Problem 70 - Prealgebra and Introductory Algebra

Question

Divide. $$\frac{4 a x-8 a}{c^{2}} \div \frac{2 y-x y}{c^{3}}$$

Step-by-Step Solution

Answer

The simplified answer to the division is \( \frac{4ac}{y} \) if \(x\neq2\), and it's undefined in the contrary case.

Step 1: Write the division as multiplication

Rewrite the original problem by changing division to multiplication and taking the reciprocal of the divisor. Thus, \(\frac{4 a x-8 a}{c^{2}} \div \frac{2 y-x y}{c^{3}}\) becomes \(\frac{4 a x-8 a}{c^{2}} * \frac{c^{3}}{2 y-x y}\).

Step 2: Simplify the fractions

Find common terms in the numerator and denominator of both fractions that can be factored out. In the first fraction, we can factor out \(4a\), leading to \(\frac{4a(x-2)}{c^{2}} * \frac{c^{3}}{2 y-x y}\). In the second fraction, we can factor out \(y\), thus \(\frac{4a(x-2)}{c^{2}} * \frac{c^{3}}{y(2-x)}\).

Step 3: Multiply the fractions

Now multiply the fractions together, giving \(\frac{4a(x-2)c^{3}}{c^{2}y(2-x)}\).

Step 4: Simplify the result

Simplify the result by canceling \(c^{2}\) from both top and bottom, leading to: \( \frac{4a(x-2)c}{y(2-x)} \). Lastly, if \(x\neq2\), then \(-2+x=2-x\), so you can cancel \((x-2)\) from the top with \((2-x)\) from the bottom.

Key Concepts

Factoring in Algebra

Factoring in algebra is one of the essential tools for simplifying expressions or solving equations. It's like undoing multiplication by breaking down a complex expression into simpler parts.
In the given exercise, the expression \(4ax - 8a\) can be factored by identifying the common factor, which is \(4a\). Once extracted, it simplifies to \(4a(x-2)\). Similarly, the expression \(2y - xy\) in the denominator can be factored as \(y(2-x)\) by taking \(y\) as a common factor.

• Factoring helps in breaking down expressions, making multiplication or division easier.
• Always look for common factors in the terms of the expression.
• Simplifying expressions is key to solving algebraic problems efficiently.

Remember, factoring is your friend whenever you need to simplify expressions!

Multiplication of Fractions

When multiplying fractions, the goal is to multiply the numerators together and then the denominators. However, simplification before multiplying can save time.

In the exercise, we dealt with the fractions \(\frac{4a(x-2)}{c^2}\) and \(\frac{c^3}{y(2-x)}\). By simplifying the fractions first, the further steps get more manageable. Once simplified, multiplying becomes straightforward, following these simple steps:

• Multiply the numerators: \(4a(x-2) \cdot c^3\).
• Multiply the denominators: \(c^2 \cdot y(2-x)\).

By simplifying, you reduce potential mistakes and make the operation cleaner. Multiplication of fractions is often crucial in dealing with algebraic expressions, helping in deriving a simplified form quickly and accurately.

Reciprocal in Mathematics

Understanding the concept of reciprocals is crucial when dealing with division in mathematics, especially in fractions. The reciprocal of a number or expression is simply one over that number or expression. It helps in transforming division into multiplication, making calculations easier.
In the original problem, by taking the reciprocal of the divisor \(\frac{2y-xy}{c^3}\), it becomes \(\frac{c^3}{2y-xy}\), allowing the division to be rewritten as multiplication with this new fraction.

• Reciprocal: if \(a\) is a number or an expression, its reciprocal is \(\frac{1}{a}\).
• Converting division into multiplication using reciprocals simplifies the operation considerably.
• Transforming the operation often leads to easier and faster simplification.

The takeaway is that using reciprocals makes algebraic manipulation more straightforward, by changing division into multiplication, allowing you to leverage multiplication techniques effectively.