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Matrix Word Problems

  1. Sep 14, 2009 #1
    Here is my first problem --

    1. The problem statement, all variables and given/known data

    http://img9.imageshack.us/img9/253/picture15lw.png [Broken]

    2. Relevant equations

    I know that I have to set up a matrix and multiply in order to solve this problem. However, I'm having difficulty setting up a matrix.

    3. The attempt at a solution
    x - green squares
    y - blue squares
    z - floral squares

    First, I set up an equation dealing with the price.
    3x + 3y + 5z = 558

    Then, I set up an equation dealing with the number of squares.
    x + y + z = 180 (because 15 x 12 would be 180 squares)

    I know I'm doing something wrong, but I'm not sure what. Any help would be appreciated!

    Here is my second problem --

    1. The problem statement, all variables and given/known data

    http://img198.imageshack.us/img198/2711/picture16a.png [Broken]

    2. Relevant equations

    I have this one partially solved, with the correct equations but I don't know how to take it a step further.

    3. The attempt at a solution
    I know that the given variables need to be plugged in, but I'm not sure which equations to use or how to set them up. I have the first part of the problem, but I'm not sure how to apply the equations I have to the second part.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 14, 2009 #2
    For the first problem, you didn't set up an equation that corresponds to the information that "there must be 19 times as many solid squares as floral". That would give you three equations with three unknowns and should give you your matrix.

    For the second problem, it says that (x,y) is married/single solution for year 1, and (m,s) is your married/single solution for year 2. If they give you a year and then ask for the total for the previous year, which set of variables do you think you need to fill into? Also, there's something fishy with your answer - the question says there were 145,500 one year and you get only 33,000 the year before, which is a pretty big jump. Are you sure you filled it in right?
     
  4. Sep 14, 2009 #3
    1)Not really sure you need a matrix for this problem, but I'll show you what i got for the answer:
    like you said for the price:
    3x+3y+5z=558
    and for the total squares: x+y+z=180

    However you forgot that there must be 19 times more floral than solid, meaning:
    (x+y)=19z

    Honestly the number of greens vs blues is insignificant, since they are the same price and there is no specification on how many of each you need. So i combined all my x+y's into one symbol, let's call it w.

    So, now we have
    1. 3w+5z=558
    2. w=19z
    3. w+z=180

    Now, you can easily substitute the w=19z (equation 2) into the w+z=180 (equation 3)
    Making equation 3 look like this:
    19z + z=180
    Add the numbers to get:
    20z=180, and thus z=9
    Now to find out the number of non-floral squares, plug the z=9 into the original equation 3. So:
    w+9=180, making w (the number of colored squares) 171.

    To check: plug the numbers into the price equation (equation 1)
    3(171) + 5(9)=558, so we got the numbers right.
    **You can pick whatever numbers you want for the solids, cause they all cost the same price so it really doesn't make a difference**
     
  5. Sep 14, 2009 #4
    Thanks for your help! I got the first problem, but I'm still having trouble with the second. I keep trying different things, and it's not coming out right. Also, the 33,100 is not the right answer in the picture.
     
  6. Sep 14, 2009 #5
    Alright, you are given a system of equations, all you have to do is solve one of them for one of the variables, then plug it into the other to find the numerical value of that variable.

    So there are 145,500 married and 34,500 single, so your equations would look like this:
    .9x+.4y=145,000
    .1x+.6y=34,500

    so, I solved the 2nd equation for x and i got (after subtracting .6y and dividing the whole right side by .1)
    x=345,000-6y

    Now plug that into the first equation, which results in:
    .9(345,000-6y)+.4y=145,500
    Which simplifies to:
    310500-5.4y+.4y= 145,500
    Put all the constants on the left side of the equation, and put all the y-values on the right side of the equation, which yields:
    165,000=5y
    y=33,000 (the # of single people)
    Now plug that y-value back into one of the original equations to get the x-value, which will be the number of married people.
     
  7. Sep 14, 2009 #6
    Thanks! I understand that part now. Do you know what I'd have to do to find out how many married/single people there were 2 years ago?
     
  8. Sep 14, 2009 #7
    Take those numbers that you got for x(married) and y(single), and make those the m and s values. Then do the exact same process that you did to find the one year previous. It is exactly the same except your equations are equal to different numbers.

    So your equations would look like this:
    .9x+.4y=(whatever y-value you got for the last equation)
    .1x+.6y=33,000 (the x-value for the last equation)

    The trick to these is solving one of the equations for either x or y then substituting it into the other equation. The rest is just simple algebra.
     
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