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Chapter 7: Problem 119

Find the missing rational expression. $$\frac{4}{x-2}-\frac{2 x+8}{(x-2)(x+1)}$$

Short Answer

Expert verified
The missing rational expression is \( \frac{x-2}{(x-2)(x+1)} \).

Step by step solution

01

Make the denominators the same

In order to operate between fractions, their denominators must be the same. Currently, the second fraction has \((x-2)(x+1)\) in the denominator, so the first fraction should be rewritten as \( \frac{4(x+1)}{(x-2)(x+1)} \) to have the same denominator.
02

Subtract the fractions

Now that both fractions have the same denominator of \((x-2)(x+1)\), we can subtract them: \( \frac{4(x+1) - (2x+8)}{(x-2)(x+1)} .\)
03

Simplify the subtraction result

Simplify the numerator: \( \frac{4(x+1) - (2x+8)}{(x-2)(x+1)} = \frac{4x+4-2x-8}{(x-2)(x+1)} .\) This simplifies further to \( \frac{2x-4}{(x-2)(x+1)} .\)
04

Simplify result further

Notice that the numerator \(2x - 4\) is divisible by 2. To simplify further, divide both terms in the numerator by 2 to get \( \frac{x-2}{(x-2)(x+1)} .\)

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Key Concepts

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

Subtracting Fractions
Rational expressions often involve fractions, which require careful handling when subtracting. To subtract fractions, their denominators need to be identical. Think of subtracting fractions like combining slices of pizza. You can only subtract if they're cut into the same-sized pieces. In the given exercise, the first fraction \( \frac{4}{x-2} \) and the second fraction \( \frac{2x+8}{(x-2)(x+1)} \) have different denominators.
  • To handle this, we turn both fractions into ones with matching denominators.
  • This way, the subtraction becomes manageable and straightforward.
Once they are under a shared denominator, you simply subtract the numerators and keep the common denominator. Remember, the key to subtracting fractions is matching the denominators first, then moving on to the numerators.
Simplifying Algebraic Expressions
Once you've subtracted the fractions, the next step is to simplify the resulting algebraic expression. Simplification makes things easier and clearer to handle.
  • The expression \( \frac{4(x+1) - (2x+8)}{(x-2)(x+1)} \) first needs the numerators subtracted.
  • This means distributing and combining like terms.
Break it down step-by-step: - Distribute 4 in the first part: \( 4(x+1) \) becomes \( 4x + 4 \).- Subtract the second part: \( 2x + 8 \).- Combine the like terms: \( 4x + 4 - 2x - 8 \) to get \( 2x - 4 \).Simplifying expressions leads to a cleaner, more elegant formula that is often easier to interpret and further manipulate.
Common Denominator
Finding a common denominator is essential when working with multiple fractions. It allows for straightforward arithmetic operations like addition or subtraction. In this context, the common denominator means finding a shared base that all fractions can be expressed in terms of.
  • For the equation \( \frac{4}{x-2} - \frac{2x+8}{(x-2)(x+1)} \), the second fraction already has the denominator \((x-2)(x+1)\).
  • Multiply the first fraction's numerator by \(x+1\) to match this format.
By rewriting \( \frac{4}{x-2} \) as \( \frac{4(x+1)}{(x-2)(x+1)} \), both fractions now have the common denominator \((x-2)(x+1)\). This uniformity is crucial because it allows you to focus solely on the numerators for subtraction.

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Most popular questions from this chapter

perform the indicated operation or operations. Simplify the result, if possible. $$\frac{6 b^{2}-10 b}{16 b^{2}-48 b+27}+\frac{7 b^{2}-20 b}{16 b^{2}-48 b+27}-\frac{6 b-3 b^{2}}{16 b^{2}-48 b+27}$$

What is a proportion? Give an example with your description.

Add or subtract as indicated. Simplify the result, if possible. $$\frac{x-1}{6}+\frac{x+2}{3}$$

denominators are opposites, or additive inverses. Add or subtract as indicated. Simplify the result, if possible. $$\frac{2 x-y}{x-y}+\frac{x-2 y}{y-x}$$

Anthropologists and forensic scientists classify skulls using $$ \frac{L+60 W}{L}-\frac{L-40 W}{L} $$ where \(L\) is the skull's length and \(W\) is its width. a. Express the classification as a single rational expression. b. If the value of the rational expression in part (a) is less than \(75,\) a skull is classified as long. A medium skull has a value between 75 and \(80,\) and a round skull has a value over \(80 .\) Use your rational expression from part (a) to classify a skull that is 5 inches wide and 6 inches long.
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