Find all ordered pairs $(a, b)$ in $a+frac{10b}{a^2+b^2} = 5;;,b+frac{10a}{a^2+b^2}=4$












0












$begingroup$



Find all ordered pairs $(a, b)$ of complex numbers with $a^2+b^2neq 0,$ and



$displaystyle a+frac{10b}{a^2+b^2} = 5;;,b+frac{10a}{a^2+b^2}=4$




$bf{My; Solution}::$ Using Complex number..



Given $displaystyle a+frac{10b}{a^2+b^2}=5..............(1)$ and $displaystyle b+frac{10a}{a^2+b^2}=4...................(2)times i$



Now Add $bf{(1)}$ and $bf{(2);,}$ We get



$displaystyle (a+ib)+frac{10}{(a+ib)cdot (a-ib)}cdot (b+ia) = 5+4i$



So $displaystyle (a+ib)+frac{10icdot (a-ib)}{(a+ib)cdot (a-ib)} = 5+4iRightarrow (a+ib)+frac{10i}{(a+ib)}=5+4i$



Now Let $(a+ib)=z;,$ Then equation is $displaystyle z+frac{10i}{z}=5+4iRightarrow z^2-(5+4i)z+10i=0$



Now Solving the above equation we get $displaystyle z = frac{(5+4i)pm sqrt{(5+4i)^2-40i}}{2}=frac{(5+4i)pm sqrt{25-16}}{2}=frac{(5+4i)pm 3}{2}=4+2i;,1+2i$



So $z=x+iy=4+2iRightarrow (x,y)=(4,2)$ and $z=x+iy=1+2iRightarrow (x,y)=(1,2)$



My Question is , can we solve it without Using Complex number



If yes, The please explain here



Thanks






algebra-precalculus systems-of-equations







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  • $begingroup$
    We have $$10b=(5-a)(a^2+b^2), 10a=(4-b)(a^2+b^2)$$ On division, $$frac ba=frac{5-a}{4-b}implies a^2-5a=b^2-4b$$ $$iff4a^2-20a=4b^2-16biff(2a-5)^2-(2b-2)^2=9$$ $$implies2a=3sectheta+5,2b=2+3tantheta$$
    $endgroup$
    – lab bhattacharjee
    Aug 24 '14 at 8:43
















0












$begingroup$



Find all ordered pairs $(a, b)$ of complex numbers with $a^2+b^2neq 0,$ and



$displaystyle a+frac{10b}{a^2+b^2} = 5;;,b+frac{10a}{a^2+b^2}=4$




$bf{My; Solution}::$ Using Complex number..



Given $displaystyle a+frac{10b}{a^2+b^2}=5..............(1)$ and $displaystyle b+frac{10a}{a^2+b^2}=4...................(2)times i$



Now Add $bf{(1)}$ and $bf{(2);,}$ We get



$displaystyle (a+ib)+frac{10}{(a+ib)cdot (a-ib)}cdot (b+ia) = 5+4i$



So $displaystyle (a+ib)+frac{10icdot (a-ib)}{(a+ib)cdot (a-ib)} = 5+4iRightarrow (a+ib)+frac{10i}{(a+ib)}=5+4i$



Now Let $(a+ib)=z;,$ Then equation is $displaystyle z+frac{10i}{z}=5+4iRightarrow z^2-(5+4i)z+10i=0$



Now Solving the above equation we get $displaystyle z = frac{(5+4i)pm sqrt{(5+4i)^2-40i}}{2}=frac{(5+4i)pm sqrt{25-16}}{2}=frac{(5+4i)pm 3}{2}=4+2i;,1+2i$



