Problem 59
Question
For the following problems, add or subtract the rational expressions. $$ \frac{3 a+5}{(a+4)(a-1)}-\frac{2 a-1}{a-1} $$
Step-by-Step Solution
Verified Answer
Question: Subtract the given rational expressions and simplify:
$$
\frac{3a+5}{a + 4} - \frac{2a-1}{a - 1}
$$
Answer:
$$
\frac{-2a^2+10a+1}{(a+4)(a-1)}
$$
1Step 1: Identify the Common Denominator
The common denominator for these two rational expressions will be the least common multiple (LCM) of their denominators, which are \((a+4)(a-1)\) and \(a-1\). Since the first denominator already contains \(a-1\), the LCM is simply \((a+4)(a-1)\).
2Step 2: Rewrite the Expressions with the Common Denominator
Rewrite both fractions with a common denominator, \((a+4)(a-1)\), in order to perform the subtraction:
$$
\frac{3a+5}{(a+4)(a-1)}-\frac{(2a-1)(a+4)}{(a-1)(a+4)}
$$
3Step 3: Subtract the Numerators
Now that both fractions have the common denominator, subtract the numerators and keep the denominator the same:
$$
\frac{(3a+5) - [(2a-1)(a+4)]}{(a+4)(a-1)}
$$
4Step 4: Distribute and Simplify the Numerator
Distribute the terms in the numerator and simplify the expression:
$$
\frac{(3a+5) - [(2a^2+7a-4)]}{(a+4)(a-1)}
$$
$$
\frac{-2a^2+7a-4+3a+5}{(a+4)(a-1)}
$$
$$
\frac{-2a^2+10a+1}{(a+4)(a-1)}
$$
5Step 5: Final Answer
The final answer, after adding and simplifying the given rational expressions, is:
$$
\frac{-2a^2+10a+1}{(a+4)(a-1)}
$$
Key Concepts
Subtracting Rational ExpressionsCommon DenominatorPolynomial SimplificationLeast Common Multiple
Subtracting Rational Expressions
Rational expressions are similar to fractions but involve polynomials in the numerator and the denominator. Subtracting rational expressions is much like subtracting fractions in arithmetic. We first ensure that both expressions have a common denominator. Once the common denominator is established, the key step is to subtract the numerators while keeping the denominator the same.
In the exercise, we started off with two expressions: \( \frac{3a+5}{(a+4)(a-1)} \) and \( \frac{2a-1}{a-1} \). Notice they initially do not share a common denominator. To subtract them effectively, the first task is to align their denominators. This is where the concept of least common multiple comes into play.
In the exercise, we started off with two expressions: \( \frac{3a+5}{(a+4)(a-1)} \) and \( \frac{2a-1}{a-1} \). Notice they initially do not share a common denominator. To subtract them effectively, the first task is to align their denominators. This is where the concept of least common multiple comes into play.
Common Denominator
A common denominator is essential when combination operations, such as subtraction, are performed on rational expressions. A common denominator is preferably the least common multiple (LCM) of the individual denominators. It ensures that each rational expression is comparable on the same scale.
In this instance, the common denominator is \((a+4)(a-1)\), which was already part of the first rational expression. The process involved multiplying the second expression's missing factor \((a+4)\) to its numerator and denominator, making them consistent. Thus, we successfully converted \(\frac{2a-1}{a-1}\) to \(\frac{(2a-1)(a+4)}{(a+4)(a-1)}\). Now, they have a shared base, the common denominator, enabling straightforward subtraction.
In this instance, the common denominator is \((a+4)(a-1)\), which was already part of the first rational expression. The process involved multiplying the second expression's missing factor \((a+4)\) to its numerator and denominator, making them consistent. Thus, we successfully converted \(\frac{2a-1}{a-1}\) to \(\frac{(2a-1)(a+4)}{(a+4)(a-1)}\). Now, they have a shared base, the common denominator, enabling straightforward subtraction.
Polynomial Simplification
When subtracting rational expressions, simplifying the resulting expression is crucial. After aligning the denominators, focus shifts to the numerators. Subtracting the numerators leads to a new polynomial expression that often requires simplification to its simplest form.
In this exercise, after keeping the common denominator the same, the subtraction of the numerators is performed: \((3a+5) - [(2a-1)(a+4)]\). We need to expand the terms within the brackets and combine like terms. Notice that every step of this expansion involves careful arithmetic operations which convert the subtraction into a final simplified form: \(-2a^2 + 10a + 1\). This simplification step ensures clarity in the resulting expression, making it concise and easy to interpret.
In this exercise, after keeping the common denominator the same, the subtraction of the numerators is performed: \((3a+5) - [(2a-1)(a+4)]\). We need to expand the terms within the brackets and combine like terms. Notice that every step of this expansion involves careful arithmetic operations which convert the subtraction into a final simplified form: \(-2a^2 + 10a + 1\). This simplification step ensures clarity in the resulting expression, making it concise and easy to interpret.
Least Common Multiple
The least common multiple (LCM) plays a pivotal role in subtracting rational expressions, acting as the bridge to unify separate expressions under a common denominator. The LCM of the expression's denominators determines what each fraction needs to be multiplied by to align them.
Here, the denominators \((a+4)(a-1)\) and \(a-1\) were evaluated. Since \((a+4)(a-1)\) already contained \(a-1\), the LCM was determined to be \((a+4)(a-1)\). By achieving this through multiplication and adjustment, the individual rational expressions are harmonized, closing the gap between discontinuous terms. Identifying this LCM ensures that no part of the expressions is overlooked, allowing seamless operation execution such as subtraction in this exercise.
Here, the denominators \((a+4)(a-1)\) and \(a-1\) were evaluated. Since \((a+4)(a-1)\) already contained \(a-1\), the LCM was determined to be \((a+4)(a-1)\). By achieving this through multiplication and adjustment, the individual rational expressions are harmonized, closing the gap between discontinuous terms. Identifying this LCM ensures that no part of the expressions is overlooked, allowing seamless operation execution such as subtraction in this exercise.
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Problem 59
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