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  3. Math problem -- help/hints please (NOW SOLVED: includes solution)

Math problem -- help/hints please (NOW SOLVED: includes solution)

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  • S Sumal V

    Is the answer -1?

    K Offline
    K Offline
    Kevin Bewley
    wrote on last edited by
    #3

    Nope.. it's a fraction (I don't want to give the answer straight away here, as it's the flaw in my method I'm trying to spot) It's ONLY an A Level problem; you'd think at the age of 38 I'd find this stuff elementary. But no, I was smarter at maths when I was 18!?? :doh:

    1 Reply Last reply
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    • K Kevin Bewley

      Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!

      D Offline
      D Offline
      dan sh
      wrote on last edited by
      #4

      The bracket you have added to 2x + 1 is not correct.

      "Fear no factor", Prime Numbers.

      K 1 Reply Last reply
      0
      • K Kevin Bewley

        Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!

        J Offline
        J Offline
        J4amieC
        wrote on last edited by
        #5

        Have you plugged 1/10 back into the original equation to check whether you're right? Thats how I always remember checking my answers during A-Level maths.

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        • D dan sh

          The bracket you have added to 2x + 1 is not correct.

          "Fear no factor", Prime Numbers.

          K Offline
          K Offline
          Kevin Bewley
          wrote on last edited by
          #6

          Sadly not 7/11 - but you're so damn close to the real answer that you probably have done it correctly.. I'm not sure I understand the mocking Indians reference either. But no, I genuinely want to know what I'm doing wrong. I've done a whole chapter of these and got them all correct - it's just that this one is incorrect.

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          0
          • K Kevin Bewley

            Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!

            I Offline
            I Offline
            Ingo
            wrote on last edited by
            #7

            So, if I'm not wrong: (x-4)/6 - 2x+1 = (3x-4)/2 ! -1 x/6 - 4/6 - 2x = 3x/2 - 2 - 1 ! +4/6 x/6 - 2x = 3x/2 - 3 + 4/6 x/6 - 12x/6 = 9x/6 - 18/6 + 4/6 ! -9x/6 -11x6 - 9x/6 = -14/6 ! *(-1) 20x/6 = 14/6 !*6 20x = 14 ! /20 x = 14/20 (The ! stands for the pipe followed by the operation I did in that step) Hope that is correct. :rolleyes:

            ------------------------------ Author of Primary ROleplaying SysTem How do I take my coffee? Black as midnight on a moonless night. War doesn't determine who's right. War determines who's left.

            K 1 Reply Last reply
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            • K Kevin Bewley

              Sadly not 7/11 - but you're so damn close to the real answer that you probably have done it correctly.. I'm not sure I understand the mocking Indians reference either. But no, I genuinely want to know what I'm doing wrong. I've done a whole chapter of these and got them all correct - it's just that this one is incorrect.

              D Offline
              D Offline
              dan sh
              wrote on last edited by
              #8

              I was wrong in my first attempt. I have changed the reply now. 7/11 is a shop supposedly, mostly run by Indians in US (I have never been to US so don't know). This is often used in stand up comic shows for jokes. That's what I was referring to.

              "Fear no factor", Prime Numbers.

              1 Reply Last reply
              0
              • K Kevin Bewley

                Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!

                M Offline
                M Offline
                Manfred Rudolf Bihy
                wrote on last edited by
                #9

                x-4 + 6(-2x + 1) ... etc. The error came in when you introduced the parenthesis around the second term on the left hand side. It changed the meaning of negative sign to include the one, which is not correct if you look closely at the original. Cheers!

                "With sufficient thrust, pigs fly just fine."

                Ross Callon, The Twelve Networking Truths, RFC1925

                K 1 Reply Last reply
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                • K Kevin Bewley

                  Sadly not 7/11 - but you're so damn close to the real answer that you probably have done it correctly.. I'm not sure I understand the mocking Indians reference either. But no, I genuinely want to know what I'm doing wrong. I've done a whole chapter of these and got them all correct - it's just that this one is incorrect.

