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Math Help - Simultaneous equations

  1. #1
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    Post Simultaneous equations

    Hello, I've a problem with equations. It's very hard to me. Could u give the solutions with some steps?10(b^3)=8c+2(a^5)10(c^3)=8a+2(b^5)10(a^3)=8b +2(c^5)I think, that:x,y,z=0x,y,z=1but what's next?x=? y=? z=?I need help.Thanks in advance
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Helloworld View Post
    Hello, I've a problem with equations. It's very hard to me. Could u give the solutions with some steps?10(b^3)=8c+2(a^5)10(c^3)=8a+2(b^5)10(a^3)=8b +2(c^5)I think, that:x,y,z=0x,y,z=1but what's next?x=? y=? z=?I need help.Thanks in advance
    Let's write this out a little more intelligibly:
    10b^3=8c+20a^5c^3=8a+20a^3b^5=8b+2c^5

    I suppose as a first step I'd write out:
    10b^3 = 8c + 20a^5c^3 \implies 5b^3 = 4c + 10a^5c^3

    10b^3 = 8a + 20a^3b^5 \implies 5b^3 = 4a + 10a^3b^5

    10b^3 = 8b + 2c^5 \implies 5b^3 = 4b + c^5

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    5b^3 = 4c + 10a^5c^3

    5b^3 = 4a + 10a^3b^5

    5b^3 = 4b + c^5
    I presume a, b, and c are to be positive integers.

    I can't finish this for you, but I will note a couple of points that you might be able to put together.

    The first equation says that
    5b^3 = c(4 + 10a^5c^2)

    1) So 5b^3 must be divisible by c. Similarly from the second equation 5b^3 must be divisible by a.

    2) The RHS of both the first and second equations is divisible by 2. Thus b is divisible by 2.

    3) Using 2), we see from the third equation that c is also divisible by 2.

    (So note that your solution of a = 1, b = 1, and c = 1 must be incorrect.)

    See if these observations get you anywhere.

    -Dan
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  4. #4
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    Quote Originally Posted by Helloworld View Post
    Hello, I've a problem with equations. It's very hard to me. Could u give the solutions with some steps?10(b^3)=8c+2(a^5)10(c^3)=8a+2(b^5)10(a^3)=8b +2(c^5)I think, that:x,y,z=0x,y,z=1but what's next?x=? y=? z=?I need help.Thanks in advance
    Hello,

    do you mean

    \begin{array}{l}10(b^3)=8c+2(a^5) \\ 10(c^3)=8a+2(b^5) \\ 10(a^3)=8b+2(c^5) \end{array}

    Where does the x, y, z come from?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by earboth View Post
    Hello,

    do you mean

    \begin{array}{l}10(b^3)=8c+2(a^5) \\ 10(c^3)=8a+2(b^5) \\ 10(a^3)=8b+2(c^5) \end{array}

    Where does the x, y, z come from?
    Ahhhh...That makes more sense...

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by earboth View Post
    Hello,

    do you mean

    \begin{array}{l}10(b^3)=8c+2(a^5) \\ 10(c^3)=8a+2(b^5) \\ 10(a^3)=8b+2(c^5) \end{array}
    So again we may divide each of these by 2:
    \begin{array}{l}5b^3=4c+a^5 \\ 5c^3=4a+b^5 \\ 5a^3=4b+c^5 \end{array}
    I guess this would explain your (now valid) a = 1, b = 1, c = 1 solution.

    It looks to me like:
    5b^3=4c+a^5 \implies 4c = 5b^3 - a^5
    implies that a and b are either both even or both odd.

    Similarly the second equation implies that b and c are either both even or both odd.

    Similarly the third equation, etc.

    So it looks like either all of a, b, and c are even or all of a, b, and c are odd.

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    So again we may divide each of these by 2:
    \begin{array}{l}5b^3=4c+a^5 \\ 5c^3=4a+b^5 \\ 5a^3=4b+c^5 \end{array}
    I guess this would explain your (now valid) a = 1, b = 1, c = 1 solution.

    It looks to me like:
    5b^3=4c+a^5 \implies 4c = 5b^3 - a^5
    implies that a and b are either both even or both odd.

    Similarly the second equation implies that b and c are either both even or both odd.

    Similarly the third equation, etc.

    So it looks like either all of a, b, and c are even or all of a, b, and c are odd.

    -Dan
    OK, u are right. now is ok, I should divided it by 2.And what I should do now? I know, that a=b=c, but how can I check that?
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  8. #8
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    ok, now I know how can I do that Topic should be closed
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Helloworld View Post
    Topic should be closed
    click on "Thread Tools" just above your first post and then "Mark thread as solved"
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