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
    11 hours ago










  • Thanks. I got it.
    – Md Masood
    11 hours ago






  • 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
    11 hours ago












  • I first thought so.
    – Md Masood
    10 hours ago
















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
    11 hours ago










  • Thanks. I got it.
    – Md Masood
    11 hours ago






  • 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
    11 hours ago












  • I first thought so.
    – Md Masood
    10 hours ago














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 11 hours ago









Zacky

4,1351546




4,1351546










asked 11 hours ago









Md Masood

528




528








  • 6




    difference of two squares
    – Lord Shark the Unknown
    11 hours ago










  • Thanks. I got it.
    – Md Masood
    11 hours ago






  • 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
    11 hours ago












  • I first thought so.
    – Md Masood
    10 hours ago














  • 6




    difference of two squares
    – Lord Shark the Unknown
    11 hours ago










  • Thanks. I got it.
    – Md Masood
    11 hours ago






  • 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
    11 hours ago












  • I first thought so.
    – Md Masood
    10 hours ago








6




6




difference of two squares
– Lord Shark the Unknown
11 hours ago




difference of two squares
– Lord Shark the Unknown
11 hours ago












Thanks. I got it.
– Md Masood
11 hours ago




Thanks. I got it.
– Md Masood
11 hours ago




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
11 hours ago






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
11 hours ago














I first thought so.
– Md Masood
10 hours ago




I first thought so.
– Md Masood
10 hours ago










2 Answers
2






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2














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














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






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      active

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      2














      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


























        2














        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
























          2












          2








          2






          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 11 hours ago









          Mohammad Riazi-Kermani

          40.7k42058




          40.7k42058























              3














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






              share|cite|improve this answer


























                3














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






                share|cite|improve this answer
























                  3












                  3








                  3






                  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 11 hours ago









                  Thomas Shelby

                  1,242116




                  1,242116






























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