we have the expression
[tex]\frac{3x+6}{x^2-9x+20}\colon\frac{x^2+7x+10}{x^2-25}[/tex]
Simplify
3x+6=3(x+2)
x^2-25=(x+5)(x-5)
x^2-9x+20=(x-4)(x-5)
x^2+7x+10=(x+5)(x+2)
substitute in the given expression
[tex]\frac{3(x+2)}{\mleft(x-4\mright)\mleft(x-5\mright)}\colon\frac{\mleft(x+5\mright)\mleft(x+2\mright)}{\mleft(x+5\mright)\mleft(x-5\mright)}[/tex]
Multiply in cross
[tex]\frac{3(x+2)(x+5)(x-5)}{(x-4)(x-5)(x+5)(x+2)}[/tex]
Remember that the denominator can not be equal to zero
so
the domain is all real numbers except for the numbers x=4, x=5, x=-5 and x=-2
Simplify
[tex]\frac{3}{(x-4)}[/tex]
