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Math Help - Word Problem And System

  1. #1
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    Word Problem And System

    I need help on these two I can get some of the way on 1st but can't seem to get 2nd

    1. You have a 2 digit number
    a. Divide the Number by product of 2 digits get 1 remainder 16
    b. Diffrence of digits squared added to Product of digits give you number
    Find the number

    2. Solve System
    2x^2+y+3xy=12
    (x+y)^2-1/2y^2=7
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  2. #2
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    Quote Originally Posted by AHDDM View Post
    I need help on these two I can get some of the way on 1st but can't seem to get 2nd

    1. You have a 2 digit number
    a. Divide the Number by product of 2 digits get 1 remainder 16
    b. Diffrence of digits squared added to Product of digits give you number
    Find the number

    2. Solve System
    2x^2+y+3xy=12
    (x+y)^2-1/2y^2=7
    I can only solve the first. The second, I cannot (yet)---either I'm still not good at Math or the second is posted wrongly.

    The first:
    1. 10x +y

    a. (10x +y)/(xy) = 1 +16/(xy). Mutiply both sides by xy,
    10x +y = xy +16 -------------(1)

    b. I tried first x is greater than y and I didn't go anywhere, so, let y > x.
    (y-x)^2 +xy = 10x +y ----------(2)

    Eliminate the xy.
    xy from (1) is (10x +y -16), substitute into (2):
    (y-x)^2 +10x +y -16 = 10x +y
    (y-x)^2 = 16
    y-x = 4
    y = x+4 ------------(3), substitute in, say, (1):
    10x +x+4 = x(x+4) +16
    11x +4 = x^2 +4x +16
    x^2 -7x +12 = 0
    Factoring that,
    (x-3)(x-4) = 0
    x = 3 or 4

    When x = 3:
    y = x+4 = 7
    So number is 10(3) +7 = 37 -----**

    When x = 4:
    y = x+4 = 8
    So number is 10(4) +8 = 48 -----**

    Check 37 against Eq.(2):
    (y-x)^2 +xy = 10x +y ----------(2)
    (7-3)^2 +(3)(7) =? 10(3) +7
    16 +21 =? 30 +7
    37 =? 37
    Yes, so, OK

    Check 48 against Eq.(2):
    (y-x)^2 +xy = 10x +y ----------(2)
    (8-4)^2 +(4)(8) =? 10(4) +8
    16 +32 =? 40 +8
    48 =? 48
    Yes, so, OK

    [Check both 37 and 48 against Eq.(1) if you like.]

    Therefore, the number is 37 or 48. ---------------answer.
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  3. #3
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    I checked 1 with yours works thank you

    I just checked if I wrote 2 correctly and it is.
    I used a calculator to solve system and I got but don't know how it got it
    Last edited by AHDDM; August 15th 2007 at 12:21 PM.
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  4. #4
    Forum Admin topsquark's Avatar
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    This is the intersection of two hyperbolas (hyperboli?), see the graph below. Each hyperbola has an axis rotated at an unfriendly angle. It's the xy cross term that gives the majority of the problems and I have been unable to find a transform that will eliminate it from both equations at the same time. I can set up two different quartic equations to get the x values of the intersection points, and use them together to reduce the problem to a cubic, but none of the polynomials have nice zeros.

    I, like ticbol, have run out of tricks. It might be possible to do analytically, but every method I have tried just get butt-ugly very quickly.

    AHDDM, does your professor not like your class or something??

    -Dan
    Attached Thumbnails Attached Thumbnails Word Problem And System-intersection.jpg  
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  5. #5
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    Hello, AHDDM!

    I see no easy way to solve #2.

    Is there a typo?
    If both equations have y^2, they are solvable ... quite nicely.


    2)\;\;\begin{array}{cccc}2x^2+3xy+{\color{red}y^2} & = & 12 & [1]\\<br />
(x+y)^2-\frac{1}{2}y^2 & = & 7 & [2]\end{array}


    \begin{array}{cccc}\text{Simplify [2]:} & 2x^2 + 4xy + y^2 & = & 14\\<br />
\text{Subtract [1]:} & 2x^2 + 3xy + y^2 & = & 12\end{array}

    And we have: . xy \:=\:2\quad\Rightarrow\quad y \:=\:\frac{2}{x}

    Substitute into [1]: . 2x^2 + 3x\left(\frac{2}{x}\right) + \left(\frac{2}{x}\right)^2 \:=\:12\quad\Rightarrow\quad2x^2 + 6 + \frac{4}{x^2} \:=\:12

    . . 2x^4 + 6x^2 + 4 \:=\:12x^2\quad\Rightarrow\quad x^4 -3x^2 + 2 \:=\:0\quad\Rightarrow\quad(x^2-1)(x^2-2)\:=\:0

    And we have: . \begin{array}{ccccc}x^2-1\:=\:0 & \Rightarrow & x^2\:=\:1 & \Rightarrow & x \:=\:\pm1 \\<br />
x^2-2\:=\:0 & \Rightarrow & x^2\:=\:2 & \Rightarrow & x \:=\:\pm\sqrt{2}\end{array}

    . . and the corresponding y-values are: . \begin{array}{c}y \:=\:\pm2 \\ y \:=\:\pm\sqrt{2}\end{array}


    There are four solutions: . \left(\pm1,\,\pm2\right),\;\left(\pm\sqrt{2},\,\pm  \sqrt{2}\right)

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  6. #6
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    I'll see tommrow I'll see if I wrote problem down wrong. I think the way that Soroban did is correct sorry for the trouble.

    PS Topquark He does hate us
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  7. #7
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    Quote Originally Posted by AHDDM View Post
    I'll see tommrow I'll see if I wrote problem down wrong. I think the way that Soroban did is correct sorry for the trouble.

    PS Topquark He does hate us
    What? Then why not give him a frog in a box for 5 days? The boxes signed by you all.

    If He is a she, then give her a harmless snake in the box. For 3 days only. Female teachers have weaker hearts.
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