91Ó°ÊÓ

Algebraic expressions are the cornerstone of algebra and involve numbers, variables, and arithmetic operations. In our exercise, we encounter the expression \(\frac{2 y}{3 y-y^{2}} \cdot \frac{2 y^{2}-9 y+9}{8 y-12}\). To make sense of it, we start by breaking down the polynomials into their factored forms. Factoring is a technique used to simplify expressions and make the multiplication, addition, or subtraction of polynomials easier.

When given a complex expression such as \(\frac{2 y}{3 y-y^{2}}\), identifying common factors is essential. Here, \(y\) is a common factor in the denominator, which helps us to rewrite the expression as \(\frac{2y}{y(3-y)}\). Similarly, recognizing a perfect square like \(2 y^{2}-9 y+9\) helps us factor it to \(2y-3)^{2}\). By focusing on these simplifications, students can tackle polynomials with less intimidation and with a clearer step-by-step approach.

Simplifying rational expressions involves reducing them to their simplest form. This is often achieved by canceling common factors between the numerator and the denominator. In our exercise's Step 2, the simplification process entails identifying and canceling common terms. The term \(2y-3\) appears in both the numerator and denominator of the second fraction, allowing for cancellation.

Similarly, the variable \(y\) is present in both the numerator of the first fraction and the denominator of the second fraction, again allowing for cancellation. The simplified product, \(\frac{2(y-3)}{4(3-y)}\), shows the importance of this step. Simplifying not only makes calculations more accessible but also helps to reduce any rational expression to its most concise form, making it easier to interpret and use in further mathematical computations.

The multiplication of ratios, or fractions, is a practical application of arithmetic in algebra. The process requires the multiplication of the numerators across to determine the new numerator and the denominators to determine the new denominator. In our final step, the multiplication of \(2(y-3)\) and \(4(3-y)\) yields \(2(3-y)\) as their common factors have been canceled out.

It’s important to note when a negative sign is involved, as in \(y-3\) versus \(3-y\). Since these expressions represent the same numbers in reverse order, multiplying them introduces a negative sign. This results in \(2(3-y)\), simplifying the expression to the final answer \(\roughly-2(y-3)\). Understanding the multiplication of ratios is crucial as it lays the foundation for dealing with more complex equations in algebra and beyond.