                  D Offline
                  D Offline
                  dan sh
                  wrote on last edited by
                  #10

                  Incorrect addition of bracket has changed the sign of 1 in the original equation which has changed your result.

                  "Fear no factor", Prime Numbers.

                  1 Reply Last reply
                  0
                  • K Kevin Bewley

                    Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!

                    V Offline
                    V Offline
                    V 0
                    wrote on last edited by
                    #11

                    I came to the same solution, but filling in the actual value doesn't equal indeed. However, make sure the answer in the book is correct, because that's not always the case either. :) PS: I'm curious too now, if you find out the error, please share ?

                    V.

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                    • M Manfred Rudolf Bihy

                      x-4 + 6(-2x + 1) ... etc. The error came in when you introduced the parenthesis around the second term on the left hand side. It changed the meaning of negative sign to include the one, which is not correct if you look closely at the original. Cheers!

                      "With sufficient thrust, pigs fly just fine."

                      Ross Callon, The Twelve Networking Truths, RFC1925

                      K Offline
                      K Offline
                      Kevin Bewley
                      wrote on last edited by
                      #12

                      Of course!! Thanks buddy! :-)

                      M 1 Reply Last reply
                      0
                      • I Ingo

                        So, if I'm not wrong: (x-4)/6 - 2x+1 = (3x-4)/2 ! -1 x/6 - 4/6 - 2x = 3x/2 - 2 - 1 ! +4/6 x/6 - 2x = 3x/2 - 3 + 4/6 x/6 - 12x/6 = 9x/6 - 18/6 + 4/6 ! -9x/6 -11x6 - 9x/6 = -14/6 ! *(-1) 20x/6 = 14/6 !*6 20x = 14 ! /20 x = 14/20 (The ! stands for the pipe followed by the operation I did in that step) Hope that is correct. :rolleyes:

                        ------------------------------ Author of Primary ROleplaying SysTem How do I take my coffee? Black as midnight on a moonless night. War doesn't determine who's right. War determines who's left.

                        K Offline
                        K Offline
                        Kevin Bewley
                        wrote on last edited by
                        #13

                        Bang on (although the 'book answer' is 7/10 - simplest fractional answer). I'm curious about your notation though, what's with the factorial -1?

                        I C 2 Replies Last reply
                        0
                        • K Kevin Bewley

                          Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!

                          D Offline
                          D Offline
                          Dalek Dave
                          wrote on last edited by
                          #14

                          I used a different method, reducing one side to x/x = 1 and hard calculating. I got 1/10th too.

                          --------------------------------- I will never again mention that I was the poster of the One Millionth Lounge Post, nor that it was complete drivel. Dalek Dave CCC Link[^]

                          J 1 Reply Last reply
                          0
                          • K Kevin Bewley

                            Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!

                            S Offline
                            S Offline
                            Sumal V
                            wrote on last edited by
                            #15

                            Is it just -2x + 1 ie, (-2x+1) or -(2x+1)? This makes a lot of difference...

                            1 Reply Last reply
                            0
                            • D Dalek Dave

                              I used a different method, reducing one side to x/x = 1 and hard calculating. I got 1/10th too.

                              --------------------------------- I will never again mention that I was the poster of the One Millionth Lounge Post, nor that it was complete drivel. Dalek Dave CCC Link[^]

                              J Offline
                              J Offline
                              J4amieC
                              wrote on last edited by
                              #16

                              You're slipping. I used a spreadsheet and brute forced it by checking every fraction from 0.1 to 3.0

                              D 1 Reply Last reply
                              0
                              • K Kevin Bewley

                                Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!

                                J Offline
                                J Offline
                                J4amieC
                                wrote on last edited by
                                #17

                                I can tell you the answer is 7/10 but not because I can remember how to solve this from A-Level maths, but because I long ago realised such things are futile, and excel can solve this much quicker than I can.

                                D 1 Reply Last reply
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                                • K Kevin Bewley

                                  Bang on (although the 'book answer' is 7/10 - simplest fractional answer). I'm curious about your notation though, what's with the factorial -1?

                                  I Offline
                                  I Offline
                                  Ingo
                                  wrote on last edited by
                                  #18

                                  Kevin Bewley wrote:

                                  I'm curious about your notation though, what's with the factorial -1?

