Problem 70 - Elementary Algebra

Question

For the following problems, perform the multiplications and divisions. $$ \frac{-6 x^{3}}{5 y^{2}} \cdot \frac{20 y}{-2 x} $$

Step-by-Step Solution

Answer

Question: Multiply and simplify the given fractions: $$\frac{-6 x^{3}}{5 y^{2}} \cdot \frac{20 y}{-2 x}$$ Answer: After multiplying and simplifying the given fractions, the result is: $$\frac{12 x^{2}}{y}$$.

Step 1: Multiply the numerators and denominators separately

To multiply the two given fractions, we need to multiply the numerators with each other and the denominators with each other. $$ \frac{-6 x^{3}}{5 y^{2}} \cdot \frac{20 y}{-2 x} = \frac{(-6 x^{3})(20 y)}{(5 y^{2})(-2 x)} $$

Step 2: Multiply the terms

Now we need to multiply the terms in the numerators and denominators. $$ \frac{(-6 x^{3})(20 y)}{(5 y^{2})(-2 x)} = \frac{-120 x^{3}y}{-10 x y^{2}} $$

Step 3: Simplify and cancel out common terms

In this step, we will simplify the expression by canceling out common terms in the numerator and the denominator. Notice that: - Both the numerator and the denominator have a negative sign, so we can remove the negative signs. - x is common in both the numerator and the denominator, so we will cancel out x by dividing them. - We can simplify the fraction -120/10 by dividing both by 10. The simplified fraction looks like: $$ \frac{120 x^{3}y}{10 x y^{2}} = \frac{12 x^{2}}{y} $$ So, the final simplified expression is: $$ \frac{-6 x^{3}}{5 y^{2}} \cdot \frac{20 y}{-2 x} = \frac{12 x^{2}}{y} $$

Key Concepts

Multiplication of Fractions

When it comes to multiplying algebraic fractions, the process follows the basic rules of multiplication of regular fractions. The key steps involve multiplying across the numerators and then the denominators. Consider the fractions \( \frac{-6x^{3}}{5y^{2}} \) and \( \frac{20y}{-2x} \).

Here's how you can approach such problems:

• Multiply the numerators: Combine \(-6x^{3}\) and \(20y\) to get \((-6x^{3}) \times (20y) = -120x^{3}y\).
• Multiply the denominators: Combine the terms \(5y^{2}\) and \(-2x\). It results in \((5y^{2}) \times (-2x) = -10xy^{2}\).

It is crucial to maintain the signs during these calculations, as they directly impact the outcome. For instance, a negative times a negative becomes positive, which will be helpful when simplifying later. So, remember the pattern:

• Negative × Negative = Positive
• Negative × Positive = Negative

Simplifying Expressions

Simplifying expressions involves reducing them to their most efficient form, making them easier to interpret and work with. In our earlier step where we achieved \(\frac{-120x^{3}y}{-10xy^{2}}\), simplifying is about identifying common factors in the numerator and the denominator and canceling them out.

• The negative signs in both the numerator \(-120x^{3}y\) and the denominator \(-10xy^{2}\) cancel each other out, simplifying the fraction further.
• Next, observe any common algebraic terms: Here, \(x\) is seen in both the numerator and denominator. You can reduce \(x^{3} \div x = x^{2}\).

Then, focus on the coefficients: \(\frac{120}{10} = 12\).

This type of simplification refines the expression into \(\frac{12x^{2}}{y}\), making it tidy and straightforward.

Polynomial Division

Polynomial division involves simplifying expressions where a polynomial in the numerator (which can involve multiple terms) is divided by the polynomial in the denominator. While our exercise focused on fractions, polynomial division can involve similar methods of reducing terms.

Imagine you are given an expression like \(\frac{120x^{3}y}{10xy^{2}}\). You want to simplify by treating it akin to polynomial long division:

• Identify terms common in polynomials: Here, \(x\) and \(y\) appears in both the numerator and the denominator.
• For the \(x\) terms, dividing powers involves subtracting exponents: Thus, \(x^{3} \div x = x^{2}\).
• Similarly, \(y \div y^{2} = \frac{1}{y}\) reducing the expression.

Through polynomial division, we effectively cancelled variables to derive \(\frac{12x^{2}}{y}\), an efficient expression that a learner can easily interpret and use in further calculations or problem-solving scenarios.