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  3. Ever get stuck on a formula?

Ever get stuck on a formula?

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  • J jerryr4

    It's trivial https://codeproject.global.ssl.fastly.net/script/Forums/Images/smiley\_wink.gif, even after almost 60 years away from Grammar School in England. The steps are, using C formulae: a / (a + b) = c / d Multiply both sides by (a + b) a = (a + b) * c / d Collect the terms in a a - a * c / d = b * c / d Collect the coefficients of a a * (1 - c / d) = b * c / d Divide both sides by (1 - c / d) a = b * c / d / (1 - c / d) Simplify a = b * c / (d - c) Hope that's right, solved before you would have found the program In those good old days you got the algebra beaten into you Show-off Jerry

    R Offline
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    Rage
    wrote on last edited by
    #13

    ... and c needs to be not equal to d ...

    Do not escape reality : improve reality !

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    • K kmoorevs

      Last week while working on a new report, I needed a formula that would calculate a shortage/overage for a variable and had the equation worked out as follows: a/(a+b) = c/d where I was solving for the variable a. It's been 30 years since I had algebra, and I struggled with a half dozen attempts at getting 'a' by itself on the left side, but nothing checked out. On a hunch, I did a quick google search and found this: https://www.mathway.com/Algebra[^] All you do is type in your formula, then it asks which variable you want to solve for. Brilliant!

      "Go forth into the source" - Neal Morse

      G Offline
      G Offline
      Gary Wheeler
      wrote on last edited by
      #14

      a/(a+b) = c/d
      a = c(a+b)/d
      a = ca/d + cb/d
      a - ca/d = cb/d
      ad - ac = cb
      a(d-c) = cb
      a = cb/(d-c)

      A slightly different version from Jerry's. I had 32 credit-hours of math in college - calculus, differential equations, and matrix algebra. All of it's gone, with the space in my head now used for old movie lines. The algebra's stayed around, as it seems to come in handy now and then.

      Software Zen: delete this;

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      • J jerryr4

        It's trivial https://codeproject.global.ssl.fastly.net/script/Forums/Images/smiley\_wink.gif, even after almost 60 years away from Grammar School in England. The steps are, using C formulae: a / (a + b) = c / d Multiply both sides by (a + b) a = (a + b) * c / d Collect the terms in a a - a * c / d = b * c / d Collect the coefficients of a a * (1 - c / d) = b * c / d Divide both sides by (1 - c / d) a = b * c / d / (1 - c / d) Simplify a = b * c / (d - c) Hope that's right, solved before you would have found the program In those good old days you got the algebra beaten into you Show-off Jerry

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        W Offline
        W Balboos GHB
        wrote on last edited by
        #15

        I am so grateful for a post from someone who doesn't feel being innumerate is a way to be part of some team. Like the slope-of-a-line question. dy/dx - WTF's the difficulty with high school math? Something akin to pride as a consequence of membership in a pool of those who "can't" seems to be spreading from the TV and Radio to CP. Maybe the problem with poor code isn't only from the mass-production hordes now being unleashed? Alas.

        Ravings en masse^

        "The difference between genius and stupidity is that genius has its limits." - Albert Einstein

        "If you are searching for perfection in others, then you seek disappointment. If you are seek perfection in yourself, then you will find failure." - Balboos HaGadol Mar 2010

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        • K kmoorevs

          Last week while working on a new report, I needed a formula that would calculate a shortage/overage for a variable and had the equation worked out as follows: a/(a+b) = c/d where I was solving for the variable a. It's been 30 years since I had algebra, and I struggled with a half dozen attempts at getting 'a' by itself on the left side, but nothing checked out. On a hunch, I did a quick google search and found this: https://www.mathway.com/Algebra[^] All you do is type in your formula, then it asks which variable you want to solve for. Brilliant!

          "Go forth into the source" - Neal Morse

          K Offline
          K Offline
          Kirk 10389821
          wrote on last edited by
          #16

          As a math minor I suddenly understand the disdain for cliffs notes... LOL

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          • K kmoorevs

            Last week while working on a new report, I needed a formula that would calculate a shortage/overage for a variable and had the equation worked out as follows: a/(a+b) = c/d where I was solving for the variable a. It's been 30 years since I had algebra, and I struggled with a half dozen attempts at getting 'a' by itself on the left side, but nothing checked out. On a hunch, I did a quick google search and found this: https://www.mathway.com/Algebra[^] All you do is type in your formula, then it asks which variable you want to solve for. Brilliant!

            "Go forth into the source" - Neal Morse

            M Offline
            M Offline
            Marc Clifton
            wrote on last edited by
            #17

            kmoorevs wrote:

            a/(a+b) = c/d where I was solving for the variable a.

