Dividing one equation by another












1














This is from Higher Algebra by Hall and Knight. enter image description here

I don't understand how this is done. Can you explain?






algebra-precalculus







share|cite|improve this question




















  • 6




    difference of two squares
    – Lord Shark the Unknown
    Dec 26 at 17:05










  • Thanks. I got it.
    – Md Masood
    Dec 26 at 17:06






  • 2




    To me the explanation given in the book seems rather unclear. It really should have mentioned that we are using the difference of two squares factorization. Otherwise it's not clear why dividing each member of (2) by the corresponding member of (1) would be a valid thing to do. I'm a bit afraid that other students who read this (not you) might think that if $a+b = x$ and $c+d= y$ then $a/c + b/d = x/y$, but of course that's usually not correct.
    – littleO
    Dec 26 at 17:10












  • I first thought so.
    – Md Masood
    Dec 26 at 17:53










  • Now try to derive the cartesian equation for an ellipse. :-)
    – John Joy
    Dec 28 at 0:48
















1














This is from Higher Algebra by Hall and Knight. enter image description here

I don't understand how this is done. Can you explain?






algebra-precalculus







share|cite|improve this question




















  • 6




    difference of two squares
    – Lord Shark the Unknown
    Dec 26 at 17:05










  • Thanks. I got it.
    – Md Masood
    Dec 26 at 17:06






  • 2




    To me the explanation given in the book seems rather unclear. It really should have mentioned that we are using the difference of two squares factorization. Otherwise it's not clear why dividing each member of (2) by the corresponding member of (1) would be a valid thing to do. I'm a bit afraid that other students who read this (not you) might think that if $a+b = x$ and $c+d= y$ then $a/c + b/d = x/y$, but of course that's usually not correct.
    – littleO
    Dec 26 at 17:10












  • I first thought so.
    – Md Masood
    Dec 26 at 17:53










  • Now try to derive the cartesian equation for an ellipse. :-)
    – John Joy
    Dec 28 at 0:48














1












1








1







This is from Higher Algebra by Hall and Knight. enter image description here

I don't understand how this is done. Can you explain?






algebra-precalculus







share|cite|improve this question















This is from Higher Algebra by Hall and Knight. enter image description here

I don't understand how this is done. Can you explain?






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 at 17:07









Zacky

4,7001646




4,7001646










asked Dec 26 at 17:03









Md Masood

528




528








  • 6




    difference of two squares
    – Lord Shark the Unknown
    Dec 26 at 17:05










  • Thanks. I got it.
    – Md Masood
    Dec 26 at 17:06






  • 2




    To me the explanation given in the book seems rather unclear. It really should have mentioned that we are using the difference of two squares factorization. Otherwise it's not clear why dividing each member of (2) by the corresponding member of (1) would be a valid thing to do. I'm a bit afraid that other students who read this (not you) might think that if $a+b = x$ and $c+d= y$ then $a/c + b/d = x/y$, but of course that's usually not correct.
    – littleO
    Dec 26 at 17:10












  • I first thought so.
    – Md Masood
    Dec 26 at 17:53










  • Now try to derive the cartesian equation for an ellipse. :-)
    – John Joy
    Dec 28 at 0:48














  • 6




    difference of two squares
    – Lord Shark the Unknown
    Dec 26 at 17:05










  • Thanks. I got it.
    – Md Masood
    Dec 26 at 17:06






  • 2




    To me the explanation given in the book seems rather unclear. It really should have mentioned that we are using the difference of two squares factorization. Otherwise it's not clear why dividing each member of (2) by the corresponding member of (1) would be a valid thing to do. I'm a bit afraid that other students who read this (not you) might think that if $a+b = x$ and $c+d= y$ then $a/c + b/d = x/y$, but of course that's usually not correct.
    – littleO
    Dec 26 at 17:10












  • I first thought so.
    – Md Masood
    Dec 26 at 17:53










  • Now try to derive the cartesian equation for an ellipse. :-)
    – John Joy
    Dec 28 at 0:48








