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    ikryspy's Avatar
    ikryspy Posts: 24, Reputation: 5
    New Member
     
    #1

    Nov 6, 2008, 03:27 PM
    Algebra II (system of equation word problems with 3 variables.)
    I have two problems that I don't understand, so if anybody can help, that would be wonderful.

    We are doing word problems with system equations with three variables
    THE SET UP:
    a.___variables__
    b.__equation 1__
    __equation 2__
    __equation 3__
    c.__x=,y=,z=___

    THIS IS WHAT I HAVE SO FAR; OBVIOUSLY NOT MUCH.

    1.)Mrs. Thompson has one-dollar, five-dollar, and ten-dollar bills that total $171. She has the same number of five-dollar bills as one-dollar bills and ten-dollar bills put together. If she has 30 bills in all how many bills of each denomination does she have?
    a.) ___x=$1, y=$5, z=$10___
    b.)__x+y+z=171___________
    _______________________
    _______________________
    c.)_______________________


    2.) John has 31 coins consisting of quarters,dimes, and nickles. If he had one more dime, he would have just as many dimes as he has nickles and quarters put together. If the total value of John's coins is $4.70, how many coins of each kind does he have?
    a.___q=quarters,d=dimes,n=nickles__
    b.__q+d+n=31____________________
    _______________________________
    __.25q+.1d+.05n=4.70____________
    c.________________________________


    PLEASE HELP!
    BTW what I have there I'm not exactly sure its right --- thanks.
    eeseely's Avatar
    eeseely Posts: 100, Reputation: 3
    Junior Member
     
    #2

    Nov 11, 2008, 06:34 PM
    Here's some help with your 2nd problem:

    If he had one more Dime, he would have just as many Dimes as he has Nickels and
    Quarters.

    Using N, D and Q for number of Nickels, Dimes and Quarters, this statement translates to:

    D = N+Q-1 (1)

    Also, N+D+Q = 31 (2)

    Substituting for D in (2) yields: N+(N+Q-1)+Q = 31

    2(N+Q) = 32 and N+Q = 16

    Then, from (1) above, D = 15

    5N+10D+25Q = 470

    5N+(10)(15)+25Q = 470

    5N+150+25Q = 470

    5N+25Q = 320

    N+5Q = 64 (3)

    From (1), N = 15-Q+1 = 16-Q

    Substituting for N in (3) yields:

    16-Q+ 5Q = 64

    4Q = 48

    Q = 12; D = 15; N = 4
    eeseely's Avatar
    eeseely Posts: 100, Reputation: 3
    Junior Member
     
    #3

    Nov 12, 2008, 08:00 AM

    Here's some help on your Problem 1.:

    Mrs. Thompson has one-dollar, five-dollar, and ten-dollar bills that total $171.

    $1x+$5y+$10z = $171 (1)

    She has the same number of five dollar bills as one dollar bills and ten dollar bills put together.
    This is represented by:

    y = x+z (2)

    If she has 30 bills in all, how many bills of each denomination does she have?

    x+y+z = 30 (3)

    Let x, y, z be the number of $1-, $5- and $10-bills, respectfully.

    Can you take it from there?
    ikryspy's Avatar
    ikryspy Posts: 24, Reputation: 5
    New Member
     
    #4

    Nov 12, 2008, 05:01 PM
    Thank you sooo! Much, I didn't think I was ever going to figure this out. Your help is really appreciated.

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