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    tywatson's Avatar
    tywatson Posts: 2, Reputation: 1
    New Member
    		  
    #1

    Jan 19, 2008, 10:35 AM
    What equation do I use to solve this?
    Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate?
    PolluxCastor's Avatar
    PolluxCastor Posts: 117, Reputation: 5
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    #2

    Jan 19, 2008, 10:56 AM
    He should have invested it all at 11%, he would have made $660.
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #3

    Jan 19, 2008, 02:01 PM
    Let x = the amount invested at 9%, then we have .09x

    Then the amount invested at 11% would be 0.11(6000-x)

    This must equal 624.

    So, we have the equation:



    Solve for x.
    Ronniesha's Avatar
    Ronniesha Posts: 1, Reputation: 1
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    #4

    Jan 23, 2008, 10:48 PM
    4y-14=10
    tywatson's Avatar
    tywatson Posts: 2, Reputation: 1
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    #5

    Jan 24, 2008, 01:31 PM
    x + y = 6000
    .09x + .11y = 624

    Multiply by .09 and subtract

    .09x + .11y = 624
    - .09x - .09y = 540
    __________________
    .02y = 84

    y = 84 / .02

    y = 4200

    x + 4200 = 6000
    x = 6000-4200
    x = 1800

    Amount invested at 9% = 1800
    Amount invested at 11% = 4200
    PolluxCastor's Avatar
    PolluxCastor Posts: 117, Reputation: 5
    Junior Member
    		  
    #6

    Jan 24, 2008, 05:21 PM
    Yay!
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