Simplify the expression (2 - (a - b)/(a + b)) times (3 - (a + 2b)/(a + b))^-1 times (2a + b) - 3b.

Step-by-step solution

1. 1

   Plan the route. Two brackets each contain a fraction over $a+b$. Combine each bracket into a single fraction, then use the fact that the $-1$ exponent means "take the reciprocal" so the second fraction flips and the two combine into one quotient. The trailing $-3b$ waits until the end.

2. 2

   Combine the first bracket over $a+b$.

   $2 - \frac{a-b}{a+b} = \frac{2(a+b) - (a-b)}{a+b} = \frac{2a+2b-a+b}{a+b} = \frac{a+3b}{a+b}$

3. 3

   Combine the second bracket the same way.

   $3 - \frac{a+2b}{a+b} = \frac{3(a+b) - (a+2b)}{a+b} = \frac{3a+3b-a-2b}{a+b} = \frac{2a+b}{a+b}$

4. 4

   Apply the negative exponent and cancel $a+b$. Since $\left(\dfrac{2a+b}{a+b}\right)^{-1} = \dfrac{a+b}{2a+b}$, the product of the two brackets is

   $\frac{a+3b}{a+b} \cdot \frac{a+b}{2a+b} = \frac{a+3b}{2a+b}$

5. 5

   Multiply by $(2a+b)$ — the denominator cancels exactly. This is the payoff of the design:

   $\frac{a+3b}{2a+b}\,(2a+b) = a+3b$

6. 6

   Subtract the last term.

   $(a+3b) - 3b = a$

   Every trace of $b$ disappears, which is a satisfying and easily checked outcome.

7. 7

   Verify with numbers. Take $a = 2$, $b = 1$, so $a+b = 3$. First bracket: $2 - \tfrac13 = \tfrac53$. Second bracket: $\left(3 - \tfrac43\right)^{-1} = \left(\tfrac53\right)^{-1} = \tfrac35$. Then $\tfrac53 \cdot \tfrac35 \cdot 5 - 3 = 5 - 3 = 2 = a$ ✓. Restrictions: $a+b \neq 0$ and $2a+b \neq 0$.