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Math Help - Please help me with the solution for this problem

  1. #1
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    Question Please help me with the solution for this problem

    The sum of 2 numbers is 36. If the larger number is divided by the smaller number, the
    qoutient is 2 and the remainder is 3. Find the numbers.

    Anwers: 20 and 11

    Can you show me the solution for this problem.........
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  2. #2
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    By the way the solution you posted cannot be correct since 11 + 20 = 31 not 36.
    Here is how to solve the problem.
    x + y = 36

    Let's assume x\ge y then
    x = 2y + 3
    Which come from
    Quote Originally Posted by vendonz View Post
    If the larger number is divided by the smaller number, the
    qoutient is 2 and the remainder is 3.
    Now, we can substitute 2y + 3 in for x which give
    2y + 3 + y = 36 \Rightarrow 3y = 33 \Rightarrow y = 11
    Finally, substitute this back in
    x = 2\cdot 11 + 3 = 25

    Answer: 25 and 11
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  3. #3
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    Yeah sir your right its 25 and 11 sorry my fault ,, can you explain it to me more simple im a slow learner ,, can you use x as our substitution
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  4. #4
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    hehehe ,, i got it now ^_^ ,, how about this sir i confuse to this, Solve for x:
    x-3-2(6-2x) = 2(2x-5)

    I Got this ;
    x-3-2(6-2x) = 2(2x-5)
    x-3-2(6-2x) = 4x-10
    x(6-2x)-4x = -10+5
    6x-2x-4x = -5
    -1(-x = -5)
    x = 5
    is this right?? the answer for this is 5
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  5. #5
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    For the second problem it is best to multiply everything out first.
    x-3-2(6-2x) = 2(2x-5)
    x-3-12+4x = 4x-10
    Then get all the x's on one side and the number on the other.
    x + 4x - 4x = -10 + 3 + 12
    x + 4x - 4x = 5

    Now you got the correct answer, but you had made some mistakes. Specifically the following
    x-3-2(6-2x) = 4x-10
    x(6-2x)-4x = -10+5
    I am unsure how you got to that second line.
    But, then you made another mistake that corrected the first one
    6x-2x-4x = -5 would actually give 0x = -5 which obviously doesn't make since. Somehow you have -1(-x = -5).
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  6. #6
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    Question

    I solved that a while ago,, but thanks for the effort this is the problem that im solving for now can you help me again???

    The length of th rectangular swimming pool is twice its width. The pool is surrounded by by a cement walk 4 feet wide. If the area of the walk is 784 square feet. Find the dimension of the pool.

    This is asking for width and length
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  7. #7
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    I would suggest creating a new thread if you have any further questions.

    The length of th rectangular swimming pool is twice its width.
    L_1 = 2w_1

    The pool is surrounded by by a cement walk 4 feet wide.
    L_2 = L_1 + 8
    w_2 = w_1 + 8
    Since there is an extra four feet on all sides of the pool.

    The area of the walk is 784 square feet. Here The best thing to do is compare the area made by L_2\cdot w_2 to that of the area of the pool ( A_{pool} = L_1\cdot w_1). So we have
    L_2\cdot w_2 = 784 + A_{pool}
    L_2\cdot w_2 = 784 + L_1\cdot w_1
    \left(L_1 + 8\right)\left(w_1 + 8\right) = 784 + L_1\cdot w_1
    Now, you will want to substitute what we know for L_1 and then solve.
    I will let you complete the problem.
    Last edited by lvleph; July 25th 2010 at 09:14 PM.
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  8. #8
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    Thanks for the help ....
    I got Width: 60
    Length: 120
    Is this correct???
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  9. #9
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    Close. Substitute 2w_1 for L_1 and you get
    \left(2w_1 + 8\right)\left(w_1 + 8\right) = 784 + 2w_1\cdot w_1
    Now expand
    2w_1^2 + 8w_1 + 16w_1 + 64 = 784 + 2w_1^2
    Get all w_1 to one side and numbers to the other
    24w_1 = 720 \Rightarrow w_1 = 30
    Substitute the value of w_1 into the formula for L_1
    L_1 = 2w_1 = 2\cdot 30 = 60
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