Problem 14 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Perform the indicated operations and simplify. \(\frac{25 y^{4}}{12 y} \cdot \frac{3 y^{2}}{5 y^{3}}\)

Step-by-Step Solution

Answer

The simplified expression is \(\frac{5y^{2}}{4}\).

Step 1: Multiply the numerators and denominators separately

We start by multiplying the numerators together and the denominators together: \[\frac{25 y^{4}}{12 y} \cdot \frac{3 y^{2}}{5 y^{3}} = \frac{25 y^{4} \cdot 3 y^{2}}{12 y \cdot 5 y^{3}}\]

Step 2: Apply the rule of exponents when multiplying powers

To multiply variables raised to different powers, we add the exponents. \[ \frac{25 y^{4} \cdot 3 y^{2}}{12 y \cdot 5 y^{3}}= \frac{25 \cdot 3 \cdot y^{(4 + 2)}}{12 \cdot 5 \cdot y^{1} \cdot y^{3}} \]

Step 3: Simplify the expression further

Now we can simplify the expression by performing the arithmetic operations: \[ \frac{25 \cdot 3 \cdot y^{(4 + 2)}}{12 \cdot 5 \cdot y^{1} \cdot y^{3}} = \frac{75y^{6}}{60y^{4}} \]

Step 4: Cancel the common factors

Identify the common factors in the numerators and denominators and cancel them: \[ \frac{75y^{6}}{60y^{4}} = \frac{15 \cdot 5y^{6}}{12 \cdot 5y^{4}} \] The common factors are 5 and \(y^{4}\). We can cancel these factors to simplify the fraction: \[ \frac{15 \cdot \cancel{5}y^{6}}{12 \cdot \cancel{5}y^{4}} = \frac{15y^{(6-4)}}{12} \]

Step 5: Perform final simplification

Now we can perform the final simplification by subtracting the exponents as indicated and simplifying the fraction: \[ \frac{15y^{(6-4)}}{12} = \frac{15y^{2}}{12} \] We notice that 15 and 12 have a common factor of 3. We can simplify further by dividing both the numerator and denominator by 3: \[ \frac{15y^{2}}{12} = \frac{5y^{2}}{4} \] The simplified expression is \(\frac{5y^{2}}{4}\).

Key Concepts

Exponents

Exponents represent repeated multiplication of the same number. In algebra, it helps to simplify expressions involving repeated factors. For instance, when we have a variable like \(y\) raised to a power of 4, written as \(y^4\), it means \(y\) is multiplied by itself four times: \(y \cdot y \cdot y \cdot y\).
When multiplying numbers with the same base, we add their exponents. For example, \(y^4\cdot y^2\) becomes \(y^{4+2}\) i.e., \(y^6\). This rule makes multiplication of variables efficient and manageable, leading to simpler expressions.

Fraction Multiplication

Multiplying fractions involves simple steps. First, we multiply the numerators (top numbers) together, and then the denominators (bottom numbers) together. For example, to multiply the fractions \(\frac{nameA}{nameB} \cdot \frac{C}{D}\), we calculate the product as \(\frac{nameA \cdot C}{nameB \cdot D}\).
This rule helps to easily combine fractions. It's essential to properly multiply the terms before attempting simplification. Handling the whole fraction without separating numerators and denominators makes it clearer and avoids confusion.

Common Factors

Common factors are numbers or expressions that divide exactly into both the numerator and the denominator of a fraction. Identifying and canceling out these common factors is key in simplifying fractions. For the expression \(\frac{15 \cdot 5y^{6}}{12 \cdot 5y^{4}}\), the common number \(5\) and the variable term \(y^4\) can be divided out.
Cancelling common factors simplifies the expression, reducing calculation complexity. Once cancelled, we see the fraction in its lowest terms, simplifying the arithmetic to calculate the final answer.

Mathematical Operations

Mathematical operations, such as addition, subtraction, multiplication, and division, form the language of algebra. Multiplication, in particular, streamlines solving expressions with fractions and exponents. Performing each operation step systematically ensures accuracy and clarity.
In algebraic simplification, operations must follow logical steps: simplify within the expression, cancel factors, and rearrange terms where necessary. Always maintaining the order promotes understanding and reduces errors. This structured approach is crucial for proficiently working through complex expressions and arriving at the correct simplified form.