So $z=x+iy=4+2iRightarrow (x,y)=(4,2)$ and $z=x+iy=1+2iRightarrow (x,y)=(1,2)$



My Question is , can we solve it without Using Complex number



If yes, The please explain here



Thanks






algebra-precalculus systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    We have $$10b=(5-a)(a^2+b^2), 10a=(4-b)(a^2+b^2)$$ On division, $$frac ba=frac{5-a}{4-b}implies a^2-5a=b^2-4b$$ $$iff4a^2-20a=4b^2-16biff(2a-5)^2-(2b-2)^2=9$$ $$implies2a=3sectheta+5,2b=2+3tantheta$$
    $endgroup$
    – lab bhattacharjee
    Aug 24 '14 at 8:43














0












0








0


2



$begingroup$



Find all ordered pairs $(a, b)$ of complex numbers with $a^2+b^2neq 0,$ and



$displaystyle a+frac{10b}{a^2+b^2} = 5;;,b+frac{10a}{a^2+b^2}=4$




$bf{My; Solution}::$ Using Complex number..



Given $displaystyle a+frac{10b}{a^2+b^2}=5..............(1)$ and $displaystyle b+frac{10a}{a^2+b^2}=4...................(2)times i$



Now Add $bf{(1)}$ and $bf{(2);,}$ We get



$displaystyle (a+ib)+frac{10}{(a+ib)cdot (a-ib)}cdot (b+ia) = 5+4i$



So $displaystyle (a+ib)+frac{10icdot (a-ib)}{(a+ib)cdot (a-ib)} = 5+4iRightarrow (a+ib)+frac{10i}{(a+ib)}=5+4i$



Now Let $(a+ib)=z;,$ Then equation is $displaystyle z+frac{10i}{z}=5+4iRightarrow z^2-(5+4i)z+10i=0$



Now Solving the above equation we get $displaystyle z = frac{(5+4i)pm sqrt{(5+4i)^2-40i}}{2}=frac{(5+4i)pm sqrt{25-16}}{2}=frac{(5+4i)pm 3}{2}=4+2i;,1+2i$



So $z=x+iy=4+2iRightarrow (x,y)=(4,2)$ and $z=x+iy=1+2iRightarrow (x,y)=(1,2)$



My Question is , can we solve it without Using Complex number



If yes, The please explain here



Thanks






algebra-precalculus systems-of-equations







share|cite|improve this question











$endgroup$





Find all ordered pairs $(a, b)$ of complex numbers with $a^2+b^2neq 0,$ and



$displaystyle a+frac{10b}{a^2+b^2} = 5;;,b+frac{10a}{a^2+b^2}=4$




$bf{My; Solution}::$ Using Complex number..



Given $displaystyle a+frac{10b}{a^2+b^2}=5..............(1)$ and $displaystyle b+frac{10a}{a^2+b^2}=4...................(2)times i$



Now Add $bf{(1)}$ and $bf{(2);,}$ We get



$displaystyle (a+ib)+frac{10}{(a+ib)cdot (a-ib)}cdot (b+ia) = 5+4i$



So $displaystyle (a+ib)+frac{10icdot (a-ib)}{(a+ib)cdot (a-ib)} = 5+4iRightarrow (a+ib)+frac{10i}{(a+ib)}=5+4i$



Now Let $(a+ib)=z;,$ Then equation is $displaystyle z+frac{10i}{z}=5+4iRightarrow z^2-(5+4i)z+10i=0$



Now Solving the above equation we get $displaystyle z = frac{(5+4i)pm sqrt{(5+4i)^2-40i}}{2}=frac{(5+4i)pm sqrt{25-16}}{2}=frac{(5+4i)pm 3}{2}=4+2i;,1+2i$



So $z=x+iy=4+2iRightarrow (x,y)=(4,2)$ and $z=x+iy=1+2iRightarrow (x,y)=(1,2)$



My Question is , can we solve it without Using Complex number



If yes, The please explain here



Thanks






algebra-precalculus systems-of-equations




algebra-precalculus systems-of-equations






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share|cite|improve this question