            It's a lot easier to see the solution when one writes it like:

            a c
            ------ = ---
            a + b d

            I'm a firm believer that visualizing (or expressing) the problem correctly is 90% of the solution. Given the above, cross multiply to get:

            ad = c (a+b)

            then:

            ad = ca + cb

            ad - ca = cb

            a(d-c) = cb

               cb
            

            a = ---------
            d-c

            Latest Article - A Concise Overview of Threads Learning to code with python is like learning to swim with those little arm floaties. It gives you undeserved confidence and will eventually drown you. - DangerBunny Artificial intelligence is the only remedy for natural stupidity. - CDP1802

            K 1 Reply Last reply
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            • M Marc Clifton

              kmoorevs wrote:

              a/(a+b) = c/d where I was solving for the variable a.

              It's a lot easier to see the solution when one writes it like:

              a c
              ------ = ---
              a + b d

              I'm a firm believer that visualizing (or expressing) the problem correctly is 90% of the solution. Given the above, cross multiply to get:

              ad = c (a+b)

              then:

              ad = ca + cb

              ad - ca = cb

              a(d-c) = cb

                 cb
              

              a = ---------
              d-c

              Latest Article - A Concise Overview of Threads Learning to code with python is like learning to swim with those little arm floaties. It gives you undeserved confidence and will eventually drown you. - DangerBunny Artificial intelligence is the only remedy for natural stupidity. - CDP1802

              K Offline
              K Offline
              kmoorevs
              wrote on last edited by
              #18

              Thanks Marc, Given the answer I was able to see where I was going wrong! :laugh: It's a good reminder that math is a skill that diminishes with nonuse. Using calculators and spreadsheets for a couple of decades is taking it's toll. :doh: Maybe we should have a MQOTW? :)

              "Go forth into the source" - Neal Morse

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              • J jerryr4

                It's trivial https://codeproject.global.ssl.fastly.net/script/Forums/Images/smiley\_wink.gif, even after almost 60 years away from Grammar School in England. The steps are, using C formulae: a / (a + b) = c / d Multiply both sides by (a + b) a = (a + b) * c / d Collect the terms in a a - a * c / d = b * c / d Collect the coefficients of a a * (1 - c / d) = b * c / d Divide both sides by (1 - c / d) a = b * c / d / (1 - c / d) Simplify a = b * c / (d - c) Hope that's right, solved before you would have found the program In those good old days you got the algebra beaten into you Show-off Jerry

                J Offline
                J Offline
                jackbrownii
                wrote on last edited by
                #19

                Probably took me the same amount of time by a slightly different route: a/(a+b)=c/d Multiply both sides by (a+b) and by d ad = c(a+b) Expand ad = ac + bc Subtract ac from both sides ad - ac = bc Collect the coefficients of a a(d-c) = bc Divide both sides by (d-c) a = bc/(d-c) Oh, if only all the equations were linear. ;P

                J 1 Reply Last reply
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                • J jackbrownii

                  Probably took me the same amount of time by a slightly different route: a/(a+b)=c/d Multiply both sides by (a+b) and by d ad = c(a+b) Expand ad = ac + bc Subtract ac from both sides ad - ac = bc Collect the coefficients of a a(d-c) = bc Divide both sides by (d-c) a = bc/(d-c) Oh, if only all the equations were linear. ;P

                  J Offline
                  J Offline
                  jerryr4
                  wrote on last edited by
                  #20

                  Probably took me the same amount of time by a slightly different route: You're right, I thought of this after having written the route. Jerry

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                  • J jerryr4

                    It's trivial https://codeproject.global.ssl.fastly.net/script/Forums/Images/smiley\_wink.gif, even after almost 60 years away from Grammar School in England. The steps are, using C formulae: a / (a + b) = c / d Multiply both sides by (a + b) a = (a + b) * c / d Collect the terms in a a - a * c / d = b * c / d Collect the coefficients of a a * (1 - c / d) = b * c / d Divide both sides by (1 - c / d) a = b * c / d / (1 - c / d) Simplify a = b * c / (d - c) Hope that's right, solved before you would have found the program In those good old days you got the algebra beaten into you Show-off Jerry

                    U Offline
                    U Offline
                    User 11319743
                    wrote on last edited by
                    #21

                    I got a slightly different (if not simpler) answer. a/(a+b) = c/d a = c/d*(a+b) a*d/c = a+b a*d/c -a = b a*(d/c -1) = b a=b/(d/c-1) Which if you multiply the numerator & denominator by c you'll get the answer the machine reports.

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                    • K kmoorevs

                      Last week while working on a new report, I needed a formula that would calculate a shortage/overage for a variable and had the equation worked out as follows: a/(a+b) = c/d where I was solving for the variable a. It's been 30 years since I had algebra, and I struggled with a half dozen attempts at getting 'a' by itself on the left side, but nothing checked out. On a hunch, I did a quick google search and found this: https://www.mathway.com/Algebra[^] All you do is type in your formula, then it asks which variable you want to solve for. Brilliant!

                      "Go forth into the source" - Neal Morse

                      M Offline
                      M Offline
                      maze3
                      wrote on last edited by
                      #22

                      oh wow, thanks. helped prove that the following does simplify (A / (B * C) ) * B simplifies to A / C

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