Subtract and simplify: (6x + 5)/(x^2 - 8x + 12) - 6/(x - 2).

Step-by-step solution

1. 1

   Factor every denominator first. Never look for a common denominator before factoring — the second denominator is usually already a factor of the first, as it is here:

   $x^{2}-8x+12=(x-2)(x-6),$

   since $(-2)+(-6)=-8$ and $(-2)(-6)=12$.

2. 2

   Read off the least common denominator. The denominators are $(x-2)(x-6)$ and $(x-2)$, so the LCD is simply $(x-2)(x-6)$ — the first denominator already is it. Only the second fraction needs building up, by the missing factor $(x-6)$:

   $\frac{6}{x-2}=\frac{6(x-6)}{(x-2)(x-6)}.$

3. 3

   Subtract over the common denominator, keeping brackets.

   $\frac{6x+5}{(x-2)(x-6)}-\frac{6(x-6)}{(x-2)(x-6)}=\frac{(6x+5)-6(x-6)}{(x-2)(x-6)}.$

4. 4

   Expand the numerator carefully. The minus sign multiplies both terms of $6(x-6)$:

   $(6x+5)-(6x-36)=6x+5-6x+36=41.$

   The $6x$ terms cancel, leaving a pure constant — a strong hint the problem was built to end this way.

5. 5

   Write the simplified result and note the restrictions.

   $\frac{41}{(x-2)(x-6)}=\frac{41}{x^{2}-8x+12},\qquad x\neq 2,\ x\neq 6.$

   Since $41$ is prime and shares no factor with either bracket, this is fully simplified.

6. 6

   Check with a value. At $x=3$ the original is $\frac{23}{9-24+12}-\frac{6}{1}=\frac{23}{-3}-6=-\frac{41}{3}$, and the answer gives $\frac{41}{(1)(-3)}=-\frac{41}{3}$. They match.