Multiplying and simplifying expressions The expression is:
$$\frac{24a^4b^2c^3}{25xy^2z^5} \cdot \frac{15x^3y^3z^3}{16a^2b^2c^2}$$
What I did was subtract the exponents of the numerator to the exponents of the denominator. I did a cross multiplication too. 
What confuses me is the whole numbers and how i should deal with the negatives.
Please clear it for me.
 A: First, we'll express the product as a single fraction, and use the commutativity of multiplication to "rearrange" terms:
$$\frac{24a^4b^2c^3}{25xy^2z^5} \cdot \frac{15x^3y^3z^3}{16a^2b^2c^2} = \frac{(3\cdot \color{red}{\bf 8})\cdot( 3\cdot \color{blue}{\bf 5})\;a^4b^2c^3\; x^3y^3 z^3}{(5\cdot \color{blue}{\bf 5})\cdot(2\cdot \color{red}{\bf 8})\;a^2b^2c^2\;xy^2z^5}$$
Now, we can cancel common integer factors that appear in both numerator and denominator, and multiply the remaining integer factors:
$$\frac{(3\cdot \color{red}{\bf 8})\cdot( 3\cdot \color{blue}{\bf 5})\;a^4b^2c^3\; x^3y^3 z^3}{(5\cdot \color{blue}{\bf 5})\cdot(2\cdot \color{red}{\bf 8})\;a^2b^2c^2\;xy^2z^5}= \frac{9a^4b^2c^3x^3y^3z^3}{10\;a^2 b^2 c^2 x^2z^5}$$
Now, using the fact that $\dfrac {k^n}{k^m} = k^{n-m}$ for $n \geq m$, and that $\dfrac {k^n}{k^m} = \dfrac 1{k^{m-n}}$ for $n \lt m$, then simplifying, we have:
$$\begin{align} \frac{9a^4b^2c^3x^3y^3z^3}{10\;a^2 b^2 c^2 x^2z^5} & = \frac{9 a^{4-2}b^{2-2}c^{3-2}x^{3-1}y^{3-2}}{10z^{5-3}}\\ \\ & = \frac{9a^2b^0cx^2y}{10z^2} \\ \\ & = \frac{9a^2cx^2y}{10z^2}\end{align}$$

Putting all steps together, we have $$\frac{24a^4b^2c^3}{25xy^2z^5} \cdot \frac{15x^3y^3z^3}{16a^2b^2c^2} = \frac{9a^2cx^2y}{10z^2}$$
A: Subtract exponents of denominator from exponents of numerator.If they are positive put them in the numerator and if they are negative you can either put them in the denominator with positive exponents or put them in the numerator with negative exponents?
Why did you cross-multiply?
A simple example
$$\frac{x^3}{a^2}\frac{a^5}{x^5}$$
$$\text{exponent of }x =3-5=-2$$
$$\text{exponent of }a= 5-2=3$$
So the expression simplified would be
$$\frac{a^3}{x^2}$$ or$${a^3x^{-2}}$$
As for the whole numbers , in the numerator you have 24 which is $4\times 6$ and 15 which is $5 \times 3$ .In the denominator , you have 25 which is $5 \times 5$ and 16, which is $4 \times 4$. Cross off( remove) the commons from numerator and denominator.
