Results 1 to 5 of 5

Math Help - Number and alegbra

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    12

    Number and alegbra

    Jane has 15 coins which include $2-coins and $5-coins only. If the total value of $2-coins is less than that of $5-coins by $33, how many $5-coins does Jane have?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    461
    Hello

    Quote Originally Posted by econlover View Post
    Jane has 15 coins which include $2-coins and $5-coins only. If the total value of $2-coins is less than that of $5-coins by $33, how many $5-coins does Jane have?
    let x be the amount of $2-coins

    y the amount of $5-coins

    Jane has 15 coins, therefor
    x+y = 15

    Suppose you have x=6 $2-coins, then its worth is 6*2 =$12, and you have y=9 coins (actually this is the solution) then you have 9*5 = $45. You know the difference between the worth of all your $2 coins and 5$-coins is 33$.
    Hence
    2x = 5y-33
    (2 and 5 because of the worth of x and y)

    You need to solve the equations
    x+y = 15
    2x = 5y-33

    Regards,
    Rapha
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2009
    Posts
    12
    Thanks for the above explanation Rapha. I do have the same equations with yours but I am stuck on how to take the next step through these equations
    x+y=15
    2x+33=5y
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    Posts
    461
    Quote Originally Posted by econlover View Post
    Thanks for the above explanation Rapha.
    You're welcome

    Quote Originally Posted by econlover View Post
    I do have the same equations with yours but I am stuck on how to take the next step through these equations
    x+y=15
    2x+33=5y
    this is the first equation: x+y=15, solving for x => x = 15-y

    now we know x, therefor:

    2(15-y)+33 = 5y

    and calculate this

    30 - 2y +33 = 5y // +2y
    63 = 7y // :7

    63/7 = y

    9 = y

    Now our x depends on y, because we got x = 15-y. We know y exactly and thus

    x = 15-9 = 6

    Done.

    Rapha
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2009
    Posts
    12
    Ah...thanks Rapha! I finally got it !
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Are these Alegbra Equations Prime?
    Posted in the Algebra Forum
    Replies: 8
    Last Post: April 25th 2010, 11:07 AM
  2. Alegbra
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 17th 2009, 03:15 PM
  3. alegbra
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 20th 2008, 08:04 AM
  4. [SOLVED] Please Help!!Alegbra
    Posted in the Algebra Forum
    Replies: 5
    Last Post: October 12th 2007, 03:09 PM
  5. Alegbra
    Posted in the Algebra Forum
    Replies: 8
    Last Post: February 21st 2006, 02:00 PM

Search Tags


/mathhelpforum @mathhelpforum