Problem 33
Question
For exercises \(25-68\), evaluate or simplify. $$ \frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{x+3}} $$
Step-by-Step Solution
Verified Answer
\( \frac{x+3}{2x+3} \)
1Step 1 - Identify the Given Expression
The given expression is \[ \frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{x+3}} \. \]
2Step 2 - Simplify the Denominator
The denominator is \[ \frac{1}{x} + \frac{1}{x+3} \]. To add these fractions, find a common denominator: \[ \frac{1}{x} + \frac{1}{x+3} = \frac{(x+3) + x}{x(x+3)} = \frac{2x+3}{x(x+3)} \.\]
3Step 3 - Substituting Back into the Expression
Substitute the simplified denominator back into the original expression: \[ \frac{\frac{1}{x}}{\frac{2x+3}{x(x+3)}} \. \]
4Step 4 - Divide the Fractions
Simplify the complex fraction by multiplying by the reciprocal: \[ \frac{\frac{1}{x}}{\frac{2x+3}{x(x+3)}} = \frac{1}{x} \times \frac{x(x+3)}{2x+3} = \frac{x(x+3)}{x(2x+3)} \. \]
5Step 5 - Cancel Common Factors
Cancel the common factor \( x \) in the numerator and denominator: \[ \frac{x(x+3)}{x(2x+3)} = \frac{x+3}{2x+3} \] Social Trick: Ensure the cancelled factor \( x eq 0 \) to avoid division by zero.
Key Concepts
algebraic fractionscommon denominatorscomplex fractionssimplification process
algebraic fractions
Algebraic fractions are fractions where the numerator, denominator, or both are algebraic expressions. These are often polynomial expressions. Simplifying algebraic fractions involves methods similar to simplifying numerical fractions. You need to factor and cancel common terms. This is essential to simplify without changing the value of the fraction.
For example, consider the fraction \(\frac{2x^2}{4x}\). To simplify this fraction, we first identify common factors. Both the numerator and the denominator have a common factor of \(2x\). Dividing both by \(2x\), we get a simplified form, which is \(\frac{x}{2}\).
Understanding how to manage algebraic fractions is crucial when working with more complex expressions, as you will repeatedly use these principles.
For example, consider the fraction \(\frac{2x^2}{4x}\). To simplify this fraction, we first identify common factors. Both the numerator and the denominator have a common factor of \(2x\). Dividing both by \(2x\), we get a simplified form, which is \(\frac{x}{2}\).
Understanding how to manage algebraic fractions is crucial when working with more complex expressions, as you will repeatedly use these principles.
common denominators
When adding or subtracting fractions, finding a common denominator is vital. This process allows you to combine the fractions into a single fraction. A common denominator is a shared multiple of the denominators of the fractions you're working with.
Take the example from our problem: \(\frac{1}{x} + \frac{1}{x+3}\). To add these fractions, you need a common denominator. Here, it would be \(x(x+3)\), which is the product of the individual denominators. Rewrite each fraction with the common denominator:
Knowing how to find common denominators is essential for simplifying algebraic fractions accurately and efficiently.
Take the example from our problem: \(\frac{1}{x} + \frac{1}{x+3}\). To add these fractions, you need a common denominator. Here, it would be \(x(x+3)\), which is the product of the individual denominators. Rewrite each fraction with the common denominator:
- \(\frac{1}{x} = \frac{x+3}{x(x+3)}\)
- \(\frac{1}{x+3} = \frac{x}{x(x+3)}\)
Knowing how to find common denominators is essential for simplifying algebraic fractions accurately and efficiently.
complex fractions
Complex fractions are fractions that contain other fractions in the numerator, denominator, or both. Simplifying complex fractions involves several steps, including finding a common denominator, simplifying the inner fractions, and then dealing with the overall fraction.
Consider the given problem: \(\frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{x+3}}\). To handle this, we first simplify the denominator. As previously discussed, we find the common denominator and combine the fractions: \(\frac{1}{x} + \frac{1}{x+3} = \frac{2x+3}{x(x+3)} \). Now our complex fraction looks like this: \(\frac{\frac{1}{x}}{\frac{2x+3}{x(x+3)}}\).
The next step is to eliminate the complex structure by multiplying by the reciprocal. This means we multiply \(\frac{1}{x}\) by the reciprocal of the denominator: \(\frac{x(x+3)}{2x+3}\). Simplifying, we get \(\frac{1}{x} \times \frac{x(x+3)}{2x+3} = \frac{x(x+3)}{x(2x+3)}\). Finally, cancel the common factor \(x\), leaving us with \( \frac{x+3}{2x+3} \). Complex fractions can look daunting, but breaking them into simpler steps makes the process manageable.
Consider the given problem: \(\frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{x+3}}\). To handle this, we first simplify the denominator. As previously discussed, we find the common denominator and combine the fractions: \(\frac{1}{x} + \frac{1}{x+3} = \frac{2x+3}{x(x+3)} \). Now our complex fraction looks like this: \(\frac{\frac{1}{x}}{\frac{2x+3}{x(x+3)}}\).
The next step is to eliminate the complex structure by multiplying by the reciprocal. This means we multiply \(\frac{1}{x}\) by the reciprocal of the denominator: \(\frac{x(x+3)}{2x+3}\). Simplifying, we get \(\frac{1}{x} \times \frac{x(x+3)}{2x+3} = \frac{x(x+3)}{x(2x+3)}\). Finally, cancel the common factor \(x\), leaving us with \( \frac{x+3}{2x+3} \). Complex fractions can look daunting, but breaking them into simpler steps makes the process manageable.
simplification process
The simplification process is crucial when working with algebraic fractions, as it helps you reduce expressions to their simplest form. This typically involves the following steps:
Remember, the goal of simplification is to make expressions easier to work with while preserving their value. Mastery of these steps allows you to handle increasingly complex algebraic fractions.
- Identify and factor all expressions whenever possible.
- Find common denominators to combine fractions when necessary.
- Cancel common factors in the numerator and denominator.
- Be cautious of restrictions like not dividing by zero.
Remember, the goal of simplification is to make expressions easier to work with while preserving their value. Mastery of these steps allows you to handle increasingly complex algebraic fractions.
Other exercises in this chapter
Problem 32
For exercises \(5-48\), simplify. $$ \frac{v^{2}}{2 v-14}-\frac{9 v-14}{2 v-14} $$
View solution Problem 32
For exercises 7-32, simplify. $$ \frac{z^{2}-4 z+6}{z^{2}-9} \cdot \frac{z^{2}+7 z+12}{z^{2}-1} $$
View solution Problem 33
For exercises 27-34, evaluate. $$ \frac{15}{19}-\frac{8}{11} $$
View solution Problem 33
For exercises \(5-48\), simplify. $$ \frac{x^{2}}{x^{2}-x-12}-\frac{2 x+15}{x^{2}-x-12} $$
View solution