multiply x in the first bracket by -y in the second bracket. This gives -xy
    Then, multiply -y in the first bracket by x in the second bracket. This gives -xy
    Finally, multiply -y in the first bracket by -y in the second bracket. This gives +y^2

    If you learned it like in Archie Meade's posts, then you might better understand it the other way.
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  13. #28
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    No I understand that but from where did we get that x^2??? because we've (x-y)(x-y)
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  14. #29
    MHF Contributor Unknown008's Avatar
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    Hmm... okay, multiplying x by x, gives x^2

    Like:
    x \times x = x^2

    2 \times 2 = 2^2

    y^2 \times y^2 = (y^2)^2

    Do you understand how the power works? Usually, what you need to remember for these is that when you get two things similar being multiplied, you put a little 2 at the upper right side of a number or symbol.

    If you had:
    a \times a \times a = a^3

    x \times x \times x \times x = x^4

    Is that good?

    Or I did not understand what you did not understand?
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  15. #30
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    I sad that I know that xD I just don't understand from where did the x and the y appears:

    x^2+xy+xy+y^2
    and we had (y-x)(y-x)
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