Manipulating this $frac{x-y}{z-y}$ to $1+frac{x-z}{z-y}$












4












$begingroup$


There is probably a very easy explanation for this that is lost on me. Came across a formula that was manipulated into another form and it was presented as a given, so I am trying to figure out how that was done.



Original:



$frac{x-y}{z-y}=1+frac{x-z}{z-y}$



So I began to break it down:



$frac{x-y}{z-y}=frac{x}{z-y}-frac{y}{z-y}$



$frac{x-y}{z-y}=frac{x}{z-y}-(frac{z}{z-y}-frac{y}{y})$ <-- maybe this is wrong, but it works and I don't know why.



Opening up the bracket:



$frac{x-y}{z-y}=frac{x}{z-y}-frac{z}{z-y}+1$



$frac{x-y}{z-y}=1+frac{x-z}{z-y}$



As I typed through all this, I see that $frac{y}{z-y}=frac{z}{z-y}-1$ is just true, and I get it when I simplify. I don't get how someone could see that to begin and wish to expand a formula like that.



Thank you for your time and patience with my high school level math question.






algebra-precalculus







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    There is probably a very easy explanation for this that is lost on me. Came across a formula that was manipulated into another form and it was presented as a given, so I am trying to figure out how that was done.



    Original:



    $frac{x-y}{z-y}=1+frac{x-z}{z-y}$



    So I began to break it down:



    $frac{x-y}{z-y}=frac{x}{z-y}-frac{y}{z-y}$



    $frac{x-y}{z-y}=frac{x}{z-y}-(frac{z}{z-y}-frac{y}{y})$ <-- maybe this is wrong, but it works and I don't know why.



    Opening up the bracket:



    $frac{x-y}{z-y}=frac{x}{z-y}-frac{z}{z-y}+1$



    $frac{x-y}{z-y}=1+frac{x-z}{z-y}$



    As I typed through all this, I see that $frac{y}{z-y}=frac{z}{z-y}-1$ is just true, and I get it when I simplify. I don't get how someone could see that to begin and wish to expand a formula like that.



    Thank you for your time and patience with my high school level math question.






    algebra-precalculus







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      There is probably a very easy explanation for this that is lost on me. Came across a formula that was manipulated into another form and it was presented as a given, so I am trying to figure out how that was done.



      Original:



      $frac{x-y}{z-y}=1+frac{x-z}{z-y}$



      So I began to break it down:



      $frac{x-y}{z-y}=frac{x}{z-y}-frac{y}{z-y}$



      $frac{x-y}{z-y}=frac{x}{z-y}-(frac{z}{z-y}-frac{y}{y})$ <-- maybe this is wrong, but it works and I don't know why.



      Opening up the bracket:



      $frac{x-y}{z-y}=frac{x}{z-y}-frac{z}{z-y}+1$



      $frac{x-y}{z-y}=1+frac{x-z}{z-y}$



      As I typed through all this, I see that $frac{y}{z-y}=frac{z}{z-y}-1$ is just true, and I get it when I simplify. I don't get how someone could see that to begin and wish to expand a formula like that.



      Thank you for your time and patience with my high school level math question.






      algebra-precalculus







      share|cite|improve this question









      $endgroup$




      There is probably a very easy explanation for this that is lost on me. Came across a formula that was manipulated into another form and it was presented as a given, so I am trying to figure out how that was done.



      Original:



      $frac{x-y}{z-y}=1+frac{x-z}{z-y}$



      So I began to break it down:



      $frac{x-y}{z-y}=frac{x}{z-y}-frac{y}{z-y}$



      $frac{x-y}{z-y}=frac{x}{z-y}-(frac{z}{z-y}-frac{y}{y})$ <-- maybe this is wrong, but it works and I don't know why.



      Opening up the bracket:



      $frac{x-y}{z-y}=frac{x}{z-y}-frac{z}{z-y}+1$



      $frac{x-y}{z-y}=1+frac{x-z}{z-y}$



      As I typed through all this, I see that $frac{y}{z-y}=frac{z}{z-y}-1$ is just true, and I get it when I simplify. I don't get how someone could see that to begin and wish to expand a formula like that.



