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Math Help - quadratic equation problems

  1. #1
    Newbie white's Avatar
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    Exclamation quadratic equation problems

    hello.need help on the 2 questions

    tickets for a concert are available at two prices. the more expensive ticket is $30 more than the cheaper one. Find the cost of each type of ticket if a group can buy 10 more of the cheaper tickets than the expensive ones for $1800
    the more expensive ticket is 30 + x
    less expensive ticket is x
    10 more of the cheaper tickets than the expensive ones for $1800
    10 + x ... ?

    the members of a club hire a bus for $2100, seven members withdraw from the club and the remaining member have to pay $10 more each to cover the cost. How many members originally agreed to go on the bus.
    let x be number of people

    so x= 2100,
    x-7= ... ?


    thank you guys
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  2. #2
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    Quote Originally Posted by white View Post
    hello.need help on the 2 questions

    tickets for a concert are available at two prices. the more expensive ticket is $30 more than the cheaper one. Find the cost of each type of ticket if a group can buy 10 more of the cheaper tickets than the expensive ones for $1800
    the more expensive ticket is 30 + x
    less expensive ticket is x
    10 more of the cheaper tickets than the expensive ones for $1800
    10 + x ... ?

    [snip]
    Keep x as you defined it. Let y = number of expensive tickets bought for $1800. Then:

    Cheap tickets: 1800 = (10 + y)x => 1800 = 10x + xy .... (1)

    Expensive tickets: 1800 = y(x + 30) => 1800 = xy + 30y .... (2)

    Equate (1) and (2): 10x + xy = xy + 30y => x = 3y .... (3)

    Substitute (3) into (2): 1800 = 3y^2 + 30y => y^2 + 10y - 600 = 0 => (y + 30)(y - 20) = 0

    etc.
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  3. #3
    Flow Master
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    Quote Originally Posted by white View Post
    [snip]
    the members of a club hire a bus for $2100, seven members withdraw from the club and the remaining member have to pay $10 more each to cover the cost. How many members originally agreed to go on the bus.
    let x be number of people

    so x= 2100,
    x-7= ... ?


    thank you guys
    Original cost per member = \frac{2100}{x}.

    New cost per member = \frac{2100}{x} + 10 = \frac{2100 + 10x}{x}.

    New number of members = x - 7.

    Therefore:

    2100 = (x - 7) \left( \frac{2100 + 10x}{x}\right)

     \Rightarrow 2100x = (x - 7)(2100 + 10x)

    \Rightarrow 210x = (x - 7)(210 + x).

    Expand the right hand side, simplify, re-arrange into the form quadratic = 0 and solve for x.

    For checking purposes: I get x = 42.
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