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I am trying to help my daughter with her Algebra... PLEASE HELP?

Please don't give us the answer if you know it, but please give us an idea of how to set this problem up.

We sold gift wrap. $4 for solid $6 for print. Total rolls sold was 480

we collected $2340. How many rolls of each kind of gift wrap were sold?

Thanks,

8 Answers

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  • 1 decade ago
    Favorite Answer

    Hi,

    S = # of solid

    P = # of print

    S + P = 480

    4S + 6P = 2340

    Multiply the first equation by -4 and add the equations to solve.

    -4(S + P = 480)

    4S + 6P = 2340

    -4S - 4P = -1920

    4S + 6P = 2340

    ------------------------

    2P = 420

    P = 210

    If P = 210 and S + P = 480, then S = 270

    270 rolls of solid and 210 rolls of print <==ANSWER

    I hope that helps!! :-)

  • Jeff A
    Lv 4
    1 decade ago

    Okay first you have to set up some equations that are hidden in the word problem:

    $4 for each roll of solid gift wrap

    we can define x to be the number of solid gift wrap rolls sold

    $6 for each roll of printed gift wrap

    we can define 480 - x to be the number of of printed rolls of gift wrap sold since 480 rolls were sold altogther, and if x of them were solid, the rest must be printed.

    that gives us the equation

    4x+6(480-x) = 2340 since 2340 is the total cost of money brought in by selling wrapping paper

    no we can simplify to get:

    4x+6*480-6x = 2340

    6*480-2x=2340

    and just continue to solve for x. After you solve for x make sure you look at the question again to see if you answered it, always do this! So in this problem they don't care what x is, after all we made that up. They want to know how much of each type of wrapping paper was sold, so x was how much solid was sold and 480-x was how many printed rolls were sold.

    good luck!

  • Anonymous
    1 decade ago

    let a be the number of $4 rolls, and b be the number of $6 rolls

    a + b = 480--> the # of 4 dollar rolls plus the # of 6 dollar rolls = 480

    4a+6b = 2340 --> the total cost of the rolls is $2340

    rearrange the first equation a + b = 480 to a = 480 -b

    substitute a = 480 -b into 4a+6b = 2340

    4a+6b = 2340

    4(480 - b) + 6b = 2340

    1920 - 4b + 6b = 2340

    2b = 2340 - 1920

    2b = 420

    b = 420/2 = 210

    substitute b= 210 into any equation above, the most easy one to do would be a = 480 -b

    a = 480 -b

    a = 480 - 210

    a = 270

    so there are 270 $4 rolls, and 210 $6 rolls

    Source(s): grade 11 student
  • 1 decade ago

    Let x = the number of solid

    Let y = the number of print

    If you sold 480, then x + y = 480. (Solids + Print)

    If the solid cost $4 and the print $6, then solid part of the money would be expressed as 4x ($4 times the number you sold, x)

    Same for the print ones, that's 6y. So if you add the money together, 4x + 6y = 2340.

    Solve for one variable or the other.

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  • 1 decade ago

    First let solid rolls = x and print =y

    then from there you get two equations

    4x + 6y = 2340

    x + y = 480

    from there you can solve for x (or y) then when you find x plug your answer into the second equation where you will find y.

  • Anonymous
    1 decade ago

    System of equations:

    4s + 6p = 2340

    s + p = 480

    s stands for solid

    p stands for print

    (Trust you know how to solve it)

  • 1 decade ago

    you can use a system to answer this.

    x+y=480

    4x+6y=2340

    multiply the top by -4 to eliminate the x's so you have

    -4x-4y= -1920

    4x+6y=2340

    so you add those two and you get

    2y=420

    divide by 2 and y=210

    then substitute back in one of the equations, i.e.

    x+210=480 so x=270

  • Anonymous
    1 decade ago

    use a system of equations

    4x + 6y=2340

    x + y = 480

    Source(s): I'm in pre-ap precalculus
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