Problem 106 - College Algebra

Question

Solve each equation in Exercises \(83-108\) by the method of your choice. $$\frac{1}{x}+\frac{1}{x+3}=\frac{1}{4}$$

Step-by-Step Solution

Answer

The solution for the equation is \(x=3\).

Step 1: Eliminate Fractions

Multiply the entire equation by \(4x(x+3)\) to eliminate the fractions. This yields \(4(x+3) + 4x = x(x+3)\).

Step 2: Simplify the Equation

Now open the brackets and simplify: \(4x+12+4x=x^2+3x\). Rearranging the terms, we obtain: \(x^2 -5x -12=0\).

Step 3: Solve for x

Factor the equation: \((x-3)(x+4)=0\). So the solutions are \(x=3\) and \(x=-4\). However, plugging \(x=-4\) back into the original equation, we see that it results in a denominator of zero in one of the fractions, which is not allowed. So the only acceptable solution is \(x=3\).

Key Concepts

Fraction Elimination

In algebra, solving equations that contain fractions often begins with eliminating those fractions to simplify the task. Fraction elimination in equations is a useful technique because it enables us to work with cleaner, integer-based equations, avoiding the additional complexity that fractions introduce.

Consider the equation: \(\frac{1}{x} + \frac{1}{x+3} = \frac{1}{4}\). The goal here is to eliminate the fractions by finding a common denominator. Instead of manipulating each fraction directly, you can multiply the entire equation by the least common multiple (LCM) of all denominators involved.

For the given equation, the LCM of \(x\), \(x+3\), and \(4\) is \(4x(x+3)\). By multiplying each term in the equation by \(4x(x+3)\), the fractions are wiped out, giving us:

- \(4(x+3)\) from \(\frac{1}{x}\)
- \(4x\) from \(\frac{1}{x+3}\)
- \(x(x+3)\) from \(\frac{1}{4}\)

This process simplifies the equation to manageable terms without fractions, making the subsequent steps lighter.

Quadratic Equations

Once the fractions have been eliminated, you are often left with an equation in a standard form — typically a quadratic equation. A quadratic equation is a polynomial equation of the second degree, usually in the form \(ax^2 + bx + c = 0\). These equations can be recognizable by the presence of the \(x^2\) term, which signifies a parabolic equation.

In our scenario, after simplifying, the equation \(4x + 12 + 4x = x^2 + 3x\) reduced to a quadratic form: \(x^2 - 5x - 12 = 0\). Recognizing and transforming your equation into this form is crucial since it allows us to apply specific solving techniques suited to quadratics.

These methods include:

• Factoring: Ideal if the quadratic can be easily expressed as a product of binomials.
• Using the Quadratic Formula: A universal method applicable in all scenarios.
• Completing the Square: Useful for converting the equation to a perfect square trinomial.

In our step-by-step solution, the factoring method was applied as it was straightforward for this specific equation.

Factoring

When working with quadratic equations, factoring offers a straightforward method of finding solutions by representing the equation as a product of simpler expressions.

For the equation \(x^2 - 5x - 12 = 0\), the task is to express it in the form \((x - p)(x + q) = 0\). Factoring relies on identifying two numbers whose product equals the constant term (\(-12\)) and whose sum equals the linear coefficient (\(-5\)).

In this case, the pair \((-3)\) and \(4\) meets these criteria:

• The product of \((-3)\) and \(4\) is \(-12\).
• The sum of \((-3)\) and \(4\) is \(-5\).

Hence, the equation factors neatly into \((x-3)(x+4) = 0\). Solving these factors separately gives the solutions \(x = 3\) and \(x = -4\).

However, we must return to the original equation to verify these solutions, ensuring no term results in a zero denominator. Here, \(x = -4\) becomes invalid due to the division by zero issue, leaving \(x = 3\) as the only valid solution. This vetting step is crucial in validating solutions when factoring.