Simplifying rational expressions

Question

(9a^2 b^3)/(27a^4 b^4 c)

Answer 1

Hi Nick!

(9a²b³)/27a⁴b⁴c

(I'm assuming you haven't learned about negative exponents yet so we won't go that route).

We can only divide apples by apples and oranges by oranges!  For example, b/c remains b/c.  We can't simplify it in any way.  However, b⁵/b³ or b³/b⁵ we CAN simplify.

If we're dividing two numbers which have the same bases but different exponents (for example, in b⁵ the b is the base and the 5 is the exponent), we can take off 1 from the exponent on the top (numerator) base for every 1 we take off the bottom (denominator) base.  For example, if we have b⁵/b³ we can say it's the same as b⁴/b² which is the same as b³/b¹.  That's because when we take off an exponent of the top and bottom, we're really factoring out a b/b which is 1, which doesn't change anything.  In other words, b⁵/b³ = (b/b)(b⁴/b³) = (1)(b⁴/b³) = b⁴/b³.

So, getting to our problem and remembering kiwis can only be divided by kiwis:

The 9 on in the numerator can only be divided by the 27 in the denominator, and 9/27=1/3

The a² the numerator can only be divided by the a⁴ in the denominator which gives you a⁰/a² and any base with a "0" exponent equals one, so you have 1/a²

The b³ in the numerator can only be divided by the b⁴ in the denominator which gives you1/b

There is no c in the numerator so there's no change in the c in the denominator.

So we end up with (1)(1)(1)/(3)(a²)bc = 1/3a²bc for your answer.