Problem 24 - Elementary Algebra

Question

For Problems 1-40, perform the indicated operations and express answers in simplest form. $$ \frac{2 x+1}{6 x+12}-\frac{3 x-4}{8 x+16} $$

Step-by-Step Solution

Answer

\(\frac{-x + 16}{24(x + 2)}\)

Step 1: Factor the Denominators

First, factor the denominators of both fractions. The first denominator is \(6x + 12\), which can be factored as \(6(x + 2)\). The second denominator is \(8x + 16\), which can be factored as \(8(x + 2)\).

Step 2: Determine the Least Common Denominator (LCD)

The least common denominator (LCD) of \(6(x + 2)\) and \(8(x + 2)\) is \(24(x + 2)\), since 24 is the least common multiple of 6 and 8.

Step 3: Rewrite Each Fraction with the LCD

Rewrite each fraction to express them with the LCD. Multiply the numerator and denominator of the first fraction by 4, giving \(\frac{4(2x + 1)}{24(x + 2)}\). Multiply the numerator and denominator of the second fraction by 3, resulting in \(\frac{3(3x - 4)}{24(x + 2)}\).

Step 4: Subtract the Numerators

Subtract the numerators of the rewritten fractions: \(4(2x + 1) - 3(3x - 4)\). This simplifies to \(8x + 4 - 9x + 12\).

Step 5: Simplify the Resulting Expression

Combine like terms in the expression: \(8x + 4 - 9x + 12 = -x + 16\). So the resulting fraction is \(\frac{-x + 16}{24(x + 2)}\).

Step 6: Simplify the Final Expression if Possible

The terms \(-x + 16\) and \(24(x + 2)\) have no common factors to simplify further. Therefore, the simplest form of the expression is \(\frac{-x + 16}{24(x + 2)}\).

Key Concepts

Factoring Polynomials

Polynomials are expressions consisting of variables and coefficients. Factoring these polynomials means expressing them as a product of their factors. This is an essential skill in algebra as it simplifies expressions and aids in solving equations reached during operation combinations like addition, subtraction, or division of fractions.

In the given exercise, the denominators are polynomials: \(6x + 12\) and \(8x + 16\). To factor these, we look for the greatest common factor (GCF) in each expression.

• For \(6x + 12\), both terms are divisible by 6, so it factors to \(6(x + 2)\).
• For \(8x + 16\), both terms are divisible by 8, so it factors to \(8(x + 2)\).

By factoring these denominators, we find common terms, which simplifies finding the Least Common Denominator (LCD) in the next steps.

Least Common Denominator (LCD)

The Least Common Denominator is the smallest multiple that is a common denominator of a set of fractions. It helps transform each fraction so that their denominators are identical, which is crucial for addition or subtraction of fractions.

In this problem, we found the LCD of the two fractions \(\frac{2x+1}{6(x+2)}\) and \(\frac{3x-4}{8(x+2)}\). The denominators \(6(x + 2)\) and \(8(x + 2)\) share the factor \((x + 2)\). By finding the least common multiple of the coefficients 6 and 8, which is 24, we determine the LCD to be \(24(x + 2)\).

• Multiply both the numerator and denominator of \(\frac{2x+1}{6(x+2)}\) by 4 because \(6 \times 4 = 24\).
• Multiply both the numerator and denominator of \(\frac{3x-4}{8(x+2)}\) by 3 because \(8 \times 3 = 24\).

This results in both fractions having the same denominator, \(24(x + 2)\), making them compatible for subtraction.

Simplifying Expressions

Simplifying expressions involves reducing them to their most basic form without changing their value. It helps in making complex algebraic expressions more manageable.

After rewriting the fractions with the common denominator, we subtract their numerators: \(4(2x + 1) - 3(3x - 4)\). You expand the terms in the parentheses:

• \(4(2x + 1)\) becomes \(8x + 4\).
• \(3(3x - 4)\) becomes \(9x - 12\).

Perform the subtraction: \(8x + 4 - 9x + 12\). Simplify by combining like terms which gives \(-x + 16\).

Finally, the expression \(\frac{-x + 16}{24(x + 2)}\) is in its simplest form, as the numerator \(-x + 16\) and the denominator \(24(x + 2)\) do not share any further common factors. Simplifying expressions not only makes them cleaner but also easier to interpret and utilize for further mathematical operations.