MHB Simultaneous Equations Challenge

AI Thread Summary
The discussion revolves around solving a system of equations involving variables a and b. Participants highlight the importance of recognizing that a equals negative b, which simplifies the problem. Trigonometric substitution is suggested as a useful technique for finding the solution. One participant expresses appreciation for another's contribution to the challenge. The conversation emphasizes collaboration and sharing different methods for solving mathematical problems.
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Solve the system of equations below:

$(a+\sqrt{a^2+1})(b+\sqrt{b^2+1})=1$

$b+\dfrac{b}{\sqrt{a^2-1}}+\dfrac{35}{12}=0$
 
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anemone said:
Solve the system of equations below:

$(a+\sqrt{a^2+1})(b+\sqrt{b^2+1})=1$

$b+\dfrac{b}{\sqrt{a^2-1}}+\dfrac{35}{12}=0$
From the 2nd equation we have b < 0 say - c

So from 1 we get $(\sqrt{a^2+1}+a)(\sqrt{c^2+1}-c) = 1$
as we have $(\sqrt{a^2+1}+a)(\sqrt{a^2+1}-a) = 1$
so we get $\sqrt{a^2+1}-a= \sqrt{c^2+1}-c$
as $\sqrt{a^2+1}-a$ is monotonically decreasing we get $a =c$
hence $b = - a$
now from second putting $a=\sec\,t$

$\sec\,t + \sec\,t \,\cot\,t = \dfrac{35}{12}$

or $\dfrac{\sin\,t + \cos\,t}{\sin\,t\cos\,t}= \dfrac{35}{12}$
square both sides and put $\sin\,t\cos\,t= y$

to get $\dfrac{1+2y}{y^2} = \dfrac{1225}{144}$
add 1 on both sides to get
$\dfrac{1+2y+y^2}{y^2} = \dfrac{1369}{144}$
take square root of both sides knowing that y is positive

$\dfrac{1+y}{y} = \dfrac{37}{12}$
or $y = \dfrac{12}{25}$

$\sin\,t \cos\,t = \dfrac{12}{25}$

as $(\dfrac{3}{5})^2 + (\dfrac{4}{5})^2 = 1 $

and product is $\dfrac{12}{25}$

$\sin\,t = \dfrac{3}{5} \, \cos\,t =\dfrac{4}{5}$

or

$\cos \,t = \dfrac{3}{5} \, \sin\,t =\dfrac{4}{5}$hence
$a = \dfrac{5}{4}, b = - \dfrac{5}{4}$

or
$a = \dfrac{5}{3}, b = - \dfrac{5}{3}$
 
kaliprasad said:
From the 2nd equation we have b < 0 say - c

So from 1 we get $(\sqrt{a^2+1}+a)(\sqrt{c^2+1}-c) = 1$
as we have $(\sqrt{a^2+1}+a)(\sqrt{a^2+1}-a) = 1$
so we get $\sqrt{a^2+1}-a= \sqrt{c^2+1}-c$
as $\sqrt{a^2+1}-a$ is monotonically decreasing we get $a =c$
hence $b = - a$
now from second putting $a=\sec\,t$

$\sec\,t + \sec\,t \,\cot\,t = \dfrac{35}{12}$

or $\dfrac{\sin\,t + \cos\,t}{\sin\,t\cos\,t}= \dfrac{35}{12}$
square both sides and put $\sin\,t\cos\,t= y$

to get $\dfrac{1+2y}{y^2} = \dfrac{1225}{144}$
add 1 on both sides to get
$\dfrac{1+2y+y^2}{y^2} = \dfrac{1369}{144}$
take square root of both sides knowing that y is positive

$\dfrac{1+y}{y} = \dfrac{37}{12}$
or $y = \dfrac{12}{25}$

$\sin\,t \cos\,t = \dfrac{12}{25}$

as $(\dfrac{3}{5})^2 + (\dfrac{4}{5})^2 = 1 $

and product is $\dfrac{12}{25}$

$\sin\,t = \dfrac{3}{5} \, \cos\,t =\dfrac{4}{5}$

or

$\cos \,t = \dfrac{3}{5} \, \sin\,t =\dfrac{4}{5}$hence
$a = \dfrac{5}{4}, b = - \dfrac{5}{4}$

or
$a = \dfrac{5}{3}, b = - \dfrac{5}{3}$

Very nicely done, kaliprasad!:cool:
 
anemone said:
Very nicely done, kaliprasad!:cool:

Thanks, I would like to have a look at another different solution in case you have any
 
kaliprasad said:
Thanks, I would like to have a look at another different solution in case you have any

Nope, my solution is more or less the same as yours, because the trick to solve this problem is to recognize that $a=-b$ and then we have to opt for the trigonometric substitution skill to solve for the rest.

Again, thanks so much for participating in my recent challenges at MHB, kali!
 
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