Simplifying algebraic fractions through multiplication and cancellation.

Problem:

The task is to simplify the expression:
\[
\frac{12x^2}{5y^3} \cdot \frac{20y^4}{6x^3}
\]

Step-by-Step Solution:

#### 1. Multiply the Numerators and Denominators:
When multiplying two fractions, we multiply the numerators together and the denominators together:
\[
\frac{12x^2}{5y^3} \cdot \frac{20y^4}{6x^3} = \frac{(12x^2) \cdot (20y^4)}{(5y^3) \cdot (6x^3)}
\]

Simplify the numerator and the denominator:
\[
\text{Numerator: } (12x^2) \cdot (20y^4) = 12 \cdot 20 \cdot x^2 \cdot y^4 = 240x^2y^4
\]
\[
\text{Denominator: } (5y^3) \cdot (6x^3) = 5 \cdot 6 \cdot y^3 \cdot x^3 = 30x^3y^3
\]

So, the expression becomes:
\[
\frac{240x^2y^4}{30x^3y^3}
\]

#### 2. Simplify the Coefficients:
Simplify the numerical coefficients in the fraction:
\[
\frac{240}{30} = \frac{240 \div 30}{30 \div 30} = \frac{8}{1} = 8
\]

So, the expression now is:
\[
\frac{8x^2y^4}{x^3y^3}
\]

#### 3. Simplify the Variables Using Exponent Rules:
Use the rule of exponents \( \frac{a^m}{a^n} = a^{m-n} \) to simplify the variables \( x \) and \( y \):

- For \( x \):
\[
\frac{x^2}{x^3} = x^{2-3} = x^{-1} = \frac{1}{x}
\]

- For \( y \):
\[
\frac{y^4}{y^3} = y^{4-3} = y^1 = y
\]

Substitute these back into the expression:
\[
\frac{8x^2y^4}{x^3y^3} = 8 \cdot \frac{1}{x} \cdot y = \frac{8y}{x}
\]

#### 4. Final Answer:
The simplified expression is:
\[
\boxed{\frac{8y}{x}}
\]

Explanation of the Cancellation Process:

- The coefficient \( \frac{240}{30} \) simplifies to 8.
- The \( x \)-terms \( \frac{x^2}{x^3} \) simplify to \( \frac{1}{x} \).
- The \( y \)-terms \( \frac{y^4}{y^3} \) simplify to \( y \).

Thus, the final result is:
\[
\boxed{\frac{8y}{x}}
\]