Problem 28 - Introductory Algebra for College Students

Question

Multiply as indicated. $$\frac{2 y}{3 y-y^{2}} \cdot \frac{2 y^{2}-9 y+9}{8 y-12}$$

Step-by-Step Solution

Answer

The solution to the multiplication of the fractions is \(\frac{y^2 - 3y}{6 (1-y) }\).

Step 1: Rewrite and Simplify the Problem

Rewrite the expression by factoring the numerator of the second fraction and the denominator of the first fraction. Expression can be rewritten as: \(\frac{2 y}{3 y(1-y)} \cdot \frac{(y-3)^2}{4(2 y-3)}\)

Step 2: Cancel Out Common Factors

Identify and cancel out common factors in the numerators and denominators of the fractions. Hence, we get: \(\frac{2 y}{3 (1-y)} \cdot \frac{(y-3)}{4}\)

Step 3: Perform the Multiplication

Multiply the numerators with each other and the denominators with each other to get the result: \( \frac{2 y \cdot (y-3)}{3 (1-y) \cdot 4 }\)

Step 4: Simplify the Result

Simplify further to obtain the final answer: \(\frac{(2 y \cdot (y-3))}{12 (1-y) } = \frac{y^2 - 3y}{6 (1-y) } \)