An algebra problem by Nazmus sakib

here,

$\dfrac{1}{x-6}=-12$

$x=\dfrac{71}{12}$

so, $f(-12)=\dfrac{12 \times \dfrac{71}{12}+12}{12\times \dfrac{71}{12}-8}$

$=\dfrac{83}{63}$

so,

$(a+b)=83+63=146$

Please precise "what is the value of a+b when a/b is irreducible ".

If solving, we shall think the next elements :

• IF $\frac{1}{x-6}$ =-12 THEN $\frac{12x+12}{12x-8}$ =a/b
• seeing that $\frac{1}{x-6}$ =-12 => x=71/12, we SHALL check that x=71/12 => $\frac{1}{x-6}$ =-12 (available because x=71/12 $\neq6$ )
• then x=71/12 => $\frac{1}{x-6}$ =-12 => $\frac{12x+12}{12x-8}$ =a/b so by transitivity x=71/12 => system 2x2

{ x=71/12

{ $\frac{12x+12}{12x-8}=\frac{a}{b}$

Then we get a/b=83/63.

Last check : is it irreducible ? 63=3 $\times$ 3 $\times$ 7 (prime factors), and 83 is no multiple of any of these factors.

(84=7 $\times$ 12 so 7 $\land$ 83=1 ; or other method, reductio ad absurdum : 7|63 ; if 7|83 then 7|(83-63) so 7|20 absurd ; so 83 not multiple of 7)

63/83 is irreducible.

* Calculating is nothing without thinking *