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Chapter 7: Problem 21

Find the sum or difference. \(\frac{3}{x+4}-\frac{1}{x+6}\)

Short Answer

Expert verified
The difference of the given fractions is \(\frac{2x+14}{(x+4)(x+6)}\)

Step by step solution

01

- Find the Common Denominator

The common denominator of the two fractions, \(\frac{3}{x+4}\) and \(\frac{1}{x+6}\), can be found by multiplying the two denominator expressions, giving us \((x+4)(x+6)\) as the common denominator.
02

- Rewrite the Fractions with the Common Denominator

Rewrite the fractions with the common denominator. The first fraction becomes \(\frac{3(x+6)}{(x+4)(x+6)}\) and the second fraction becomes \(\frac{1(x+4)}{(x+4)(x+6)}\).
03

- Perform the Subtraction

Now subtract the second fraction from the first one: \(\frac{3(x+6)}{(x+4)(x+6)} - \frac{1(x+4)}{(x+4)(x+6)}\).
04

- Simplify the Numerator

Expand and simplify the numerator: \(\frac{3x+18-x-4}{(x+4)(x+6)}\) leads to \(\frac{2x+14}{(x+4)(x+6)}\).
05

- Simplify Further if Possible

Look for the potential to simplify the result further. In this case, no further simplification is possible.

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Key Concepts

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

Common Denominator
When dealing with rational expressions, finding a common denominator is crucial for operations like addition and subtraction. This is because it creates a shared baseline that allows you to directly operate on the numerators. To find a common denominator, consider the denominators of both expressions involved.

For the exercise, the denominators are
  • \(x + 4\)
  • \(x + 6\)
The simplest way to find a common denominator is to multiply these two expressions, resulting in \((x+4)(x+6)\).

By forming this product, you effectively create a structure that can support both fractions. This allows them to "talk" in the same language, so to speak, setting the stage for the upcoming subtraction.
Fraction Subtraction
Once a common denominator is established, you're ready to perform subtraction. Fraction subtraction is similar to integer subtraction but focuses on the numerators since the denominators are now a match. Begin by rewriting each fraction so they both have the common denominator.

In this case,
  • Rewrite the first fraction as \( \frac{3(x+6)}{(x+4)(x+6)} \)
  • Rewrite the second fraction as \( \frac{1(x+4)}{(x+4)(x+6)}\)
Now, you subtract the second numerator from the first while keeping the common denominator intact:
\[\frac{3(x+6)}{(x+4)(x+6)} - \frac{1(x+4)}{(x+4)(x+6)}\]
Think of the denominator as a constant for this step, simplifying your task to merely dealing with what's above the line.
Expression Simplification
Expression simplification is the art of making math expressions as neat and tidy as possible. After subtracting the numerators, you鈥檒l often have an expression that can be simplified further, which is the case here. The initial step involves expanding and combining like terms within the numerator.
  • First, expand: \(3(x + 6) = 3x + 18\) and \(1(x + 4) = x + 4\).
  • After the expansion, subtract: \(3x + 18 - (x + 4)\) results in \(3x + 18 - x - 4\).
  • Combine like terms: \(3x - x = 2x\) and \(18 - 4 = 14\), giving the numerator \(2x + 14\).

Now, your fraction looks like:\[\frac{2x + 14}{(x+4)(x+6)}\]At this stage, always check for further simplification. Simplifying might involve factoring or reducing expressions, but in this case, \(2x + 14\) can't be factored in relation to the denominator, so this is your simplified result.

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Most popular questions from this chapter

In Exercises 33-40, rewrite the function in the form \(g(x)=\frac{a}{x-h}+k\). Graph the function. Describe the graph of \(g\) as a transformation of the graph of \(f(x)=\frac{a}{x}\). $$ g(x)=\frac{4 x-11}{x-2} $$

In Exercises 33-40, rewrite the function in the form \(g(x)=\frac{a}{x-h}+k\). Graph the function. Describe the graph of \(g\) as a transformation of the graph of \(f(x)=\frac{a}{x}\). $$ g(x)=\frac{9 x-3}{x+7} $$

The Doppler effect occurs when the source of a sound is moving relative to a listener, so that the frequency \(f_{\ell}\) (in hertz) heard by the listener is different from the frequency \(f_s\) (in hertz) at the source. In both equations below, \(r\) is the speed (in miles per hour) of the sound source. Moving away: Approaching: $$ f_{\ell}=\frac{740 f_s}{740+r} \quad f_{\ell}=\frac{740 f_s}{740-r} $$ a. An ambulance siren has a frequency of 2000 hertz. Write two equations modeling the frequencies heard when the ambulance is approaching and when the ambulance is moving away. b. Graph the equations in part (a) using the domain \(0 \leq r \leq 60\). c. For any speed \(r\), how does the frequency heard for an approaching sound source compare with the frequency heard when the source moves away?

You plan a trip that involves a 40 -mile bus ride and a train ride. The entire trip is 140 miles. The time (in hours) the bus travels is \(y_1=\frac{40}{x}\), where \(x\) is the average speed (in miles per hour) of the bus. The time (in hours) the train travels is \(y_2=\frac{100}{x+30}\). Write and simplify a model that shows the total time \(y\) of the trip.

In Exercises 3-10, graph the function. Compare the graph with the graph of \(f(x)=\frac{1}{x}\). $$ g(x)=\frac{-12}{x} $$
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