91Ó°ÊÓ

When you multiply fractions, you simply multiply the numerators together to form the new numerator, and the denominators together to form the new denominator. In our exercise, we are multiplying \( \frac{4x^2}{5y} \) by \( \frac{7y}{12x^4} \).

• For the numerators: \( 4x^2 \times 7y = 28x^2y \)
• For the denominators: \( 5y \times 12x^4 = 60x^4y \)

After the multiplication, you get a single fraction: \( \frac{28x^2y}{60x^4y} \). It makes things easier just to multiply straight across; you don't have to worry about finding a common denominator as you do when adding or subtracting fractions.
Once this is completed, we have to simplify the expression, which can often be done by canceling out common factors.

Simplifying expressions means reducing them to their simplest form. After multiplying the fractions, we're left with \( \frac{28x^2y}{60x^4y} \). Here, you want to identify and cancel out common factors in the numerator and the denominator.

• Both the numerator and denominator have \( x^2 \) and \( y \) as factors, so we can cancel them out. You're left with \( \frac{28}{60x^2} \).

To further reduce, you look for the greatest common divisor (GCD) of 28 and 60, which is 4. Divide both numbers by 4:

• \( 28 \div 4 = 7 \)
• \( 60 \div 4 = 15 \)

Thus, the simplified expression is \( \frac{7}{15x^2} \). Simplification helps to make expressions easier to work with—always a handy skill when dealing with algebraic fractions.

Variable restrictions in rational expressions come from the need to keep the denominator non-zero since division by zero is undefined in mathematics. When working with the expression \( \frac{28x^2y}{60x^4y} \), the variables that appear in the original expression can inform us about restrictions.
The simplification stage often shows variables that can be set to zero from canceling. However, it's critical to note that before cancelation, \( y \) and \( x \) must not be zero, because they were in the denominators originally. After canceling \( y \), it no longer appears in the denominator of the simplified expression. Thus, for this problem, the key restriction we have to note is:

• \( x eq 0 \)

This is important since \( x \) remained in the denominator in the later steps, indicating \( x \) must never be zero to avoid making any part of the fraction undefined. Always think about these restrictions as they ensure the expressions work in all mathematical scenarios where they're used.