Question

Can someone my answers

Answer 1

Answer:

Option 2 is correct that is $\frac{x+2}{x+9}$  and $x\neq-9$and $x\neq-12$

Step-by-step explanation:

We have been given with the expression $\frac{x^2-10x-24}{x^2-3x-108}$

We will simplify  the given expression by factorisation we will get $\frac{x^2-12x+2x-24}{x^2-12x+9x-108}$  

Taking x as common from first two terms in numerator and 2 from last two terms in numerator

Similarly, take x common from first two terms and 9 from last two terms in denominator we will get

$\frac{x(x-12)+2(x-12)}{x(x-12)+9(x-12)}$

After arranging the terms we will get $\frac{(x+2)(x-12)}{(x+9)(x-12)}$

Taking out the common factor which is (x-12)  from numerator and denominator it will get cancelled we will get

$\frac{(x+2)}{(x+9)}$

Hence, Option 2 is correct that is $\frac{x+2}{x+9}$  and $x\neq-9$and $x\neq-12$