Problem 49
Question
For the following exercises, simplify the rational expression. \(\frac{\frac{2 c}{c+2}+\frac{c-1}{c+1}}{\frac{2 c+1}{c+1}}\)
Step-by-Step Solution
Verified Answer
The simplified expression is \(\frac{3c^2 + 3c - 2}{(c+2)(2c+1)}\).
1Step 1: Identify and Factorize Denominators
Examine the given rational expression: \( \frac{\frac{2c}{c+2} + \frac{c-1}{c+1}}{\frac{2c+1}{c+1}} \). We need a common denominator to combine the fractions in the numerator. The denominators for the fractions in the numerator are \(c+2\) and \(c+1\).
2Step 2: Find a Common Denominator for the Numerator
The common denominator for \( \frac{2c}{c+2} + \frac{c-1}{c+1} \) is \((c+2)(c+1)\). Rewrite each fraction with this common denominator: \[ \frac{2c(c+1) + (c-1)(c+2)}{(c+2)(c+1)}. \]
3Step 3: Simplify the Numerator
Distribute and simplify the numerators:\[ 2c(c+1) = 2c^2 + 2c, \]\[ (c-1)(c+2) = c^2 + 2c - c - 2 = c^2 + c - 2. \]Combine these results:\[ 2c^2 + 2c + c^2 + c - 2 = 3c^2 + 3c - 2. \]
4Step 4: Simplify the Fraction
Plug the simplified numerator back into the expression:\[ \frac{3c^2 + 3c - 2}{(c+2)(c+1)}. \] Then divide this by the denominator of the original expression: \(\frac{2c+1}{c+1}\). Recall that dividing by a fraction is equivalent to multiplying by its reciprocal. So, the expression becomes:\[ \frac{3c^2 + 3c - 2}{(c+2)(c+1)} \times \frac{c+1}{2c+1}. \]
5Step 5: Cancel Common Factors
Cancel the \(c+1\) terms, which are common in the numerator and denominator. The expression now reduces to:\[ \frac{3c^2 + 3c - 2}{(c+2)(2c+1)}. \]
6Step 6: Finalize the Simplified Expression
The rational expression is fully simplified as it stands. No further factorization is possible without specific values of \(c\). Thus, the simplified expression is:\[ \frac{3c^2 + 3c - 2}{(c+2)(2c+1)}. \]
Key Concepts
Algebraic FractionsCommon DenominatorFactoringReciprocal
Algebraic Fractions
An algebraic fraction is essentially a fraction where the numerator and the denominator are algebraic expressions. It is similar to a regular fraction, but it involves variables in its terms. Understanding algebraic fractions is vital in simplifying rational expressions, as it lays the groundwork for operations like addition, subtraction, multiplication, and division of expressions that involve variables.
In our exercise, fractions like \( \frac{2c}{c+2} \) and \( \frac{c-1}{c+1} \) are examples of algebraic fractions. Both have polynomials in their numerators and denominators. Simplifying them involves working with these algebraic components, making it crucial to grasp the concept of algebraic fractions before proceeding with complex operations such as finding a common denominator or factoring.
In our exercise, fractions like \( \frac{2c}{c+2} \) and \( \frac{c-1}{c+1} \) are examples of algebraic fractions. Both have polynomials in their numerators and denominators. Simplifying them involves working with these algebraic components, making it crucial to grasp the concept of algebraic fractions before proceeding with complex operations such as finding a common denominator or factoring.
Common Denominator
Finding a common denominator is a key step when adding or subtracting algebraic fractions. Just as in numerical fractions, it allows us to combine fractions into a single fraction. By converting fractions to have the same denominator, we can seamlessly perform operations on them.
In the exercise, we identify the denominators \( c+2 \) and \( c+1 \) for the fractions involved in the numerator. The common denominator is \( (c+2)(c+1) \), which incorporates both individual denominators. By rewriting each fraction to have this common denominator, we can add their numerators and consolidate them into a single algebraic fraction.
In the exercise, we identify the denominators \( c+2 \) and \( c+1 \) for the fractions involved in the numerator. The common denominator is \( (c+2)(c+1) \), which incorporates both individual denominators. By rewriting each fraction to have this common denominator, we can add their numerators and consolidate them into a single algebraic fraction.
Factoring
Factoring involves rewriting a mathematical expression as a product of its factors. In the context of simplifying algebraic fractions, factoring helps break down complex expressions to make simplification easier, especially when dealing with polynomials.
For example, in our exercise, we factor expressions such as \( c^2 + c - 2 \) that result from expanding the numerator expressions. Finding common factors or recognizing patterns, like differences of squares or trinomial squares, enables us to simplify the fraction effectively. This skill is invaluable, as it aids in reducing terms and potentially canceling common factors between numerators and denominators.
For example, in our exercise, we factor expressions such as \( c^2 + c - 2 \) that result from expanding the numerator expressions. Finding common factors or recognizing patterns, like differences of squares or trinomial squares, enables us to simplify the fraction effectively. This skill is invaluable, as it aids in reducing terms and potentially canceling common factors between numerators and denominators.
Reciprocal
A reciprocal refers to flipping a fraction—making the numerator the denominator, and vice versa. This concept is especially relevant in division of fractions, as dividing by a fraction is equivalent to multiplying by its reciprocal.
In the exercise, dividing \( \frac{3c^2 + 3c - 2}{(c+2)(c+1)} \) by \( \frac{2c+1}{c+1} \) prompts us to multiply by the reciprocal instead. Therefore, the expression becomes \( \frac{3c^2 + 3c - 2}{(c+2)(c+1)} \times \frac{c+1}{2c+1} \). Understanding reciprocals allows us to simplify complex rational expressions effectively and continue with the reduction process, which ultimately helps in solving the problem.
In the exercise, dividing \( \frac{3c^2 + 3c - 2}{(c+2)(c+1)} \) by \( \frac{2c+1}{c+1} \) prompts us to multiply by the reciprocal instead. Therefore, the expression becomes \( \frac{3c^2 + 3c - 2}{(c+2)(c+1)} \times \frac{c+1}{2c+1} \). Understanding reciprocals allows us to simplify complex rational expressions effectively and continue with the reduction process, which ultimately helps in solving the problem.
Other exercises in this chapter
Problem 48
The Gross Domestic Product (GDP) for the United States in the first quarter of 2014 was \(\$ 1.71496 \times 10^{13}\). Rewrite the GDP in standard notation.
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For the following exercises, simplify the expression. \(18 y-2(1+7 y)\)
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For the following exercises, factor the polynomials. \(5 z(2 z-9)^{-\frac{3}{2}}+11(2 z-9)^{-\frac{1}{2}}\)
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For the following exercises, multiply the polynomials. \((4 t-x)(t-x+1)\)
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