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Math Help - Another word problem (systems of linear equations)

  1. #1
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    Another word problem (systems of linear equations)

    im not asking anyone to do the problem for me, just help me out get me on the right foot. im very bad with word problems.

    heres the problem: ellen wishes to mix candy worth 1.80 per lb with candy worth 2.40 per lb to form 48 lbs of a mixture worth $2.00 per lb. how many lbs of the more expensive candy should she use?

    if you can help me write out the problem and explain how you got it that way i should be fine solving the rest of the problem on my own.

    thank you very much for any and all help
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by NecroWinter View Post
    im not asking anyone to do the problem for me, just help me out get me on the right foot. im very bad with word problems.

    heres the problem: ellen wishes to mix candy worth 1.80 per lb with candy worth 2.40 per lb to form 48 lbs of a mixture worth $2.00 per lb. how many lbs of the more expensive candy should she use?

    if you can help me write out the problem and explain how you got it that way i should be fine solving the rest of the problem on my own.

    thank you very much for any and all help
    Again, you are searching for pounds, so let them be variables. let x be the # of lbs of the cheaper candy used, and y be the # of lbs of the more expensive candy.

    the total is 48 lbs, so: x + y = 48

    the cost of each is 1.8x and 2.4y respectively (do you see why?)

    and you want the total cost to be, 2*48 = 96 lbs, hence, you want 1.8x + 2.4y = 96

    so you get the simultaneous system of equations

    x + y = 48 .......................(1)
    1.8x + 2.4y = 96 ...............(2)

    Solve this system for y
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  3. #3
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    Quote Originally Posted by NecroWinter View Post
    im not asking anyone to do the problem for me, just help me out get me on the right foot. im very bad with word problems.

    heres the problem: ellen wishes to mix candy worth 1.80 per lb with candy worth 2.40 per lb to form 48 lbs of a mixture worth $2.00 per lb. how many lbs of the more expensive candy should she use?

    if you can help me write out the problem and explain how you got it that way i should be fine solving the rest of the problem on my own.

    thank you very much for any and all help
    let x = number of pounds of cheaper candy

    y = number of pounds of more expensive candy

    x + y = 48

    x(1.80) + y(2.40) = 48(2.00)

    solve the system for x and y ... the question is asking for the value of y.
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  4. #4
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    thank you all very much, i solved it and got the correct answer i was looking for

    i was not thinking about the 2*48 = 96 part, and that was the problem i encountered. thanks for explaining it and ill keep this in mind for future issues.
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