Problem 51 - Elementary and Intermediate Algebra

Question

Divide, and then simplify, if possible. \(\frac{x^{2}+7 x}{5 x-10} \div \frac{(x+7)^{2}}{15 x-30}\)

Step-by-Step Solution

Answer

The simplified form is \(\frac{3x}{x+7}\).

Step 1: Understand Division of Fractions

When dividing fractions such as \(\frac{a}{b} \div \frac{c}{d}\), rewrite it as multiplication by the reciprocal: \(\frac{a}{b} \times \frac{d}{c}\). This means you invert the second fraction and change the division to multiplication.

Step 2: Rewrite the Expression

Rewrite the given division problem as a multiplication by the reciprocal: \(\frac{x^2 + 7x}{5x - 10} \times \frac{15x - 30}{(x+7)^2}\).

Step 3: Factor the Numerators and Denominators

Factor the expressions wherever possible:- \(x^2 + 7x\) factors to \(x(x + 7)\).- \(5x - 10\) factors to \(5(x - 2)\).- \(15x - 30\) factors to \(15(x - 2)\).

Step 4: Write the Factored Expression

Substitute the factored forms into the expression:\[\frac{x(x + 7)}{5(x - 2)} \times \frac{15(x - 2)}{(x+7)^2}\]

Step 5: Cancel Common Factors

Cancel the common factors in the numerators and denominators. The term \((x+7)\) appears in both the numerator and one of the denominators, and \((x-2)\) is present in both denominators. After cancelation, the expression becomes:\[\frac{x \cdot 15}{5(x+7)}\]

Step 6: Simplify the Expression

Simplify the expression by performing the multiplication and reducing fractions:\[\frac{15x}{5(x+7)} = \frac{3x}{x+7}\].

Step 7: Final Simplified Form

The expression cannot be simplified further since \(x + 7\) and \(3x\) do not have any common factors. Thus, the final simplified form is \(\frac{3x}{x+7}\).

Key Concepts

Factorization

Factorization is a technique used in algebra to simplify expressions or solve equations by writing them as a product of their factors. It involves expressing a polynomial as a product of its linear or simpler polynomial factors. In the original exercise, factorization is a crucial part of solving the problem, helping to simplify the algebraic fractions involved.

To factor a polynomial, look for the greatest common factor (GCF) in its terms. For instance, take the expression \(x^2 + 7x\). Here, \(x\) is a common factor. Factor it out to get:

• \(x(x + 7)\)

Similarly, for the expression \(5x - 10\), the GCF is \(5\). Factoring gives:

• \(5(x - 2)\)

The same method applies to the expression \(15x - 30\) with a GCF of \(15\):

• \(15(x - 2)\)

Discovering these common factors and rewriting expressions in terms of them simplifies calculations and helps identify terms that can be canceled.

Simplifying Fractions

Simplifying fractions involves reducing them to their simplest form where the numerator and denominator share no common factors other than 1. In algebraic fractions, this often involves canceling out terms that appear in both the numerator and the denominator.

In the original problem, once the expressions have been factored,

• \(x(x + 7)\)
• \(5(x - 2)\)
• \(15(x - 2)\)
• \((x+7)^2\)

common factors are observed. The step where we substitute these factored forms shows this:

\(\frac{x(x + 7)}{5(x - 2)} \times \frac{15(x - 2)}{(x+7)^2}\)

We identify the common factors: \((x+7)\) and \((x-2)\). These can be canceled from both the numerator and denominator, allowing us to simplify the fraction further. By removing shared factors, the fractions become easier to manage and solve.

Reciprocal Multiplication

Reciprocal multiplication is a method used to simplify division of fractions by converting it into multiplication. When we have two fractions, the division of one fraction by another is equivalent to multiplying by the reciprocal of the second fraction.

In this exercise, we start with \(\frac{x^{2}+7x}{5x-10} \div \frac{(x+7)^{2}}{15x-30}\). To solve it, we use reciprocal multiplication. Invert the second fraction and change the operation to multiplication:

• \(\frac{x^{2}+7x}{5x-10} \times \frac{15x-30}{(x+7)^2}\)

This step is crucial as it converts what could be a complex division problem into a simpler multiplication problem.

Once we have this multiplication of fractions expression, we can factorize, simplify, and eventually solve more straightforwardly. This technique is vital in algebra to handle fractions efficiently and avoid errors in computation or cancellation.