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

Step-by-step solution

1. 1

   Work from the inside out. The exponent $-1$ applies to the whole bracket, so that bracket has to be reduced to a single fraction before the reciprocal is taken. Nothing else can be simplified until this is done.

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

2. 2

   Expand the numerator of the inner bracket.

   $3(a+b)-(a+2b)=3a+3b-a-2b=2a+b$

   so the bracket equals $\dfrac{2a+b}{a+b}$. The appearance of $2a+b$ here - the same factor that multiplies the whole term - is the signal that everything is about to cancel.

3. 3

   Invert the bracket. A power of $-1$ on a fraction simply swaps numerator and denominator:

   $\left(\frac{2a+b}{a+b}\right)^{-1}=\frac{a+b}{2a+b}$

4. 4

   Cancel the whole product.

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

   Both $a+b$ and $2a+b$ cancel outright, so a three-factor product collapses to the single binomial $a-b$.

5. 5

   Finish the outer arithmetic.

   $2-(a-b)-3b=2-a+b-3b=2-a-2b$

   The minus sign in front of the bracket flips both signs inside - the last place an error can creep in.

6. 6

   Verify numerically. At $(a,b)=(3,5)$ the original expression evaluates to $-11$ and $2-a-2b=2-3-10=-11$ ✓. At $(a,b)=(2.2,-1.7)$ both give $3.2$ ✓ (valid since $a+b\neq0$ and $2a+b\neq0$).