      Thank you for your time and patience with my high school level math question.






      algebra-precalculus




      algebra-precalculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 6 at 19:09









      TheCuriousOneTheCuriousOne

      233




      233






















          5 Answers
          5






          active

          oldest

          votes


















          6












          $begingroup$

          Suppose that we have the expression $dfrac{x-y}{z-y}$ and we want the numerator to be independent of $y$. Then the only way to do this is to make part of the numerator like the denominator, so that both cancel: $$frac{x-y}{z-y}=frac{z-y+x-z}{z-y}=frac{z-y}{z-y}+frac{x-z}{z-y}=1+frac{x-z}{z-y}$$ This is sometimes useful in integration, factorisation and other related topics.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Appreciate the explanation and reasoning behind why someone would want to do this. Removing a variable from the numerator, makes sense. Thank you.
            $endgroup$
            – TheCuriousOne
            Jan 6 at 19:29










          • $begingroup$
            I'm glad this was useful for you. Another application could be partial fractions, but I suspect that it may be slightly advanced.
            $endgroup$
            – TheSimpliFire
            Jan 6 at 19:31



















          2












          $begingroup$

          $
          frac{x-y}{z-y} = frac{z - y + x - z}{z-y} = 1 + frac{x - z}{z-y}
          $






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Hint: Write $$frac{x-y}{z-y}=frac{z-y+x-z}{z-y}$$ I think it is a nulladdition






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you for the hint.
              $endgroup$
              – TheCuriousOne
              Jan 6 at 19:27



















            1












            $begingroup$

            You are certainly right to say that
            $$frac{y}{z-y}=frac{z}{z-y}-frac{y}{y}$$
            Since
            $$frac{z}{z-y}-frac{y}{y}=frac{z}{z-y}-1=frac{z}{z-y}-frac{z-y}{z-y}=frac{y}{z-y}$$






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              A super easy way to get what you want is to just add it in (then subtract it so that you don't change the expression's value).



              $$begin{align}
              frac{x-y}{z-y} + (1+frac{x-z}{z-y}) - (1+frac{x-z}{z-y}) &= 1+frac{x-z}{z-y}+frac{(x-y)-(z-y)-(x-z))}{z-y}\
              &=1+frac{x-z}{z-y}\
              end{align}$$
              It's especially useful for doing messy trig proofs, e.g..
              $$begin{align}
              frac{2sin^2 x - 5sin x + 2}{sin x - 2} &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - (2sin x - 1) + (2sin x - 1)\
              &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - frac{(2sin x - 1)(sin x - 2)}{sin x - 2} + (2sin x - 1)\
              &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - frac{2sin^2 x -5sin x + 2}{sin x - 2} + (2sin x - 1)\
              &= 0+ (2sin x - 1)\
              &= 2sin x - 1\
              end{align}$$






              share|cite|improve this answer









              $endgroup$













                Your Answer





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                5 Answers
                5






                active

                oldest

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                5 Answers
                5






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                6












                $begingroup$

                Suppose that we have the expression $dfrac{x-y}{z-y}$ and we want the numerator to be independent of $y$. Then the only way to do this is to make part of the numerator like the denominator, so that both cancel: $$frac{x-y}{z-y}=frac{z-y+x-z}{z-y}=frac{z-y}{z-y}+frac{x-z}{z-y}=1+frac{x-z}{z-y}$$ This is sometimes useful in integration, factorisation and other related topics.






                share|cite|improve this answer









                $endgroup$









                • 1




                  $begingroup$
                  Appreciate the explanation and reasoning behind why someone would want to do this. Removing a variable from the numerator, makes sense. Thank you.
                  $endgroup$
                  – TheCuriousOne
                  Jan 6 at 19:29










                • $begingroup$
                  I'm glad this was useful for you. Another application could be partial fractions, but I suspect that it may be slightly advanced.
                  $endgroup$
                  – TheSimpliFire
                  Jan 6 at 19:31
















