Question

3x^2-3x ----------- -3 x^2-9x please give me the simplified answer + explanation

Answer 1

You want us to simplify $\frac{3x^2-3x}{x^2-9x}-3$


first, factor top and bottom
$3x^2-3x=3x(x-1)$
$x^2-9x=x(x-9)$

so we have
$\frac{3x(x-1)}{x(x-9)}-3=$
$\frac{3(x-1)}{x-9}-3$
we want to combine fractions so multiply 3 by $\frac{x-9}{x-9}$
$\frac{3(x-1)}{x-9}-\frac{3(x-9)}{x-9}=$
$\frac{3(x-1)-3(x-9)}{x-9}=$
$\frac{3x-3-3x+27}{x-9}=$
$\frac{24}{x-9}$


the simplified answer is $\frac{24}{x-9}$