Problem 49 - Algebra and Trigonometry

Question

The given equation is either linear or equivalent to a linear equation. Solve the equation. \(\frac{3}{x+4}=\frac{1}{x}+\frac{6 x+12}{x^{2}+4 x}\)

Step-by-Step Solution

Answer

The equation has no valid solution as \( x = -4 \) leads to undefined expressions.

Step 1: Confirm if the equation is linear

We need to simplify the given equation \( \frac{3}{x+4} = \frac{1}{x} + \frac{6x+12}{x^2+4x} \). Notice that all terms involve some form of rational expression. We should simplify and eliminate the fractions to check its linear nature.

Step 2: Simplify the equation

The equation is \( \frac{3}{x+4} = \frac{1}{x} + \frac{6x+12}{x^2+4x} \). Factor the expression in the last term on the right-hand side: \( \frac{6x+12}{x^2+4x} = \frac{6(x+2)}{x(x+4)} \). So, the equation becomes \( \frac{3}{x+4} = \frac{1}{x} + \frac{6(x+2)}{x(x+4)} \).

Step 3: Find common denominator

The common denominator for the fractions is \( x(x+4) \). Rewrite and combine terms as fractions with this common denominator: \( \frac{3x}{x(x+4)} = \frac{x+4}{x(x+4)} + \frac{6(x+2)}{x(x+4)} \).

Step 4: Combine terms and clear the denominator

Equating the numerators since denominators are the same, we get: \( 3x = (x+4) + 6(x+2) \). Expand and simplify the right-hand side: \( x + 4 + 6x + 12 = 7x + 16 \).

Step 5: Solve the linear equation

With the simplified equation \( 3x = 7x + 16 \), solve for \( x \). Subtract \( 7x \) from both sides: \( 3x - 7x = 16 \), which simplifies to \( -4x = 16 \). Divide both sides by \(-4\) to find \( x = -4 \).

Step 6: Verify solution

Substitute \( x = -4 \) back into the original equation. Note that when \( x = -4 \), both \( \frac{3}{x+4} \) and \( \frac{6(x+2)}{x^2+4x} \) become undefined. Therefore, \( x = -4 \) is not a valid solution as it causes division by zero. Thus, no solution exists.

Key Concepts

Understanding Rational Expressions

Rational expressions are like fractions, but they contain polynomials in their numerators and denominators. For example, in the equation \( \frac{3}{x+4} = \frac{1}{x} + \frac{6x+12}{x^2+4x} \), each term is a rational expression. The key is to identify these expressions and understand how they behave during algebraic manipulation.

• Numerator: The top part of the rational expression, which can be a number, a variable, or a polynomial.
• Denominator: The bottom part, which influences the domain as it cannot be zero, or else the expression will be undefined.

This understanding helps in simplifying them and finding a common denominator, which is crucial to solving equations involving rational expressions.

Finding a Common Denominator

A common denominator is needed when combining rational expressions. Just like adding standard fractions, you can't simply add them straight across without a common denominator. In this equation, \( \frac{3}{x+4} = \frac{1}{x} + \frac{6x+12}{x^2+4x} \), we factor the denominator in preparation to find a common denominator.

• The common denominator helps convert each expression so their denominators match, which allows us to equate numerators directly.
• In our example, \( x(x+4) \) becomes the common denominator after factoring, which allows us to rewrite each fraction accordingly.

By finding this common denominator, we facilitate the process of simplifying the equation and solving for the unknown variable.

Simplifying Equations

Simplifying an equation involves reducing it to its simplest form, making it easier to solve. When you have a common denominator, like \( x(x+4) \), in rational expressions, you can focus on simplifying the numerators. This was done in the example as we equated and expanded the numerators:

• Simplifying helps unravel complex terms and reduces the workload by focusing on solving a simpler form of the equation.
• The goal is to isolate the variable, as seen when \( 3x = 7x + 16 \) simplifies further to \( -4x = 16 \), making it straightforward to solve.

Unchecked simplification errors can lead to incorrect results. Always verify by substituting back into the original equation to catch any errors or undefined terms, ensuring the solution is valid.