Problem 42 - Algebra and Trigonometry with Analytic Geometry

Question

Solve the equation. $$\frac{3}{2 x+5}+\frac{4}{2 x-5}=\frac{14 x+3}{4 x^{2}-25}$$

Step-by-Step Solution

Answer

There are no solutions to the equation.

Step 1: Identify a Common Denominator

To solve this equation, we need a common denominator for the terms. Notice that the expression on the right, \(4x^2 - 25\), can be factored as \((2x+5)(2x-5)\). This means \(2x+5\) and \(2x-5\) are factors on the denominators on the left side as well. This helps us combine under one common denominator: \((2x+5)(2x-5)\).

Step 2: Rewrite Each Fraction with the Common Denominator

Rewrite each fraction so that they all have the denominator \((2x+5)(2x-5)\). For \(\frac{3}{2x+5}\), multiply both the numerator and denominator by \(2x-5\). For \(\frac{4}{2x-5}\), multiply both the numerator and denominator by \(2x+5\). The right side is already perfect with this common denominator.

Step 3: Simplify the Equation

Rewrite the equation with the common denominator: \(\frac{3(2x-5) + 4(2x+5)}{(2x+5)(2x-5)} = \frac{14x+3}{(2x+5)(2x-5)}\). Simplify the numerators: \(3(2x-5) = 6x - 15\) and \(4(2x+5) = 8x + 20\).

Step 4: Combine and Simplify Numerators

Combine the terms in the numerator: \((6x - 15) + (8x + 20) = 14x + 5\). Now the equation becomes \(\frac{14x + 5}{(2x+5)(2x-5)} = \frac{14x + 3}{(2x+5)(2x-5)}\).

Step 5: Solve the Numerator Equation

Since the denominators are equal, set the numerators equal to each other: \(14x + 5 = 14x + 3\). Subtract \(14x\) from both sides: \(5 = 3\).

Step 6: Determine the Conclusion

The equation \(5 = 3\) is a contradiction; thus, there are no solutions where this equation holds true.

Key Concepts

Common Denominator

Finding a common denominator is one of the crucial steps when dealing with rational equations. It allows you to combine fractions into a single unified equation, making it easier to solve. In our exercise, the first task was to identify the common denominator. The equation involved terms with different denominators, specifically \(2x+5\) and \(2x-5\). The denominator on the right side, \(4x^2 - 25\), can be factored into \( (2x+5)(2x-5)\). This is crucial because recognizing the factored form lets us treat all parts of the equation under a single denominator.

So, to bring all fractions under a common denominator, we rewrite each fraction with \( (2x+5)(2x-5)\) as the denominator. This means you have to multiply the numerator and denominator of \( \frac{3}{2x+5}\) by \(2x-5\) and \( \frac{4}{2x-5}\) by \(2x+5\). The goal is to have each fraction expressed over this common product, allowing us to combine them seamlessly in subsequent steps.

Factoring Quadratics

Factoring quadratics is an essential skill when solving rational equations, especially when identifying common denominators. Factoring involves expressing a quadratic polynomial as a product of linear factors. In the exercise, the quadratic expression \(4x^2 - 25\) could be seen as a difference of squares.

Recall the identity for the difference of squares: \(a^2 - b^2 = (a-b)(a+b)\). Applying this identity, we factor \(4x^2-25\) as \( (2x-5)(2x+5)\).

• This step is pivotal for the subsequent manipulations as it impacts how terms are handled across the equation.
• By factoring, you permit all terms of the equation to fit seamlessly into a unified frame, the common denominator.

This makes combining terms much more straightforward and sets the stage for simplified equation solving.

Equation Solving

Solving the equation involves isolating terms and manipulating them to find values that satisfy the equation. After ensuring all terms in our rational equation are rewritten using a common denominator, we can focus on the numerators. With the equation transformed, \( \frac{14x + 5}{(2x+5)(2x-5)} = \frac{14x + 3}{(2x+5)(2x-5)}\), the denominators are the same. Therefore, we equate the numerators:

\(14x + 5 = 14x + 3\)

• Setting numerators equal is valid here due to the uniform denominators.
• This means the equation's solution revolves on solving the simplified linear expression.

Yet, in this specific example, solving \(14x + 5 = 14x + 3\) results in a contradiction.

No Solution Equations

Sometimes, while solving equations, you may encounter a situation where no solution exists. This can happen, as we've seen in our calculation, when manipulating the numerators ends in a contradiction. When we subtracted \(14x\) from both sides, we were left with \(5 = 3\).

A contradiction like this demonstrates an impossibility and confirms there are no values for \(x\) that satisfy the original rational equation.

• Such scenarios teach us to interpret each step critically, ensuring we account for all possible results.
• This insight provides more profound understanding on cases where equations don't always yield tangible solutions.

"No solution" acts as a valid conclusion, showing that not all mathematical problems resolve neatly during solving.