Simplify the expression (2 - (a - b)/(a + b)) times (3 - (a + 2b)/(a + b)) to the power -1, times (2a + b), minus 3b.

Step-by-step solution

1. 1

   Read the structure before touching anything. There are three factors multiplied together and then $-3b$ subtracted at the end. The middle factor carries an exponent of $-1$, which simply means reciprocal - it is not a subtraction and it does not distribute over the bracket. So the plan is: simplify each bracket into a single fraction, invert the second one, multiply, and only then deal with the trailing $-3b$.

2. 2

   Turn the first bracket into one fraction. Write the $2$ with denominator $a+b$ so the two terms can be combined:

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

   The minus sign in front of the fraction must reach both terms of $a-b$, which is where the $+b$ comes from.

3. 3

   Do the same with the second bracket, then invert it. Using $3=\dfrac{3(a+b)}{a+b}$:

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

   The exponent $-1$ turns this upside down:

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

4. 4

   Multiply the three factors and cancel. Everything now lines up for cancellation - $a+b$ cancels against $a+b$, and $2a+b$ cancels against the third factor:

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

   This cancellation is exactly why the problem was built with those particular numerators.

5. 5

   Subtract the leftover term. The final $-3b$ wipes out the $3b$ that survived:

   $a+3b-3b=a.$

   So the entire expression is just $a$. It is worth stating the restrictions the original form carried: we need $a+b\neq 0$ (the fractions) and $2a+b\neq 0$ (the $-1$ exponent needs a nonzero base). A numerical spot check with $a=-\tfrac{37}{5}$, $b=\tfrac{38}{9}$ returns $-\tfrac{37}{5}$, matching $a$.