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    Don’t you have to equate the sides given to 22cm and then rearrange to find x and y. Then substitute it back into the formula? I honestly don’t know if that will work but it’s worth a try 😂
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    (Original post by frankabagnale)
    It's a rectangle, so opposite sides are the same.

    This yields 2x+7y = x-2y

    Secondly, considering the perimeter yields

    (x-2y) + (2x+7y) + 16 = 22

    These are two equations in two unknowns, so you can solve for x,y.
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    Since it is a rectangle, x - 2y = 2x + 7y

    Also, x - 2y + 2x + 7y = 22 - 2x8
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    (Original post by frankabagnale)
    You again. The most useless person. You state the obvious and show no working to the actual answer. Also, i already knew this, as stated in the reply above. Give it a go (for once). It doesn't work.
    NO ONE HELP HIM OUT, he sounds like a rude ****
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    (Original post by frankabagnale)
    You again. The most useless person. You state the obvious and show no working to the actual answer. Also, i already knew this, as stated in the reply above. Give it a go (for once). It doesn't work.
    So then do it correctly for once if you know how. It works just fine.
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    (Original post by frankabagnale)
    I've tried, hence I'm asking in the student room?
    Being rude to those who can help you is like shooting yourself in your foot. Bit stupid.

    Good luck with the problem.
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    (Original post by frankabagnale)
    Doesn't work.

    Neither does equating the two widths and finding x in terms of y or vise versa and then subbing that back in.

    I'm clueless.
    The two sides with x and y on them are equal since it's a regular rectangle with 2 pairs of equally lengthed sides. You also then add all 4 sides to get something equall to 22. Solve the 2 equations simultaenously.
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    Try what others said.
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    It's a rectangle so the opposite sides are equal. Therefore:

    x - 2y = 2x + 7y

    Rearrange for x:

    x = -9y

    Perimeter is 22. Therefore:

    x - 2y + 16 + 2x + 7y = 22

    Simplify

    3x + 5y - 6 = 0

    Substitute in x = -9y

    -22y = 6
    y = -3/11

    Use this to find x, then sub x and y back into x - 2y or 2x + 7y to find the dimensions.

    Dimensions are 3 by 8
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    (Original post by frankabagnale)
    When i that y= (minus) 3/11

    Even disregarding the minus, subbing that back in to find x, x comes to 27/11. It doesnt work when i test these in the original formula of the perimeter
    that should be right, could you retry testing it? what are the actual answers?
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    (Original post by frankabagnale)
    When i that y= (minus) 3/11

    Even disregarding the minus, subbing that back in to find x, x comes to 27/11. It doesnt work when i test these in the original formula of the perimeter
    Those are correct.. Maybe try substituting into (x-2y) and (2x+7y) to find the sides?
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    The people who made simultaneous equations were correct. Working on from that...

    x = 27/11
    y = -3/11

    Let's find the length of the short side now.

    x - 2y
    = 27/11 - 2(-3/11)
    = (27 + 6)/11
    = 33/11
    x = 3cm
    y = 8cm

    To check our solution, we calculate the perimeter.
    3 + 3 + 8 + 8 = 22cm
    I'm right!
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    (Original post by frankabagnale)
    Cheers pal, i was so confused all along because i disregarded the minus of -3/11 when i tried to find x. I guess i was thinking lengths can't be minus. My bad

    THANKS ALL
    No problem
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    (Original post by frankabagnale)
    I've tried, hence I'm asking in the student room?
    The Student Room is not a place to get people to do your homework for you. Be kind and respectful to people on here and in turn they will help you.
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    (Original post by frankabagnale)
    I've tried, hence I'm asking in the student room?
    RDKGames is "the most useless person"? I think you meant one of the most helpful! I often seem to find myself repping them in various threads. In fact, so many times I've clicked on a thread to try to help someone, only to find that RDKGames has already done a brilliant job of it! As for actually "giving the question a go (for once)", they probably have, but simply gave advice on how to approach it - rather than just ignoring the TSR guidelines and posting a full solution.

    If you're getting frustrated with the advice of RDKGames, then you must really just want to be spoon-fed the answer. Good luck with the exam (when you won't have anyone to tell you the answers).

    If you're so adamant that a method doesn't work just because of something not seeming right at a point, then work it through to the final answer, and see if the answer is actually reasonable. Just because y is negative doesn't mean that the sides are negative too - they each also have an x-component to make them positive.

    TL;DR: We're all more than happy to help you, but please don't be sarcastic with those trying to do so, or throw words like "useless" about at those who are legitimately helping.

    Spoiler:
    Show


    Rant over. Have a nice day!

    Please don't take any offence from anything that I've said, and I hope that you get to the solution soon.

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    (Original post by mupsman2312)
    RDKGames is "the most useless person"? I think you meant one of the most helpful! I often seem to find myself repping them in various threads. In fact, so many times I've clicked on a thread to try to help someone, only to find that RDKGames has already done a brilliant job of it! As for actually "giving the question a go (for once)", they probably have, but simply gave advice on how to approach it - rather than just ignoring the TSR guidelines and posting a full solution.

    If you're getting frustrated with the advice of RDKGames, then you must really just want to be spoon-fed the answer. Good luck with the exam (when you won't have anyone to tell you the answers).

    If you're so adamant that a method doesn't work just because of something not seeming right at a point, then work it through to the final answer, and see if the answer is actually reasonable. Just because y is negative doesn't mean that the sides are negative too - they each also have an x-component to make them positive.

    TL;DR: We're all more than happy to help you, but please don't be sarcastic with those trying to do so, or throw words like "useless" about at those who are legitimately helping.

    Spoiler:
    Show



    Rant over. Have a nice day!

    Please don't take any offence from anything that I've said, and I hope that you get to the solution soon.


    Absolute fire right there :headfire:
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    Its so easy lmao, can you not just do 22-16 for the two lengths = 6
    Then /2 for the width = 3. Technically isnt it?

    So its 8 by 3
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    (Original post by Purplebottle)
    Its so easy lmao, can you not just do 22-16 for the two lengths = 6
    Then /2 for the width = 3. Technically isnt it?

    So its 8 by 3
    Yeah, that's what I was thinking as well
    I don't know why they put the equations there if all they wanted was the dimensions
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    (Original post by Purplebottle)
    Its so easy lmao, can you not just do 22-16 for the two lengths = 6
    Then /2 for the width = 3. Technically isnt it?

    So its 8 by 3
    I just facepalmed myself. Hard. Doh!
 
 
 
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