((a+3b)/(a+b−1))+((a+3b−1)/(a+b−3))=4 a+b=?


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Question Number 197683 by sulaymonnorboyev140 last updated on 26/Sep/23

$\frac{a + 3 b}{a + b - 1} + \frac{a + 3 b - 1}{a + b - 3} = 4$ $a + b = ?$
Commented by mr W last updated on 26/Sep/23

$n o u n i q u e s o l u t i o n f o r a + b .$ $p l e a s e c h e c k y o u r q u e s t i o n!$
	Answered by Frix last updated on 26/Sep/23
	$This is equal to ((2(a−((11)/4))^2 )/3)−((2(b+(3/4))^2 )/3)=1∧b≠1−a∧b≠3−a Which is the equation of a hyperbola without the points ((3/2), −(1/2)) and (4,−1) a+b∈R\{2}$
	$This is equal to$ $\frac{2 {(a - \frac{11}{4})}^{2}}{3} - \frac{2 {(b + \frac{3}{4})}^{2}}{3} = 1 \land b \neq 1 - a \land b \neq 3 - a$ $Which is the equation of a hyperbola without$ $the points (\frac{3}{2}, - \frac{1}{2}) and (4, - 1)$ $a + b \in R ∖ {2}$
	Answered by ajfour last updated on 26/Sep/23

	$a + b = p a - b = q$ $\frac{2 p - q}{p - 1} + \frac{2 p - q - 1}{p - 3} = 4$ $\Rightarrow (2 p - q) (p - 3) + (2 p - q - 1) (p - 1)$ $= 4 (p - 1) (p - 3)$ $(16 - 6 - q - 2 - q - 1) p$ $= 12 + 3 q + q + 1$ $\Rightarrow (7 - 2 q) p = 13 + 4 q$ $p = a + b = \frac{13 + 4 q}{7 - 2 q} = \frac{27 - (14 - 4 q)}{7 - 2 q}$ $a + b = \frac{27}{7 - 2 q} - 2$
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