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Find all ordered pairs $(a, b)$ in $a+frac{10b}{a^2+b^2} = 5;;,b+frac{10a}{a^2+b^2}=4$
$begingroup$
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
edited Jan 18 at 7:52
Michael Rozenberg
105k1892198
asked Aug 24 '14 at 8:27
juantheronjuantheron
34.3k1148143
$endgroup$
add a comment |
$begingroup$
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
edited Jan 18 at 7:52
Michael Rozenberg
105k1892198
asked Aug 24 '14 at 8:27
juantheronjuantheron
34.3k1148143
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 18 at 7:52
Michael Rozenberg
105k1892198
asked Aug 24 '14 at 8:27
juantheronjuantheron
34.3k1148143
$endgroup$
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
edited Jan 18 at 7:52
Michael Rozenberg
105k1892198
asked Aug 24 '14 at 8:27
juantheronjuantheron
34.3k1148143
edited Jan 18 at 7:52
Michael Rozenberg
105k1892198
asked Aug 24 '14 at 8:27
juantheronjuantheron
34.3k1148143
edited Jan 18 at 7:52
Michael Rozenberg
105k1892198
edited Jan 18 at 7:52
Michael Rozenberg
105k1892198
edited Jan 18 at 7:52
Michael Rozenberg
105k1892198
105k1892198
asked Aug 24 '14 at 8:27
juantheronjuantheron
34.3k1148143
asked Aug 24 '14 at 8:27
juantheronjuantheron
34.3k1148143
asked Aug 24 '14 at 8:27
juantheronjuantheron
34.3k1148143
34.3k1148143
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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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.
answered Jan 18 at 7:50
Michael RozenbergMichael Rozenberg
105k1892198
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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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.
answered Jan 18 at 7:50
Michael RozenbergMichael Rozenberg
105k1892198
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 18 at 7:50
Michael RozenbergMichael Rozenberg
105k1892198
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 18 at 7:50
Michael RozenbergMichael Rozenberg
105k1892198
$endgroup$
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.
answered Jan 18 at 7:50
Michael RozenbergMichael Rozenberg
105k1892198
answered Jan 18 at 7:50
Michael RozenbergMichael Rozenberg
105k1892198
answered Jan 18 at 7:50
Michael RozenbergMichael Rozenberg
105k1892198
answered Jan 18 at 7:50
Michael RozenbergMichael Rozenberg
105k1892198
105k1892198
add a comment |
add a comment |
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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