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Math Help - systems of equations (word problem)

  1. #1
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    systems of equations (word problem)

    Tasty Bakery sells three kinds of muffins: chocolate chip muffins at 40 cents each, oatmeal muffins at 45 cents each, and cranberry muffins at 50 cents each. Charles buys some of each kind and chooses three times as many cranberry muffins as chocolate chip muffins. If he spends $8.85 on 19 muffins, how many chocolate chip muffins does he buy?
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  2. #2
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    let x be the number of choc-chip muffins, y be the number of oatmeal muffins and z be the number of cranberry muffins.
    you know:
    x+y+z = 19
    z = 3x
    .4x+.45y+.5z=8.85

    Can you get it from here?
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  3. #3
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    Quote Originally Posted by badgerigar View Post
    let x be the number of choc-chip muffins, y be the number of oatmeal muffins and z be the number of cranberry muffins.
    you know:
    x+y+z = 19
    z = 3x
    .4x+.45y+.5z=8.85

    Can you get it from here?
    I got that far as a matter of fact, but I just couldn't solve it.

    4x + y + 0z = 19

    0.4x + 0.45y + 0.50z = 8.85

    I tried using rref method in TI-84, where you enter the equations as a matrix and it didn't work. I am use to solving 3 variable equations when there are 3 equations, not two.
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  4. #4
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    Quote Originally Posted by Power View Post
    I got that far as a matter of fact, but I just couldn't solve it.

    4x + y + 0z = 19

    0.4x + 0.45y + 0.50z = 8.85

    I tried using rref method in TI-84, where you enter the equations as a matrix and it didn't work. I am use to solving 3 variable equations when there are 3 equations, not two.
    You're working with the wrong equations. They have been clearly given:

    x+y+z = 19
    z = 3x
    .4x+.45y+.5z=8.85

    The last, I'd multiply by 100 and simplify if solving the system by hand:

    8x + 9y + 10z = 177.

    So the system is:

    x + y + z = 19 .... (1)

    z = 3x .... (2)

    8x + 9y + 10z = 177 .... (3)

    Sub (2) into (1) and (3) to get two equations in z and y:

    4x + y = 19 .... (1')

    38x + 9y = 177 .... (3')

    Solve these two simultaneously for x and y. Elimination method is probably simplest - (3') - 9x(1'):

    2x = 6 => x = 3.

    Therefore y = .....

    Therefore z = .....
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  5. #5
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    Yup, now I got it, I didnt know that z = 3x was one of the three equations. Now I can have set of three equations and use rref to solve faster.

    x = 3
    y = 7
    z = 9

    Thanks
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  6. #6
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    Quote Originally Posted by Power View Post
    Yup, now I got it, I didnt know that z = 3x was one of the three equations. Now I can have set of three equations and use rref to solve faster.

    x = 3
    y = 7
    z = 9

    Thanks
    A word to the wise ...... You should make sure you can solve them by-hand as well as using your TI-84 .... Otherwise you might experience a power failure down the track .......
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  7. #7
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    Thanks for that advice mate, I know that in future (on college tests) I am going to have to do stuff w/o a calculator but I am just not good at mental math kind of stuff, especially when it comes to fractions. Like i dont know how to add/subtract/multiply/divide fractions...if you find a goood site to teach, then let me know.

    Tuesday is my exam, so I am planning to do it all on my calculator for now. But I will try to avoid it..
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