Problem 39
Question
For the following problems, replace \(N\) with the proper quantity. $$ \frac{2 m}{m-1}=\frac{N}{(m-1)(m+2)} $$
Step-by-Step Solution
Verified Answer
Question: Replace the variable N in the equation: $$\frac{2m}{m-1} = \frac{N}{(m-1)(m+2)}$$
Answer: N = 2m^2 + 4m
1Step 1: Identify the common denominator
First, we need to find the common denominator for the two fractions. Since both have m-1, (m-1)(m+2) can serve as the common denominator.
2Step 2: Rewrite the left side with the new denominator
Now that we have identified the common denominator, we can rewrite the left side of the equation with it. We must multiply both the numerator and the denominator of the left side by (m+2) so that the denominators match:
$$
\frac{2 m}{m-1} \cdot \frac{m+2}{m+2} = \frac{2 m (m + 2)}{(m - 1)(m + 2)}
$$
3Step 3: Simplify the left side
Next, we will simplify the numerator of the left side by distributing the 2m:
$$
\frac{2 m (m + 2)}{(m - 1)(m + 2)} = \frac{2 m^2 + 4 m}{(m - 1)(m + 2)}
$$
4Step 4: Compare the numerators
Now that both sides of the equation have the same denominator, we can find the value of N by comparing the numerators:
$$
N = 2 m^2 + 4 m
$$
Thus, the correct answer to this problem is:
$$
\frac{2 m}{m-1}=\frac{2 m^2 + 4 m}{(m-1)(m+2)}
$$
Key Concepts
Common DenominatorSimplify NumeratorEquation Solving
Common Denominator
Understanding the concept of a common denominator is crucial when working with fractions, particularly algebraic fractions. A common denominator refers to a shared multiple of all the denominators in a given set of fractions. In the context of algebra, it can often involve variables.
The idea is to have the same bottom number (denominator) for every fraction involved to simplify the process of addition, subtraction, or comparison. This step is foundational because, without a common denominator, it becomes difficult to directly compare or combine fractions.
For instance, if you have two algebraic fractions with denominators of \(m-1\) and \(m+2\), their common denominator would be the product of these two, \(m-1)(m+2)\). This is because each of the original denominators divides evenly into the new common denominator, ensuring that we have equivalent fractions that can then be easily compared or combined.
When finding a common denominator, look for the least common multiple (LCM) of the denominators if they are numerical. For algebraic expressions, factor each denominator if necessary, and then multiply each unique factor the greatest number of times it occurs in any of the denominators. This will give you the least common denominator.
The idea is to have the same bottom number (denominator) for every fraction involved to simplify the process of addition, subtraction, or comparison. This step is foundational because, without a common denominator, it becomes difficult to directly compare or combine fractions.
For instance, if you have two algebraic fractions with denominators of \(m-1\) and \(m+2\), their common denominator would be the product of these two, \(m-1)(m+2)\). This is because each of the original denominators divides evenly into the new common denominator, ensuring that we have equivalent fractions that can then be easily compared or combined.
When finding a common denominator, look for the least common multiple (LCM) of the denominators if they are numerical. For algebraic expressions, factor each denominator if necessary, and then multiply each unique factor the greatest number of times it occurs in any of the denominators. This will give you the least common denominator.
Simplify Numerator
Once we have established the common denominator, our next task is usually to simplify the numerator. Simplifying the numerator makes the expression cleaner and often easier to work with, especially when solving equations.
When simplifying the numerator of an algebraic fraction, as in the given exercise, what we actually do is multiply out any parentheses and combine like terms if possible. In the example provided, the numerator is simplified by distributing \(2m\) into the parenthesis \(m+2\), resulting in \(2m^2 + 4m\).
Here's why simplification is essential:
To simplify a numerator with multiple terms, distribute any multipliers on the outside of parentheses across the terms inside, combine any like terms, and if possible, factor the resulting expression. This is a powerful tool in algebra that can turn complex-looking expressions into more manageable forms.
When simplifying the numerator of an algebraic fraction, as in the given exercise, what we actually do is multiply out any parentheses and combine like terms if possible. In the example provided, the numerator is simplified by distributing \(2m\) into the parenthesis \(m+2\), resulting in \(2m^2 + 4m\).
Here's why simplification is essential:
- It reveals the underlying structure of the algebraic expression.
- It reduces computational mistakes in further steps of solving the problem.
- It aids in revealing potential factors that could be used for canceling out terms with the denominator.
To simplify a numerator with multiple terms, distribute any multipliers on the outside of parentheses across the terms inside, combine any like terms, and if possible, factor the resulting expression. This is a powerful tool in algebra that can turn complex-looking expressions into more manageable forms.
Equation Solving
Solving equations is a fundamental skill in algebra that involves finding the values of the unknown variables that make the equation true. The process often requires a series of steps that includes simplifying expressions, isolating the variable, and performing the same operations on both sides of the equation to maintain balance.
To solve equations with algebraic fractions, we have to apply the principle of maintaining equal values on both sides. In the example at hand, once we obtain the same common denominator for both sides, we essentially have fractions that, mathematically, must be equivalent. This allows us to focus solely on the numerators, as the denominators are already equal.
By setting the simplified numerators equal to each other, we derive an equation that we can solve through further simplification or, if needed, more advanced methods like factoring, completing the square, or using the quadratic formula.
In the textbook exercise, after simplifying and attaining the same denominator, we find that \(N = 2m^2 + 4m\), which is the simplified form of the numerator after the previous steps. We then set this expression as equivalent to the numerator on the other side of the equation, which effectively solves for \(N\). Equation solving often involves this strategic maneuvering to isolate the variable in question, leading to the solution.
To solve equations with algebraic fractions, we have to apply the principle of maintaining equal values on both sides. In the example at hand, once we obtain the same common denominator for both sides, we essentially have fractions that, mathematically, must be equivalent. This allows us to focus solely on the numerators, as the denominators are already equal.
By setting the simplified numerators equal to each other, we derive an equation that we can solve through further simplification or, if needed, more advanced methods like factoring, completing the square, or using the quadratic formula.
In the textbook exercise, after simplifying and attaining the same denominator, we find that \(N = 2m^2 + 4m\), which is the simplified form of the numerator after the previous steps. We then set this expression as equivalent to the numerator on the other side of the equation, which effectively solves for \(N\). Equation solving often involves this strategic maneuvering to isolate the variable in question, leading to the solution.
Other exercises in this chapter
Problem 39
For the following problems, fill in the missing term. $$ -\frac{2 x+7}{5 x-1}=\frac{\underline{\phantom{xx}}}{5 x-1} $$
View solution Problem 39
For the following problems, perform the multiplications and divisions. $$ x^{3} \cdot \frac{4 a b}{x} $$
View solution Problem 39
For the following problems, add or subtract the rational expressions. $$ \frac{b+7}{b-6}-\frac{b-1}{b-6} $$vvvvvvvvvvv
View solution Problem 39
For the following problems, reduce each rational expression to lowest terms. $$ \frac{14(x-4)^{3}(x-10)^{6}}{-7(x-4)^{2}(x-10)^{2}} $$
View solution