Problem 6 - Beginning and Intermediate Algebra

Question

Find each product and simplify if possible. $$ -\frac{9 x^{3} y^{2}}{18 x y^{5}} \cdot y^{3} $$

Step-by-Step Solution

Answer

The simplified product is \( -\frac{1}{2}x^{2} \).

Step 1: Simplify the Expression

Start by simplifying the fraction \(-\frac{9 x^{3} y^{2}}{18 x y^{5}}\). Find common factors in the numerator and the denominator. Notice that \(-9\) and \(18\) have a common factor of \(9\), \(x^{3}\) and \(x^{1}\) have \(x\) in common, and \(y^{2}\) and \(y^{5}\) have \(y^{2}\) in common. Divide each part by its common factor to simplify: \[-\frac{9x^{3}y^{2}}{18xy^{5}} = -\frac{1}{2} \cdot x^{2} \cdot y^{-3}.\]

Step 2: Multiply by the Remaining Expression

Now that we have simplified \(-\frac{1}{2}x^{2}y^{-3}\), multiply it by the remaining expression \(y^{3}\). The expression becomes \[-\frac{1}{2}x^{2}y^{-3} \cdot y^{3}.\]

Step 3: Apply the Laws of Exponents

When multiplying powers of the same base, add their exponents. In the expression \(y^{-3} \cdot y^{3}\), add the exponents: \(-3 + 3 = 0\). Thus, \(y^{-3} \cdot y^{3} = y^{0} = 1\). The expression now becomes \[-\frac{1}{2}x^{2} \cdot 1 = -\frac{1}{2}x^{2}.\]

Step 4: Write the Final Simplified Expression

After multiplication and simplification, the final simplified expression is \[-\frac{1}{2}x^{2}.\]

Key Concepts

Simplifying Fractions

When simplifying fractions, we essentially reduce the fraction to its simplest form. This means that the numerator and the denominator have no common factors other than 1. Let's break this down:

• First, identify any common factors the numerator and denominator share. Here, we have 9 in \(-9x^{3}y^{2}\) and 18 in \(18xy^{5}\). The greatest common factor is 9.
• Similarly, for the variables, both \(x^{3}\) and \(x\) include \(x\) as a common factor. Thus, we can divide both by \(x\).
• For the \(y\) terms, \(y^{2}\) and \(y^{5}\) have \(y^{2}\) as a common factor, which can be eliminated from both the numerator and the denominator.

Thus, after dividing through by these common factors, the expression \(-\frac{9x^{3}y^{2}}{18xy^{5}}\) simplifies to \(-\frac{1}{2}x^{2}y^{-3}\). This simplified form makes the expression more manageable for further calculations.

Laws of Exponents

The laws of exponents are fundamental to manipulating and simplifying expressions involving powers. Here's a quick guide on the most relevant rules for this exercise:

• When multiplying expressions with the same base, such as \(y^{-3}\) and \(y^{3}\), add the exponents together: \((-3) + 3 = 0\).
• The result \(y^{0}\) equals 1, thanks to the zero exponent rule, as any number raised to the power of zero is always 1.
• When you have negative exponents, it indicates a reciprocal. For example, \(y^{-3} = \frac{1}{y^{3}}\). However, when added to a positive exponent of the same number, the negative and positive will essentially "cancel" each other out, leading to a zero exponent as seen here.

By applying the laws of exponents to our expression \(-\frac{1}{2}x^{2}y^{-3} \cdot y^{3}\), we significantly simplify the multiplication of terms.

Multiplying Variables

Multiplying variables involves combining like terms, using the laws of exponents to simplify such operations. Here's how we proceed:

• In our expression \(-\frac{1}{2}x^{2}y^{-3} \cdot y^{3}\), notice \(y^{-3}\) and \(y^{3}\) are like terms with the same base, which means you can add their exponents.
• As calculated, \(y^{-3} \cdot y^{3} = y^{0} = 1\). This basically removes \(y\) from the equation, helping simplify it further.
• For the \(x\) term, there's only one such term, \(x^{2}\), in the expression, which remains unaffected by any operations. So, the term \(x^{2}\) remains as it is.

After multiplying and applying the laws of exponents correctly, our expression is reduced to its simplest form, \(-\frac{1}{2}x^{2}\), demonstrating how variable multiplication can be straightforward when using the right rules.