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  • Algebra problem "HELP"

    I need help with an algebra problem, I can't figure it out and I've called at least 10 friends with no success, can someone on here figure it out, because I can't, so here it is;

    1/x + 5/y = 2/z , solve for y

    and please show me how you got the problem solved in dummy terms, I've tried it several different ways with no success.
    Thanks, for any help that you can give

  • #2
    ummm 136
    Last edited by Truckweld; 05-16-2005, 11:11 PM.

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    • #3
      At this point, I'm baffled, confused, and dazed,
      all at the same time.

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      • #4
        its so easy....
        x=2 y=10 and z=2

        1/2 + 5/10 = 1
        and 2/2 = 1

        so 1/2 + 5/10 = 2/2

        come on give me a harder math question
        Last edited by 98sahara; 05-16-2005, 11:31 PM.

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        • #5
          Originally posted by 98sahara
          its so easy....
          x=2 y=10 and z=2

          1/2 + 5/10 = 1
          and 2/2 = 1

          so 1/2 + 5/10 = 2/2

          come on give me a harder math question
          I don't know if that is exactly what he was looking for because x=1, y=5, z=1 fits also. This is solving for Y. I think


          Y = 5 / (2/z - 1/x)
          Last edited by KCK; 05-17-2005, 12:19 AM.

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          • #6
            "no tv, no sleep, just studying make me go something something"

            'go crazy'

            "don't mind if i Do... waka waka zoup swip zonks bonkers

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            • #7
              sorry but x,y, and z arent on my calculator!

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              • #8
                I've sworn off math after my Differential Equations final, but I'll see what I can do...

                If you want to solve for 'y' in terms of 'x' and 'z' :

                Original- 1/x + 5/y = 2/z
                1st- 5/y = 2/z - 1/x
                2nd- 5 = y * ( 2/z - 1/x )
                3rd- y = 5 / ( 2/z - 1/x )

                I'm sure there is a way to reduce that to something a bit less ugly, but my brain is fried, and I've never been good at common denominators. Maybe someone else can do that for you.

                -Josh

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                • #9
                  Originally posted by KCK
                  I don't know if that is exactly what he was looking for because x=1, y=5, z=1 fits also. This is solving for Y. I think


                  Y = 5 / (2/z - 1/x)
                  Close... maybe...??? I get... (THE FOLLOWING EQUATION IS WRONG... see posts below)

                  Y = (2/Z - 1/X) /5

                  of course depending on the level of math you are at this may not be the final answer... now you could substitute in this "value" for Y into the original equation to get a new equation giving you the following equation (once simplified some)...

                  Z^2 - XZ + 1/(4X^2) = 1/4

                  Often you would be given another equation (or two) referred to as a "system of equations" that you would use to solve for actual numeric values for the variables. As others have found... there are multiple possible "actual numeric values" and without further information these values can not be determined...

                  ok so at least that is how I remember algebra... which could be really wrong since it has been more than 15 years since I last did this stuff...

                  by the way... if you find the right answer... let us know...
                  Last edited by jeffm3434; 05-17-2005, 03:44 AM.

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                  • #10
                    jeff-

                    if you work your answer backwards

                    Y = (2/Z - 1/X) /5
                    5y = 2/z - 1/x
                    1/x + 5y = 2/z

                    that doesn't get the original equation. I stand by my answer (agree w/ KCK).

                    I do agree that depending on the level of math and/or teacher preference, that answer may not be sufficient. However, without values for 'x' and 'z' any explicit answer would be arbitrary anyway (i.e. everyone would have a different answer because you can put any number in for 'x' and 'z')

                    one possible solution:
                    x=2 z=1
                    y = 5 / (2/1 - 1/2)
                    y = 5 / (3/2)
                    y = 10/3
                    Last edited by NoneMoreBlk; 05-17-2005, 03:29 AM.

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                    • #11
                      Originally posted by NoneMoreBlk
                      jeff-

                      if you work your answer backwards

                      Y = (2/Z - 1/X) /5
                      5y = 2/z - 1/x
                      1/x + 5y = 2/z

                      that doesn't get the original equation. I stand by my answer (agree w/ KCK).

                      I do agree that depending on the level of math and/or teacher preference, that answer may not be sufficient. However, without values for 'x' and 'z' any explicit answer would be arbitrary anyway (i.e. everyone would have a different answer because you can put any number in for 'x' and 'z')

                      one possible solution:
                      x=2 z=1
                      y = 5 / (2/1 - 1/2)
                      y = 5 / (3/2)
                      y = 10/3

                      You are right and you caught me... I had just turned off my computer when I realized my mistake and was coming back to correct my mistake and by the time I got back to this topic you had me busted ... like I said... it has been way too long. Which also means my follow-up equation in my original post is bunk. My error was that when I multiplied 5/Y by 1/5 I incorrectly made that Y instead of 1/Y. My next step should have been to invert both sides yielding the same answer as you and KCK. And as you (and I alluded to)... this is the most correct answer that can be given. There are several possible actual numeric values.
                      Last edited by jeffm3434; 05-17-2005, 03:46 AM.

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                      • #12
                        RU
                        -- = Y
                        18
                        Last edited by FUBAR; 05-17-2005, 09:20 AM.

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                        • #13
                          Here's what my instructor finally ended up with after a bit of funny looks and several erasures

                          1/x + 5/y = 2/z
                          to get rid of the denominators you multiply everything by the denominators.
                          (xyz)1/x + (xyz)5/y = (xyz)2/z
                          the x's cross out in the first one leaving; yz
                          then it's + (xyz)5/y, which leaves; 5xz
                          and that = (xyz)2/z, which leaves it at; 2xy
                          which goes to y(z-2x)/z-2x = -5xz/z-2x
                          which comes down to y= -5xz/z-2x

                          now I'm not saying this is correct,
                          but it's what my Algebra Instructor got for his final answer, and he also stated there would not be a problem like this on his test either.
                          Thanks, for helping

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                          • #14
                            That looks like a much more efficient way to do it, and much prettier too. Like I said, I'm not too great with common denominators.

                            Glad I could help though, it was kinda fun not having to use some crazy matrix and 2nd order derivatives.

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                            • #15
                              y'all think way too much,

                              got waaaaaaaaaay too much free time.....




                              So please, explain to me, where, in day to day dealings with your Jobs, friends or WHEELING, where the fawk your'e gonna use that crap!

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