Work Proof... Need help!I'm having trouble figuring out a proof at work, and to spare the business jargon I put it in terms of variables. Could someone help?Prove: A − B + Y 1 − C ⋅ C + A − B + X 1 − D ⋅ D = A − B 1 − C − D ⋅ ( C + D )where: X = A − B + Y 1 − C ⋅ C Y = A − B + X 1 − D ⋅ D

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Work Proof... Need help!I&#039;m having trouble figuring out a proof at work, and to spare the business jargon I put it in terms of variables. Could someone help?Prove:            A      −      B      +      Y              1      −      C        ⋅  C  +            A      −      B      +      X              1      −      D        ⋅  D  =            A      −      B              1      −      C      −      D        ⋅  (  C  +  D  )where:  X  =            A      −      B      +      Y              1      −      C        ⋅  C  Y  =            A      −      B      +      X              1      −      D        ⋅  D

Nezveda6q · Accepted Answer

From the condition we get  (  1  −  C  )  X  −  C  Y  =  (  A  −  B  )  C and   −  D  X  +  (  1  −  D  )  Y  =  (  A  −  B  )  D or  (  1  −  C  )  (  1  −  D  )  X  −  C  (  1  −  D  )  Y  =  (  A  −  B  )  (  1  −  D  )  Cand after adding we get   (  (  1  −  C  )  (  1  −  D  )  −  D  C  )  X  =  (  A  −  B  )  C, which gives  X  =            (      A      −      B      )      C              1      −      C      −      D       and   Y  =            (      A      −      B      )      D              1      −      C      −      D      Thus,            A      −      B      +      Y              1      −      C        ⋅  C  +            A      −      B      +      X              1      −      D        ⋅  D  =            A      −      B      +                        (          A          −          B          )          D                          1          −          C          −          D                            1      −      C        ⋅  C  +            A      −      B      +                        (          A          −          B          )          C                          1          −          C          −          D                            1      −      D        ⋅  D  =  =  (  A  −  B  )      (                  1        +                  D                      1            −            C            −            D                                      1        −        C              ⋅    C    +                  1        +                  C                      1            −            C            −            D                                      1        −        D              ⋅    D    )    =            (      A      −      B      )      (      C      +      D      )              1      −      C      −      D

afermavas4h · Accepted Answer

In the future, you can prove this sort of thing by typing it into free websites like Wolframalpha.comFor instance, you&#039;re trying to prove something like   α  +  β  =  γ, so you could type into the webpage   α  +  β  −  γ. This will simplify to 0, which proves the equation you wanted.You could also plug in lots and lots of points. If you do this literally randomly (use some random number generator to plug in numbers between say 0 and 1). If the numbers are random, then for many equations, you can actually prove that they always work simply because they work when you plug in some random numbers.A third option that works for things like this would be to multiply everything in sight to get rid of all the denominators. Then just multiply out everything and things will just cancel like crazy. This is very time-consuming to do by hand however, and people are prone to mistakes. [This is what the webpage would do though]

I'm having trouble figuring out a proof at work, and to spare the business jargon I put it in terms of variables. Could someone help? Prove: (A-B+Y)/(1-C)C+(A-B+X)/(1-D)D=(A-B)/(1-C-D)(C+D) where: X=(A-B+Y)/(1-C)C Y=(A-B+X)/(1-D)*D

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