                6












                $begingroup$

                Suppose that we have the expression $dfrac{x-y}{z-y}$ and we want the numerator to be independent of $y$. Then the only way to do this is to make part of the numerator like the denominator, so that both cancel: $$frac{x-y}{z-y}=frac{z-y+x-z}{z-y}=frac{z-y}{z-y}+frac{x-z}{z-y}=1+frac{x-z}{z-y}$$ This is sometimes useful in integration, factorisation and other related topics.






                share|cite|improve this answer









                $endgroup$









                • 1




                  $begingroup$
                  Appreciate the explanation and reasoning behind why someone would want to do this. Removing a variable from the numerator, makes sense. Thank you.
                  $endgroup$
                  – TheCuriousOne
                  Jan 6 at 19:29










                • $begingroup$
                  I'm glad this was useful for you. Another application could be partial fractions, but I suspect that it may be slightly advanced.
                  $endgroup$
                  – TheSimpliFire
                  Jan 6 at 19:31














                6












                6








                6





                $begingroup$

                Suppose that we have the expression $dfrac{x-y}{z-y}$ and we want the numerator to be independent of $y$. Then the only way to do this is to make part of the numerator like the denominator, so that both cancel: $$frac{x-y}{z-y}=frac{z-y+x-z}{z-y}=frac{z-y}{z-y}+frac{x-z}{z-y}=1+frac{x-z}{z-y}$$ This is sometimes useful in integration, factorisation and other related topics.






                share|cite|improve this answer









                $endgroup$



                Suppose that we have the expression $dfrac{x-y}{z-y}$ and we want the numerator to be independent of $y$. Then the only way to do this is to make part of the numerator like the denominator, so that both cancel: $$frac{x-y}{z-y}=frac{z-y+x-z}{z-y}=frac{z-y}{z-y}+frac{x-z}{z-y}=1+frac{x-z}{z-y}$$ This is sometimes useful in integration, factorisation and other related topics.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 19:17









                TheSimpliFireTheSimpliFire

                12.3k62460




                12.3k62460








                • 1




                  $begingroup$
                  Appreciate the explanation and reasoning behind why someone would want to do this. Removing a variable from the numerator, makes sense. Thank you.
                  $endgroup$
                  – TheCuriousOne
                  Jan 6 at 19:29










                • $begingroup$
                  I'm glad this was useful for you. Another application could be partial fractions, but I suspect that it may be slightly advanced.
                  $endgroup$
                  – TheSimpliFire
                  Jan 6 at 19:31














                • 1




                  $begingroup$
                  Appreciate the explanation and reasoning behind why someone would want to do this. Removing a variable from the numerator, makes sense. Thank you.
                  $endgroup$
                  – TheCuriousOne
                  Jan 6 at 19:29










                • $begingroup$
                  I'm glad this was useful for you. Another application could be partial fractions, but I suspect that it may be slightly advanced.
                  $endgroup$
                  – TheSimpliFire
                  Jan 6 at 19:31








                1




                1




                $begingroup$
                Appreciate the explanation and reasoning behind why someone would want to do this. Removing a variable from the numerator, makes sense. Thank you.
                $endgroup$
                – TheCuriousOne
                Jan 6 at 19:29




                $begingroup$
                Appreciate the explanation and reasoning behind why someone would want to do this. Removing a variable from the numerator, makes sense. Thank you.
                $endgroup$
                – TheCuriousOne
                Jan 6 at 19:29












                $begingroup$
                I'm glad this was useful for you. Another application could be partial fractions, but I suspect that it may be slightly advanced.
                $endgroup$
                – TheSimpliFire
                Jan 6 at 19:31




                $begingroup$
                I'm glad this was useful for you. Another application could be partial fractions, but I suspect that it may be slightly advanced.
                $endgroup$
                – TheSimpliFire
                Jan 6 at 19:31











                2












                $begingroup$

                $
                frac{x-y}{z-y} = frac{z - y + x - z}{z-y} = 1 + frac{x - z}{z-y}
                $