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edited Jan 18 at 7:52









Michael Rozenberg

111k1897201




111k1897201










asked Aug 24 '14 at 8:27









juantheronjuantheron

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  • $begingroup$
    We have $$10b=(5-a)(a^2+b^2), 10a=(4-b)(a^2+b^2)$$ On division, $$frac ba=frac{5-a}{4-b}implies a^2-5a=b^2-4b$$ $$iff4a^2-20a=4b^2-16biff(2a-5)^2-(2b-2)^2=9$$ $$implies2a=3sectheta+5,2b=2+3tantheta$$
    $endgroup$
    – lab bhattacharjee
    Aug 24 '14 at 8:43


















  • $begingroup$
    We have $$10b=(5-a)(a^2+b^2), 10a=(4-b)(a^2+b^2)$$ On division, $$frac ba=frac{5-a}{4-b}implies a^2-5a=b^2-4b$$ $$iff4a^2-20a=4b^2-16biff(2a-5)^2-(2b-2)^2=9$$ $$implies2a=3sectheta+5,2b=2+3tantheta$$
    $endgroup$
    – lab bhattacharjee
    Aug 24 '14 at 8:43
















$begingroup$
We have $$10b=(5-a)(a^2+b^2), 10a=(4-b)(a^2+b^2)$$ On division, $$frac ba=frac{5-a}{4-b}implies a^2-5a=b^2-4b$$ $$iff4a^2-20a=4b^2-16biff(2a-5)^2-(2b-2)^2=9$$ $$implies2a=3sectheta+5,2b=2+3tantheta$$
$endgroup$
– lab bhattacharjee
Aug 24 '14 at 8:43




$begingroup$
We have $$10b=(5-a)(a^2+b^2), 10a=(4-b)(a^2+b^2)$$ On division, $$frac ba=frac{5-a}{4-b}implies a^2-5a=b^2-4b$$ $$iff4a^2-20a=4b^2-16biff(2a-5)^2-(2b-2)^2=9$$ $$implies2a=3sectheta+5,2b=2+3tantheta$$
$endgroup$
– lab bhattacharjee
Aug 24 '14 at 8:43










1 Answer
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$begingroup$

Easy to see that $abneq0$.



Thus, our system is equal to
$$(5-a)b=frac{10b^2}{a^2+b^2}$$ and
$$(4-b)a=frac{10a^2}{a^2+b^2},$$ which after summing gives
$$(5-a)b+(4-b)a=10$$ or
$$(2a-5)(b-2)=0.$$
The rest is smooth.






share|cite|improve this answer









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    $begingroup$

    Easy to see that $abneq0$.



    Thus, our system is equal to
    $$(5-a)b=frac{10b^2}{a^2+b^2}$$ and
    $$(4-b)a=frac{10a^2}{a^2+b^2},$$ which after summing gives
    $$(5-a)b+(4-b)a=10$$ or
    $$(2a-5)(b-2)=0.$$
    The rest is smooth.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Easy to see that $abneq0$.



      Thus, our system is equal to
      $$(5-a)b=frac{10b^2}{a^2+b^2}$$ and
      $$(4-b)a=frac{10a^2}{a^2+b^2},$$ which after summing gives
      $$(5-a)b+(4-b)a=10$$ or
      $$(2a-5)(b-2)=0.$$
      The rest is smooth.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Easy to see that $abneq0$.



        Thus, our system is equal to
        $$(5-a)b=frac{10b^2}{a^2+b^2}$$ and
        $$(4-b)a=frac{10a^2}{a^2+b^2},$$ which after summing gives
        $$(5-a)b+(4-b)a=10$$ or
        $$(2a-5)(b-2)=0.$$
        The rest is smooth.






        share|cite|improve this answer









        $endgroup$



        Easy to see that $abneq0$.



        Thus, our system is equal to
        $$(5-a)b=frac{10b^2}{a^2+b^2}$$ and
        $$(4-b)a=frac{10a^2}{a^2+b^2},$$ which after summing gives
        $$(5-a)b+(4-b)a=10$$ or
        $$(2a-5)(b-2)=0.$$
        The rest is smooth.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 18 at 7:50









        Michael RozenbergMichael Rozenberg

        111k1897201




        111k1897201






























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