Problem 57 - Intermediate Algebra

Question

Simplify each complex fraction. $$ \frac{\frac{2}{a+1}+\frac{1}{a+3}}{\frac{2 a}{a^{2}+4 a+3}} $$

Step-by-Step Solution

Answer

\( \frac{3a+7}{2a} \)

Step 1: Identify the Inner Fractions

Observe the given complex fraction:\[ \frac{\frac{2}{a+1}+\frac{1}{a+3}}{\frac{2a}{a^{2}+4a+3}} \]We have two fractions in the numerator, \( \frac{2}{a+1} \) and \( \frac{1}{a+3} \), and one fraction in the denominator, \( \frac{2a}{a^2+4a+3} \).

Step 2: Simplify the Denominator

Factor the quadratic expression in the denominator:\[ a^2 + 4a + 3 = (a+1)(a+3) \]Hence, the denominator becomes:\[ \frac{2a}{(a+1)(a+3)} \]

Step 3: Simplify the Numerator

To combine the numerators, \( \frac{2}{a+1} \) and \( \frac{1}{a+3} \), find a common denominator, which is \((a+1)(a+3)\):\[ \frac{2(a+3)}{(a+1)(a+3)} + \frac{1(a+1)}{(a+1)(a+3)} = \frac{2(a+3) + (a+1)}{(a+1)(a+3)} \]Simplify the expression:\[ \frac{2a + 6 + a + 1}{(a+1)(a+3)} = \frac{3a + 7}{(a+1)(a+3)} \]

Step 4: Divide Complex Fraction

Divide the simplified numerator by the simplified denominator:\[ \frac{\frac{3a+7}{(a+1)(a+3)}}{\frac{2a}{(a+1)(a+3)}} = \frac{3a+7}{(a+1)(a+3)} \times \frac{(a+1)(a+3)}{2a} \]The \((a+1)(a+3)\) cancels out:\[ \frac{3a+7}{2a} \]

Step 5: Final Step: Simplify the Result

The simplified form of the complex fraction is obtained by reducing the following:\[ \frac{3a+7}{2a} \]

Key Concepts

Rational Expressions

Rational expressions are fractions where the numerator and the denominator are polynomials. These expressions are similar to numerical fractions but involve variables. In simplifying complex fractions, the aim is to express them in their simplest form, which makes them easier to work with. For example, - If you have a complex fraction like \( \frac{\frac{2}{a+1}+\frac{1}{a+3}}{\frac{2a}{a^{2}+4a+3}} \) as in the exercise, each of the parts is a rational expression. - The numerator involves rational expressions \( \frac{2}{a+1} \) and \( \frac{1}{a+3} \), while the denominator is \( \frac{2a}{a^2+4a+3} \).
Understanding how to work with these expressions is key to simplifying them as seen in the exercise where we obtained \( \frac{3a+7}{2a} \) as the simplified form.

Factoring Quadratics

Factoring involves breaking down a complex expression into the product of simpler factors. It is a crucial skill when dealing with quadratic expressions in rational expressions. In the exercise, the quadratic \( a^2 + 4a + 3 \) is factored into \( (a+1)(a+3) \). This factorization is necessary:- It helps to identify the common denominator quickly, which is crucial in simplifying the complex fraction.- It simplifies both the numerator and the denominator, making the cancellation step possible.
Recognizing patterns such as squares, perfect square trinomials, and difference of squares can aid in efficient factorization. Remember, once you factor a quadratic, rechecking your multiplication can confirm accuracy.

Common Denominator

A common denominator is essential for combining fractions. It is the smallest multiple that the denominators share. In the numerator of our complex fraction, we combine \( \frac{2}{a+1} \) and \( \frac{1}{a+3} \) by finding their common denominator, \((a+1)(a+3)\). Here's why this step is important:- Ensures fractions share the same "base," enabling their numerators to add directly.- Mismatching denominators would prevent simplification and combining steps.
This process of finding a common denominator illuminates the structure of the fractions involved and ultimately aids in simplifying by streamlining complex expressions into manageable parts.

Simplification of Algebraic Expressions

Simplifying algebraic expressions reduces them to their most compact and understandable form. Simplification can involve factors, cancellation, and distribution of terms. In the exercise:- The rational expressions in the numerator are combined into one: \( \frac{3a + 7}{(a+1)(a+3)} \).- Using the multiplication inverse for division, we multiplied by the reciprocal of the denominator to cancel out like terms.
By canceling \((a+1)(a+3)\) from the numerator and the denominator, and reducing to \( \frac{3a+7}{2a} \), we illustrate simplification. This final form is easier to interpret and use in further algebraic work or problems. Simplification often makes complex-looking expressions straightforward, revealing the core truth of mathematical relationships.