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  $
                  frac{x-y}{z-y} = frac{z - y + x - z}{z-y} = 1 + frac{x - z}{z-y}
                  $






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    $
                    frac{x-y}{z-y} = frac{z - y + x - z}{z-y} = 1 + frac{x - z}{z-y}
                    $






                    share|cite|improve this answer









                    $endgroup$



                    $
                    frac{x-y}{z-y} = frac{z - y + x - z}{z-y} = 1 + frac{x - z}{z-y}
                    $







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 6 at 19:19









                    Karsten LeonhardtKarsten Leonhardt

                    365




                    365























                        1












                        $begingroup$

                        Hint: Write $$frac{x-y}{z-y}=frac{z-y+x-z}{z-y}$$ I think it is a nulladdition






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          Thank you for the hint.
                          $endgroup$
                          – TheCuriousOne
                          Jan 6 at 19:27
















                        1












                        $begingroup$

                        Hint: Write $$frac{x-y}{z-y}=frac{z-y+x-z}{z-y}$$ I think it is a nulladdition






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          Thank you for the hint.
                          $endgroup$
                          – TheCuriousOne
                          Jan 6 at 19:27














                        1












                        1








                        1





                        $begingroup$

                        Hint: Write $$frac{x-y}{z-y}=frac{z-y+x-z}{z-y}$$ I think it is a nulladdition






                        share|cite|improve this answer









                        $endgroup$



                        Hint: Write $$frac{x-y}{z-y}=frac{z-y+x-z}{z-y}$$ I think it is a nulladdition







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jan 6 at 19:17









                        Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                        74.4k42865




                        74.4k42865












                        • $begingroup$
                          Thank you for the hint.
                          $endgroup$
                          – TheCuriousOne
                          Jan 6 at 19:27


















                        • $begingroup$
                          Thank you for the hint.
                          $endgroup$
                          – TheCuriousOne
                          Jan 6 at 19:27
















                        $begingroup$
                        Thank you for the hint.
                        $endgroup$
                        – TheCuriousOne
                        Jan 6 at 19:27




                        $begingroup$
                        Thank you for the hint.
                        $endgroup$
                        – TheCuriousOne
                        Jan 6 at 19:27











                        1












                        $begingroup$

                        You are certainly right to say that
                        $$frac{y}{z-y}=frac{z}{z-y}-frac{y}{y}$$
                        Since
                        $$frac{z}{z-y}-frac{y}{y}=frac{z}{z-y}-1=frac{z}{z-y}-frac{z-y}{z-y}=frac{y}{z-y}$$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          You are certainly right to say that
                          $$frac{y}{z-y}=frac{z}{z-y}-frac{y}{y}$$
                          Since
                          $$frac{z}{z-y}-frac{y}{y}=frac{z}{z-y}-1=frac{z}{z-y}-frac{z-y}{z-y}=frac{y}{z-y}$$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            You are certainly right to say that
                            $$frac{y}{z-y}=frac{z}{z-y}-frac{y}{y}$$
                            Since
                            $$frac{z}{z-y}-frac{y}{y}=frac{z}{z-y}-1=frac{z}{z-y}-frac{z-y}{z-y}=frac{y}{z-y}$$






                            share|cite|improve this answer









                            $endgroup$



                            You are certainly right to say that
                            $$frac{y}{z-y}=frac{z}{z-y}-frac{y}{y}$$
                            Since
                            $$frac{z}{z-y}-frac{y}{y}=frac{z}{z-y}-1=frac{z}{z-y}-frac{z-y}{z-y}=frac{y}{z-y}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 6 at 19:19









                            LarryLarry

                            2,27231028




                            2,27231028























                                1












                                $begingroup$

                                A super easy way to get what you want is to just add it in (then subtract it so that you don't change the expression's value).



