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8th gr. Algebra

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I have 3 math problems for homework that are stumping me. Can anyone help guide me? Thanks! Here they are:

1. A full bucket of water weighs eight kilograms. If the water weighs five times as much as the empty bucket, how much does the water weigh?

2. Helen has twice as many dimes as nickels and five more quarters than nickels. The value of her coins is $4.75. How many dimes does she have?

3. Ly has three more dimes than nickels and twice as many quarters as dimes. The value of his coins is $9.60. How many of each kind of coin does he have?

  • 8th gr. Algebra -

    Interesting-- some of your problems are similar to the ones that are stumping me! Maybe you would know some of mine. They are a few ones below yours, under Algebra by Celina. I'll try to help you.

    #1, I can't help you with. For #2 and #3, I would choose one of the coins as a variable for your equation.

    In 2, I would definitely use nickels as your variable. Here is an equation you could use to solve it:
    x = nickels

    4.75 = 2x + (x + 5)

    I'm not positive that is the right equation or not, but it might be one you can try.

    Hope I helped!!! :)

  • 8th gr. Algebra -

    #1.
    Let B = weight of the empty bucket.
    Let x = weight of the water ===========================
    B + x = 8 kg
    Weight of water is 5 times the weight of the empty bucket; therefore, x = 5B.
    Substitute 5B for x and solve for B, then x. You can go through the algebra but I have 1.333 kg for the weight of the empty bucket and 6.667 kg for the weight of the water. Check my thinking. If you have trouble, post your work and we can find the math error. I'll work on the other two problems shortly.

  • 8th gr. Algebra -

    Let n = # nickels.
    Let d = # dimes.
    Let q = # quarters.
    =====================
    Helen has twice as many dimes as nickels; therefore, 2n=d
    Helen as five more quarters than nickels; therefore, n+5 = q
    Finally, the value is 4.75 if we add all of it together; so
    0.05n + 0.10d + 0.25q = 4.75.
    Substitute n+5 for q and substitute 2n for d and solve for n, then for d and q.
    Check my thinking. Post your work if you get stuck and we can find your math error. I have # nickels = 7.

  • 8th gr. Algebra -

    I don't get what you mean but substitute numbers for other numbers.

  • 8th gr. Algebra -

    Ok. So what I got for my equation was
    n+2n+(n+5)=4.75. Is that right

  • 8th gr. Algebra -

    #3 is the same type of problem and its worked the same way. I'll leave that for you after you have finished with #2.

  • 8th gr. Algebra -

    I am done with number 2. I kinda need the next equation

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