I'm working through some textbook exercise and am unable to solve the following exercise:
Use the basic rules of algebra to simplify the following expression:
$$ab\frac{1}{a}b^2cb^{-3}$$
Within the chapter of the book in question the author has covered:
- Associative, commutative and distributive properties of an equation
- Expanding brackets
- Factoring
- Quadratic factoring
- Completing the square
The solution in the answers section is simply "$c$". The equation simplifies to $c$. I cannot for the life of me see how. The solution provided does not give the steps in between, only the final solution $c$.
How can one arrive at $c$ with all the steps in between?
Best Answer
First off, associativity means we don't need parentheses, which is good, because there are none given to us. Now, let's use the definition of exponents to get $$ ab\frac1abbc\frac1b\frac1b\frac1b $$ Then we use commutativity to arrange things alphabetically. This gives us $$ a\frac1abbb\frac1b\frac1b\frac1bc $$ The definition of fraction gives $a\frac1a=1$ (and similarly for $b$). We get $$ 1\cdot1\cdot1\cdot1\cdot c $$ And finally, by definition of $1$, this simplifies to $c$.