Simplify (9a2 - 4)/(2 - 3a) - (6a2 - 5a - 6)/(3 - 2a).

Step-by-step solution

1. 1

   Plan the route: factor everything before finding a common denominator. Both denominators are written 'backwards' relative to the factors that will appear on top, and that reversal is worth exactly a factor of $-1$. Spotting it first avoids a messy common-denominator computation.

2. 2

   Factor and reduce the first fraction. $9\alpha^2-4 = (3\alpha)^2-2^2 = (3\alpha-2)(3\alpha+2)$, while $2-3\alpha = -(3\alpha-2)$. Cancelling the matching factor: $\frac{(3\alpha-2)(3\alpha+2)}{-(3\alpha-2)} = -(3\alpha+2) = -3\alpha-2, \qquad \alpha \neq \tfrac23.$

3. 3

   Factor the second numerator. For $6\alpha^2-5\alpha-6$ find integers with product $6(-6) = -36$ and sum $-5$: those are $4$ and $-9$. Splitting and grouping: $6\alpha^2+4\alpha-9\alpha-6 = 2\alpha(3\alpha+2)-3(3\alpha+2) = (3\alpha+2)(2\alpha-3)$.

4. 4

   Reduce the second fraction. Since $3-2\alpha = -(2\alpha-3)$, $\frac{(3\alpha+2)(2\alpha-3)}{-(2\alpha-3)} = -(3\alpha+2) = -3\alpha-2, \qquad \alpha \neq \tfrac32.$ Both fractions have reduced to the same expression.

5. 5

   Subtract. $(-3\alpha-2)-(-3\alpha-2) = -3\alpha-2+3\alpha+2 = 0.$ The expression is identically zero wherever it is defined.

6. 6

   Record the excluded values and spot-check. The original expression is undefined at $\alpha=\tfrac23$ and $\alpha=\tfrac32$. Substituting $\alpha=0.3$, $1.7$, $-2$ and $5$ all give $0$ to full precision, confirming the identity.