Answer
$\dfrac{5}{4n^{2}-12n+8}-\dfrac{3}{3n^{2}-6n}=\dfrac{n+4}{4n(n-1)(n-2)}$
Work Step by Step
$\dfrac{5}{4n^{2}-12n+8}-\dfrac{3}{3n^{2}-6n}$
Factor the denominator of the first fraction and take out common factor $3n$ from the denominator of the second fraction:
$\dfrac{5}{4n^{2}-12n+8}-\dfrac{3}{3n^{2}-6n}=\dfrac{5}{4(n-1)(n-2)}-\dfrac{3}{3n(n-2)}=...$
Evaluate the substraction of the two rational expressions:
$...=\dfrac{5(3n)-3(4)(n-1)}{(4)(3n)(n-1)(n-2)}=\dfrac{15n-12n+12}{12n(n-1)(n-2)}=...$
$...=\dfrac{3n+12}{12n(n-1)(n-2)}=...$
Take out common factor $3$ from the numerator and simplify:
$...=\dfrac{3(n+4)}{12n(n-1)(n-2)}=\dfrac{n+4}{4n(n-1)(n-2)}$