                                  Oh, that should be a pipe, but I wasn't able to type it in the editor, so it's just the remark, what I did. Forgot to post that :rolleyes:

                                  ------------------------------ Author of Primary ROleplaying SysTem How do I take my coffee? Black as midnight on a moonless night. War doesn't determine who's right. War determines who's left.

                                  1 Reply Last reply
                                  0
                                  • K Kevin Bewley

                                    Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!

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                                    S Offline
                                    Sumal V
                                    wrote on last edited by
                                    #19

                                    Following your same method, but only changing the brackets to 2x-1, 1(x-4) + 6(-2x+1) = 3(3x-4) multiply out the brackets: x - 4 - 12x + 6 = 9x - 12 which yields: 20x = 14 or x = 7/10

                                    1 Reply Last reply
                                    0
                                    • K Kevin Bewley

                                      Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!

                                      J Offline
                                      J Offline
                                      Jonathan Nethercott
                                      wrote on last edited by
                                      #20

                                      Other people have pointed this out already (so I'm not going to take the credit for solving this), but trying to stick to your calculations as closely as possible:

                                      (x-4)/6 - 2x+1 = (3x-4)/2

                                      My approach was to find the lcm, which is 6 and then do this:

                                      1(x-4) - 6(2x-1) = 3(3x-4) <-- note 2x-1, not 2x+1 because previous line is -2x + 1, not - (2x + 1)

                                      multiply out the brackets:

                                      x-4 - 12x+6 = 9x-12 <-- from previous line 12x+6, not 12x-6

                                      which yields:
                                      20x = 14
                                      or x = 7/10

                                      Jon CodeWrite

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                                      • J J4amieC

                                        You're slipping. I used a spreadsheet and brute forced it by checking every fraction from 0.1 to 3.0

                                        D Offline
                                        D Offline
                                        Dalek Dave
                                        wrote on last edited by
                                        #21

                                        Piece of paper and a pen!

                                        --------------------------------- I will never again mention that I was the poster of the One Millionth Lounge Post, nor that it was complete drivel. Dalek Dave CCC Link[^]

                                        1 Reply Last reply
                                        0
                                        • K Kevin Bewley

                                          Okay, so I've dug out my old A-level maths text to try and brush up on my skills. I've managed to struggle through the 1st chapter and associated exercises but have one question that I've got wrong and I can't for the life of me figure out where I'm going wrong. So, if any of you guys can help point me in the right direction I'd be really grateful: So I have the following I need to solve for x: (x-4)/6 - 2x+1 = (3x-4)/2 My approach was to find the lcm, which is 6 and then do this: 1(x-4) - 6(2x+1) = 3(3x-4) multiply out the brackets: x-4 - 12x-6 = 9x-12 which yields: 20x = 2 or x = 1/10 However, the back of the book says I'm wrong. It's really bugging me - I feel I'm falling at first step as I'm pretty confident once I've removed the fractional component that I'm doing everything correctly. Thanks guys, Kev Correct Solution follows: (x-4)/6 - 2x+1 = (3x-4)/2 lcm is 6 so multiply out the terms: 1(x-4) + 6(-2x+1) = 3(3x-4)/2 => This is where I'd gone wrong 6(-2x+1) multiplying out: x -4 -12x +6 = 9x -12 -4 +6 +12 = 9x -x +12x 14=20x x=14/20 or 7/10 YAY!

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                                          C Offline
                                          cpkilekofp
                                          wrote on last edited by
                                          #22

                                          You dropped a sign in your third step: (x - 4) / 6 - 2x + 1 = (3x - 4) / 2 (x - 4) - 6(2x + 1) = 3(3x - 4) (x - 4) - 12x + 6 = 3(3x - 4) x - 4 - 12x + 6 = 9x - 12 -11x + 2 = 9x - 12 -20x = -14 x = 7 / 10

                                          "Seize the day" - Horace "It's not what he doesn't know that scares me; it's what he knows for sure that just ain't so!" - Will Rogers, said by him about Herbert Hoover

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