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Chapter 8: Problem 32

Write each rational expression in lowest terms. $$\frac{10 d-5}{12 d-6}$$

Short Answer

Expert verified
The simplified rational expression in lowest terms is \(\frac{5}{6}\).

Step by step solution

01

Factor the numerator and denominator

First, let's find the common factors of both the numerator and the denominator. We will also try to factor out these common terms from each part of the expression. Numerator: - For the expression \(10d - 5\), we see that both terms are divisible by 5. So, we can factor out 5: $$5(2d -1)$$ Denominator: - For the expression \(12d - 6\), both terms are divisible by 6. So, we can factor out 6: $$6(2d -1)$$
02

Simplify the expression

Now that we have factored both the numerator and the denominator, we can simplify the rational expression. We do this by canceling out the common factors, which in this case is \((2d -1)\). $$\frac{10d - 5}{12d - 6} = \frac{5(2d - 1)}{6(2d - 1)}$$ Since both the numerator and the denominator have a common factor of \((2d - 1)\), we can cancel it out to simplify the expression: $$\frac{5(2d - 1)}{6(2d - 1)} = \frac{5}{6}$$ So, the rational expression in lowest terms is \(\frac{5}{6}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring
Factoring is a crucial step when simplifying rational expressions. It involves breaking down expressions into simpler terms, making it easier to spot common factors. In this exercise, we factored both the numerator and the denominator.
  • For the numerator, \(10d - 5\), the expression can be factored by finding the greatest common factor (GCF), which is 5.
    This results in the factorization: \(5(2d - 1)\).
  • For the denominator, \(12d - 6\), the GCF is 6, leading to the factorization: \(6(2d - 1)\).
Factoring helps to make the process of simplification straightforward, allowing us to easily identify any terms that can be canceled out.
Numerator and Denominator
The terms "numerator" and "denominator" refer to the top and bottom parts of a fraction, respectively. With rational expressions, these components can often be simplified by factoring.
In our exercise:
  • The numerator is \(10d - 5\), which after factoring becomes \(5(2d - 1)\).
  • The denominator is \(12d - 6\), which simplifies through factoring to \(6(2d - 1)\).
Understanding how each part of the expression contributes to the overall fraction is key to simplifying it. Always start by addressing the numerator and denominator to see where common factors exist.
Lowest Terms
Simplifying a rational expression to its lowest terms means reducing it until no common factors are present in both the numerator and the denominator. This makes the expression as simple as possible.
After factoring, we notice that both the numerator and denominator shared a \((2d - 1)\) term. By canceling these out, we are left with the simplified expression: \(\frac{5}{6}\).
It is important to ensure that the expression is indeed in its lowest terms before considering it complete, helping to minimize complexity and potential errors in future calculations.
Common Factors
Common factors are vital when simplifying rational expressions because they are terms that appear in both the numerator and the denominator. Identifying these enables us to cancel them out, simplifying the fraction.
In this exercise, \((2d - 1)\) was a common factor. Canceling out this term from both the top and bottom reduced the fraction significantly.
Remember:
  • Identify any common factors early during the factoring process.
  • Cancel common factors to achieve the simplest form of the expression.
This step is crucial in ensuring you arrive at the most reduced version of a rational expression.

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Most popular questions from this chapter

Solve each equation. $$\frac{8}{3 g^{2}-7 g-6}+\frac{4}{g-3}=\frac{8}{3 g+2}$$

Perform the indicated operations. $$\frac{2 k+7}{k^{2}-4 k}+\frac{9 k}{2 k^{2}-15 k+28}+\frac{15}{2 k^{2}-7 k}$$

Solve each equation. $$\frac{w}{3}=\frac{2 w-5}{12}$$

Perform the indicated operations. $$\frac{n+5}{4 n^{2}+7 n-2}-\frac{n-4}{3 n^{2}+7 n+2}$$

Solve each equation. $$\frac{k^{2}}{3}=\frac{k^{2}+2 k}{4}$$
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