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    Heres the question;

    One hundred and seven people get together in teams to take part in a quiz. Each team has three, four or
    five members. There are twenty-seven teams altogether and there are twice as many teams with
    four members as there are teams with three members. How many teams have five members?
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    (Original post by CraigBackner)
    Heres the question;

    One hundred and seven people get together in teams to take part in a quiz. Each team has three, four or
    five members. There are twenty-seven teams altogether and there are twice as many teams with
    four members as there are teams with three members. How many teams have five members?
    Do you need help with this or is it just a challenge?
    If you need help, try writing the information as two equations, assigning letters to mentioned unknown quantities.
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    (Original post by IrrationalRoot)
    Do you need help with this or is it just a challenge?
    If you need help, try writing the information as two equations, assigning letters to mentioned unknown quantities.
    I've literally done that but I don't know how to solve it. Could you explain how please?
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    It's just simulataneous equations. Model it algebraically, with x, y, z representingg how many people have 3, 4 or 5 people respectfully.
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    (Original post by BobBobson)
    It's just simulataneous equations. Model it algebraically, with x, y, z representingg how many people have 3, 4 or 5 people respectfully.
    would the equation be
    3x+4y+5z=170
    and
    3x+4y+5z=27?
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    (Original post by CraigBackner)
    I've literally done that but I don't know how to solve it. Could you explain how please?
    If you have two simultaneous equations in x and y (say), one method is to solve for y in one equation and substitute that into the other equation to get an equation only in x. Another method is to multiply both equations by constants in such a way that when you subtract them, the ys or xs cancel out.
    Surprised you haven't learnt simultaneous equations in school yet. If you still don't understand it'd probably be best to look them up for a more comprehensive explanation.
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    (Original post by IrrationalRoot)
    If you have two simultaneous equations in x and y (say), one method is to solve for y in one equation and substitute that into the other equation to get an equation only in x. Another method is to multiply both equations by constants in such a way that when you subtract them, the ys or xs cancel out.
    Surprised you haven't learnt simultaneous equations in school yet. If you still don't understand it'd probably be best to look them up for a more comprehensive explanation.
    I know how to solve simultaneous equations, I just dont know how to solve this problem which involes it , thats what I need explaning
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    (Original post by CraigBackner)
    would the equation be
    3x+4y+5z=170
    and
    3x+4y+5z=27?
    The first ones right, but not the second one. The total number of Teams equals 27, so it's just the number of each type of team added together.
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    (Original post by CraigBackner)
    would the equation be
    3x+4y+5z=170
    and
    3x+4y+5z=27?
    There's also a third equations that can be made with the later part of the question
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    (Original post by CraigBackner)
    I know how to solve simultaneous equations, I just dont know how to solve this problem which involes it , thats what I need explaning
    Simple. You have teams of 3, 4 and 5. You do not know how many of each team therefore you assign variables to each and it has to sum up to 107 so 3x+4y+5z=107. Then you're told that the amount of teams is equal to 27 so x+y+z=27 and then you are ALSO told there are twice as many teams of 4 as there are of 3 so y=2x and you can simplify the two equations into 2 variables and solve it as normal.


    (Original post by CraigBackner)
    would the equation be
    3x+4y+5z=170
    and
    3x+4y+5z=27?

    That wouldn't make sense as this system of equations is inconsistent; you'd get 0=143. Geometrically, these are 2 parallel planes so no intersections obviously.
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    (Original post by BobBobson)
    There's also a third equations that can be made with the later part of the question
    okay so the two equations are
    3x+4y+5z=27
    and
    x+y+z = 27

    and then I eliminate one of the varbiles, lets say x, which will create another equation, in which I use to solve
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    (Original post by CraigBackner)
    okay so the two equations are
    3x+4y+5z=27
    and
    x+y+z = 27

    and then I eliminate one of the varbiles, lets say x, which will create another equation, in which I use to solve
    The third equation. The third equation about "there are twice as many teams with four members as there are teams with three members."
 
 
 
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