                                $$begin{align}
                                frac{x-y}{z-y} + (1+frac{x-z}{z-y}) - (1+frac{x-z}{z-y}) &= 1+frac{x-z}{z-y}+frac{(x-y)-(z-y)-(x-z))}{z-y}\
                                &=1+frac{x-z}{z-y}\
                                end{align}$$
                                It's especially useful for doing messy trig proofs, e.g..
                                $$begin{align}
                                frac{2sin^2 x - 5sin x + 2}{sin x - 2} &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - (2sin x - 1) + (2sin x - 1)\
                                &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - frac{(2sin x - 1)(sin x - 2)}{sin x - 2} + (2sin x - 1)\
                                &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - frac{2sin^2 x -5sin x + 2}{sin x - 2} + (2sin x - 1)\
                                &= 0+ (2sin x - 1)\
                                &= 2sin x - 1\
                                end{align}$$






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  A super easy way to get what you want is to just add it in (then subtract it so that you don't change the expression's value).



                                  $$begin{align}
                                  frac{x-y}{z-y} + (1+frac{x-z}{z-y}) - (1+frac{x-z}{z-y}) &= 1+frac{x-z}{z-y}+frac{(x-y)-(z-y)-(x-z))}{z-y}\
                                  &=1+frac{x-z}{z-y}\
                                  end{align}$$
                                  It's especially useful for doing messy trig proofs, e.g..
                                  $$begin{align}
                                  frac{2sin^2 x - 5sin x + 2}{sin x - 2} &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - (2sin x - 1) + (2sin x - 1)\
                                  &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - frac{(2sin x - 1)(sin x - 2)}{sin x - 2} + (2sin x - 1)\
                                  &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - frac{2sin^2 x -5sin x + 2}{sin x - 2} + (2sin x - 1)\
                                  &= 0+ (2sin x - 1)\
                                  &= 2sin x - 1\
                                  end{align}$$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    A super easy way to get what you want is to just add it in (then subtract it so that you don't change the expression's value).



                                    $$begin{align}
                                    frac{x-y}{z-y} + (1+frac{x-z}{z-y}) - (1+frac{x-z}{z-y}) &= 1+frac{x-z}{z-y}+frac{(x-y)-(z-y)-(x-z))}{z-y}\
                                    &=1+frac{x-z}{z-y}\
                                    end{align}$$
                                    It's especially useful for doing messy trig proofs, e.g..
                                    $$begin{align}
                                    frac{2sin^2 x - 5sin x + 2}{sin x - 2} &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - (2sin x - 1) + (2sin x - 1)\
                                    &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - frac{(2sin x - 1)(sin x - 2)}{sin x - 2} + (2sin x - 1)\
                                    &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - frac{2sin^2 x -5sin x + 2}{sin x - 2} + (2sin x - 1)\
                                    &= 0+ (2sin x - 1)\
                                    &= 2sin x - 1\
                                    end{align}$$






                                    share|cite|improve this answer









                                    $endgroup$



                                    A super easy way to get what you want is to just add it in (then subtract it so that you don't change the expression's value).



                                    $$begin{align}
                                    frac{x-y}{z-y} + (1+frac{x-z}{z-y}) - (1+frac{x-z}{z-y}) &= 1+frac{x-z}{z-y}+frac{(x-y)-(z-y)-(x-z))}{z-y}\
                                    &=1+frac{x-z}{z-y}\
                                    end{align}$$
                                    It's especially useful for doing messy trig proofs, e.g..
                                    $$begin{align}
                                    frac{2sin^2 x - 5sin x + 2}{sin x - 2} &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - (2sin x - 1) + (2sin x - 1)\
                                    &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - frac{(2sin x - 1)(sin x - 2)}{sin x - 2} + (2sin x - 1)\
                                    &= frac{2sin^2 x - 5sin x + 2}{sin x - 2} - frac{2sin^2 x -5sin x + 2}{sin x - 2} + (2sin x - 1)\
                                    &= 0+ (2sin x - 1)\
                                    &= 2sin x - 1\
                                    end{align}$$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 6 at 20:18









                                    John JoyJohn Joy

                                    6,20611526




                                    6,20611526






























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