6




6




difference of two squares
– Lord Shark the Unknown
Dec 26 at 17:05




difference of two squares
– Lord Shark the Unknown
Dec 26 at 17:05












Thanks. I got it.
– Md Masood
Dec 26 at 17:06




Thanks. I got it.
– Md Masood
Dec 26 at 17:06




2




2




To me the explanation given in the book seems rather unclear. It really should have mentioned that we are using the difference of two squares factorization. Otherwise it's not clear why dividing each member of (2) by the corresponding member of (1) would be a valid thing to do. I'm a bit afraid that other students who read this (not you) might think that if $a+b = x$ and $c+d= y$ then $a/c + b/d = x/y$, but of course that's usually not correct.
– littleO
Dec 26 at 17:10






To me the explanation given in the book seems rather unclear. It really should have mentioned that we are using the difference of two squares factorization. Otherwise it's not clear why dividing each member of (2) by the corresponding member of (1) would be a valid thing to do. I'm a bit afraid that other students who read this (not you) might think that if $a+b = x$ and $c+d= y$ then $a/c + b/d = x/y$, but of course that's usually not correct.
– littleO
Dec 26 at 17:10














I first thought so.
– Md Masood
Dec 26 at 17:53




I first thought so.
– Md Masood
Dec 26 at 17:53












Now try to derive the cartesian equation for an ellipse. :-)
– John Joy
Dec 28 at 0:48




Now try to derive the cartesian equation for an ellipse. :-)
– John Joy
Dec 28 at 0:48










2 Answers
2






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3














Note that (2) could be factored as



$$ (3x^2-4x+34)-(3x^2-4x-11) = (sqrt {3x^2-4x+34} -sqrt {3x^2-4x-11})(sqrt {3x^2-4x+34} +sqrt {3x^2-4x-11})=45$$



Now dividing by $(1)$ results in $(3)$






share|cite|improve this answer





























    4














    Hint: Use the identity
    $$a^2-b^2=(a-b)(a+b) $$






    share|cite|improve this answer





















      Your Answer





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      2 Answers
      2






      active

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      2 Answers
      2






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      3














      Note that (2) could be factored as



      $$ (3x^2-4x+34)-(3x^2-4x-11) = (sqrt {3x^2-4x+34} -sqrt {3x^2-4x-11})(sqrt {3x^2-4x+34} +sqrt {3x^2-4x-11})=45$$



      Now dividing by $(1)$ results in $(3)$






      share|cite|improve this answer


























        3














        Note that (2) could be factored as



        $$ (3x^2-4x+34)-(3x^2-4x-11) = (sqrt {3x^2-4x+34} -sqrt {3x^2-4x-11})(sqrt {3x^2-4x+34} +sqrt {3x^2-4x-11})=45$$



        Now dividing by $(1)$ results in $(3)$






        share|cite|improve this answer
























          3












          3








          3






          Note that (2) could be factored as



          $$ (3x^2-4x+34)-(3x^2-4x-11) = (sqrt {3x^2-4x+34} -sqrt {3x^2-4x-11})(sqrt {3x^2-4x+34} +sqrt {3x^2-4x-11})=45$$



          Now dividing by $(1)$ results in $(3)$






          share|cite|improve this answer












          Note that (2) could be factored as



          $$ (3x^2-4x+34)-(3x^2-4x-11) = (sqrt {3x^2-4x+34} -sqrt {3x^2-4x-11})(sqrt {3x^2-4x+34} +sqrt {3x^2-4x-11})=45$$



          Now dividing by $(1)$ results in $(3)$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 26 at 17:13









          Mohammad Riazi-Kermani

          40.8k42059




          40.8k42059























              4














              Hint: Use the identity
              $$a^2-b^2=(a-b)(a+b) $$






              share|cite|improve this answer


























                4














                Hint: Use the identity
                $$a^2-b^2=(a-b)(a+b) $$






                share|cite|improve this answer
























                  4












                  4








                  4






                  Hint: Use the identity
                  $$a^2-b^2=(a-b)(a+b) $$






                  share|cite|improve this answer












                  Hint: Use the identity
                  $$a^2-b^2=(a-b)(a+b) $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 26 at 17:08









                  Thomas Shelby

                  1,445216




                  1,445216






























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