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Chapter 0: Problem 27

Simplify the given expression as much as possible. $$ \frac{m+1}{2}+\frac{3}{n} $$

Short Answer

Expert verified
The simplified expression is \(\frac{(m + 1)n + 6}{2n}\).

Step by step solution

01

Find the common denominator

To find the least common denominator, we will multiply the two denominators together: 2 * n = 2n.
02

Rewrite each fraction with the common denominator

Now, we will rewrite each fraction with the common denominator (2n), being careful to multiply both the numerator and denominator by the necessary values. For the first fraction: \[ \frac{m +1}{2} = \frac{(m +1)(n)}{2n} \] For the second fraction: \[ \frac{3}{n} = \frac{3(2)}{2n} \] The given expression becomes: \[ \frac{(m +1)n}{2n} + \frac{6}{2n} \]
03

Combine the fractions and simplify

Now that both fractions have the same denominator (2n), we can combine them: \[ \frac{(m +1)n + 6}{2n} \] This is the simplified expression.

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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 algebraic fractions, it's crucial to find a common denominator so that you can add or subtract them. The common denominator is a shared multiple of all the denominators involved, allowing you to combine the fractions seamlessly.
  • For two fractions, such as \( \frac{m+1}{2} \) and \( \frac{3}{n} \), the common denominator is found by multiplying the two denominators: \( 2 \times n = 2n \).
  • Choosing the least common denominator (LCD) is important when working with multiple fractions, as it simplifies calculations. However, in this example, simply multiplying the two denominators is straightforward because there are only two of them.
Remember, the main purpose of a common denominator is to create a foundation where both fractions can be added directly, making the next steps of solving these expressions far more manageable.
Fraction Simplification
Simplifying fractions is a key step in dealing with algebraic expressions. Once you've ensured that all fractions have a common denominator, the next move is to simplify the resulting expression by combining like terms and reducing wherever possible.
  • Once you have both fractions with the common denominator \( 2n \), combine the numerators: \( (m+1)n + 6 \).
  • It is important to keep everything over the common denominator: \( \frac{(m+1)n + 6}{2n} \).
  • Simplification is complete when you cannot further reduce the terms without altering the value of the expression.
This step ensures that the expression is as concise as possible, making it easier to use in subsequent calculations.
Numerator and Denominator Multiplication
In algebraic fractions, multiplying the numerator and denominator by the same number doesn't change the fraction's value. This principle allows us to rewrite fractions with a common denominator while keeping equivalency.
  • For \( \frac{m+1}{2} \), multiply both the numerator (\(m+1\)) and the denominator by \(n\) to obtain \( \frac{(m+1)n}{2n} \).
  • Similarly, for \( \frac{3}{n} \), multiply both by \(2\) to get \( \frac{6}{2n} \).
  • This approach expands the fractions without changing their values, perfectly lining them up for addition or subtraction, as shown in the exercise solution.
Through careful multiplication, fractions align under a universal standard (the common denominator), making subsequent operations possible without altering their inherent value.

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Most popular questions from this chapter

Explain why $$ \left|\frac{a}{b}\right|=\frac{|a|}{|b|} $$ for all real numbers \(a\) and \(b\) (with \(b \neq 0\) ).

Simplify the given expression as much as possible. $$ \frac{\frac{1}{x+a}-\frac{1}{x}}{a} $$

Explain how you could show that \(51 \times 49=\) 2499 in your head by using the identity \((a+b)(a-b)=a^{2}-b^{2}\).

The intersection of two sets of numbers consists of all numbers that are in both sets. If \(A\) and \(B\) are sets, then their intersection is denoted by \(A \cap B .\) In Exercises \(31-40,\) write each intersection as a single interval. $$ [-2,8] \cap(-1,4) $$

Determine how many different values can arise by inserting one pair of parentheses into the given expression. $$ 5 \cdot 3 \cdot 2+6 \cdot 4 $$
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