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Chapter 1: Problem 63

Simplify the compound fractional expression. $$\frac{\frac{1}{x-1}+\frac{1}{x+3}}{x+1}$$

Short Answer

Expert verified
\( \frac{2}{(x-1)(x+3)} \)

Step by step solution

01

Identify the Compound Fraction

The given expression is \( \frac{\frac{1}{x-1}+\frac{1}{x+3}}{x+1} \). It is a compound fraction, which is a fraction where the numerator, the denominator, or both, contain fractions themselves.
02

Find a Common Denominator for the Numerator

To simplify the numerator \( \frac{1}{x-1} + \frac{1}{x+3} \), we need to find a common denominator. The common denominator for \( x-1 \) and \( x+3 \) is \((x-1)(x+3)\).
03

Combine the Fractions in the Numerator

Rewrite the numerator as a single fraction: \[ \frac{1}{x-1} + \frac{1}{x+3} = \frac{(x+3) + (x-1)}{(x-1)(x+3)}. \] Simplify the numerator by combining like terms: \( (x+3) + (x-1) = 2x + 2 \). Thus, the numerator becomes \( \frac{2x + 2}{(x-1)(x+3)} \).
04

Simplify the Compound Fraction

Substitute the combined numerator back into the original compound fraction: \( \frac{\frac{2x+2}{(x-1)(x+3)}}{x+1} \). This can be simplified by multiplying by the reciprocal of the denominator: \( \frac{2x+2}{(x-1)(x+3)} \times \frac{1}{x+1} = \frac{2x+2}{(x-1)(x+3)(x+1)} \).
05

Factor and Simplify

Notice that the numerator \( 2x+2 \) can be factored as \( 2(x+1) \): \[ \frac{2(x+1)}{(x-1)(x+3)(x+1)}. \] Cancel out \( x+1 \) from the numerator and denominator to simplify the expression: \[ \frac{2}{(x-1)(x+3)}. \]

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Key Concepts

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

Simplification
Simplifying compound fractions is all about making expressions easier to work with. When you encounter a compound fraction, it means that the problem involves fractions within fractions.
The goal is to transform these intricate expressions into simpler forms. To achieve simplification, follow these key guidelines:
  • Combine fractions inside the numerator or denominator as needed, turning them into a single fraction.
  • Once combined, convert these large fractions into more manageable expressions, often involving factoring or finding common denominators (which we will explain later).
  • Finally, eliminate any unnecessary components by canceling common factors in both the numerator and the denominator.
Understanding simplification allows you to see through the fraction fog, making it much clearer and easier to handle in further calculations.
Common Denominator
Finding a common denominator is like finding a common ground for fractions; it helps you combine them into a single expression. This step is crucial when simplifying compound fractions because it allows you to deal with just one fraction instead of multiple ones.
To find a common denominator, identify what the denominators of the fractions within the compound fraction have in common. Think of it as finding the least common multiple of the denominators.
  • Multiply these denominators together if they have no common factors. This ensures that each fraction within the larger fraction can be converted to an equivalent fraction with this common denominator.
  • In the given exercise, the fractions within the numerator, \(\frac{1}{x-1}\) and \(\frac{1}{x+3}\), require a common denominator of \((x-1)(x+3)\).
  • Using the common denominator simplifies the process of adding, subtracting, or otherwise combining fractions.
This process lays the essential groundwork for further simplifying or factoring the expression.
Factoring
Factoring is akin to breaking down numbers or expressions into their building blocks, which makes them easier to simplify or work with. It involves expressing a number or expression as a product of its factors. In the context of simplifying fractions, it helps in canceling out common elements in the numerator and denominator.
In the compound fraction simplification, factoring works its magic especially when the expression is in its simplest form after finding a common denominator.
  • Take a look at the numerator: If it is possible to express it using common factors, do so. For example, \(2x + 2\) becomes \(2(x+1)\).
  • This step is vital as it often reveals factors that can be canceled out with the denominator, which reduces the fraction further.
  • By removing these factors, you ensure the fraction is in its most reduced form, which in the example becomes \(\frac{2}{(x-1)(x+3)}\).
Factoring not only simplifies the expression but also makes further algebraic manipulations much easier.

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Most popular questions from this chapter

Prove the following Laws of Exponents for the case in which \(m\) and \(n\) are positive integers and \(m>n\) (a) Law \(2: \frac{a^{m}}{a^{n}}=a^{m-n}\) (b) Law \(5:\left(\frac{a}{b}\right)^{n}=\frac{a^{n}}{b^{n}}\)

Decimal Notation Write each number in decimal notation. (a) \(3.19 \times 10^{5}\) (b) \(2.721 \times 10^{8}\) (c) \(2.670 \times 10^{-8}\) (d) \(9.999 \times 10^{-9}\)

DISCUSS: Proof That \(0=1 ?\) The following steps appear to give equivalent equations, which seem to prove that \(1=0 .\) Find the error. \(\begin{aligned} x &=1 & & \text { Given } \\ x^{2} &=x & & \text { Multiply by } x \\ x^{2}-x &=0 & & \text { Subtract } x \end{aligned}$$(x-1)=0\)Factor \(\frac{(x-1)}{x-1}=\frac{0}{x-1} \quad\) Divide by \(x-1$$x=0 \quad\) Simplify\(1=0 \quad\) Given \(x=1\)

Use scientific notation, the Laws of Exponents, and a calculator to perform the indicated operations. State your answer rounded to the number of significant digits indicated by the given data. $$\frac{(73.1)\left(1.6341 \times 10^{28}\right)}{0.0000000019}$$

The Lens Equation If \(F\) is the focal length of a convex lens and an object is placed at a distance \(x\) from the lens, then its image will be at a distance \(y\) from the lens, where \(F, x,\) and \(y\) are related by the lens equation $$\frac{1}{F}=\frac{1}{x}+\frac{1}{y}$$ Suppose that a lens has a focal length of \(4.8 \mathrm{cm}\) and that the image of an object is \(4 \mathrm{cm}\) closer to the lens than the object itself. How far from the lens is the object?
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