Problem 43 - Algebra and Trigonometry with Analytic Geometry

Question

Solve the equation. $$\frac{5}{2 x+3}+\frac{4}{2 x-3}=\frac{14 x+3}{4 x^{2}-9}$$

Step-by-Step Solution

Answer

No solution; \(x = \frac{3}{2}\) causes division by zero.

Step 1: Equate Common Denominators

The given equation is \( \frac{5}{2x+3} + \frac{4}{2x-3} = \frac{14x+3}{4x^2-9} \). Notice that the denominator \( 4x^2 - 9 \) is equivalent to \((2x+3)(2x-3)\), which are the factors of a difference of squares \( a^2-b^2 = (a-b)(a+b) \). We need a common denominator for all the fractions. Here, it's \((2x+3)(2x-3)\). Now all terms share this common denominator.

Step 2: Rewrite Each Fraction with the Common Denominator

Express each term with the denominator \((2x+3)(2x-3)\): \[ \frac{5}{2x+3} = \frac{5(2x-3)}{(2x+3)(2x-3)} \] \[ \frac{4}{2x-3} = \frac{4(2x+3)}{(2x+3)(2x-3)} \] This adjusts each fraction to have the common denominator.

Step 3: Combine the Fractions on the Left Side

Combine the fractions on the left side using their common denominator:\[ \frac{5(2x-3) + 4(2x+3)}{(2x+3)(2x-3)} = \frac{14x+3}{4x^2-9} \] Simplify the numerator of the left side: \( 5(2x-3) = 10x - 15 \)\( 4(2x+3) = 8x + 12 \)The combined numerator is \( 10x - 15 + 8x + 12 = 18x - 3 \), making the new left side:\[ \frac{18x - 3}{(2x+3)(2x-3)} \]

Step 4: Eliminate Denominators by Multiplying Across the Equation

Now that both sides are expressed with the same denominator, multiply through by \((2x+3)(2x-3)\) to eliminate the denominators, yielding:\[ 18x - 3 = 14x + 3 \] This simplifies the equation down to solving just what's left from the numerators.

Step 5: Solve the Resulting Linear Equation

With the numerators, solve the equation:\[ 18x - 3 = 14x + 3 \] Subtract \(14x\) from both sides: \[ 4x - 3 = 3 \]Add \(3\) to both sides: \[ 4x = 6 \]Divide by \(4\): \[ x = \frac{6}{4} = \frac{3}{2} \]

Step 6: Validate the Solution

Substitute \( x = \frac{3}{2} \) back into the original equation to ensure there are no division by zero instances or computational mistakes. Factor \((2x+3)(2x-3)\) yields zero denominators when \(x = \pm \frac{3}{2}\), thus \(x = \frac{3}{2}\) isn't a valid solution due to division by zero. There is no valid solution.

Key Concepts

Difference of Squares

The concept of the difference of squares is a vital stepping stone in algebra, particularly when dealing with polynomials. Consider the expression given, which is written as:
\[ 4x^2 - 9 \].
This can be recognized as a special form where one square number is subtracted from another. It simplifies to:
\[ a^2 - b^2 = (a-b)(a+b) \].
For our specific instance:

• \[ a^2 = (2x)^2 \]
• \[ b^2 = 3^2 \]
• This results in \[ (2x + 3)(2x - 3) \].

This decomposition enables us to factor the quadratic expression into two linear factors. Factoring using the difference of squares simplifies computation, making it easier to find common denominators and solve equations involving polynomials.

Common Denominator

A common denominator is critical when dealing with rational expressions because it allows us to add or subtract fractions. In our equation, the fractions had denominators:
\[ 2x+3, \quad 2x-3, \quad 4x^2-9 \].
By recognizing the difference of squares, we see that:
\[ 4x^2 - 9 = (2x+3)(2x-3) \].
This tells us that the least common denominator (LCD) of all terms is effectively \((2x+3)(2x-3)\).

• Rewrite each fraction using this common denominator.
• For \(\frac{5}{2x+3}\), multiply both the numerator and the denominator by \(2x-3\).
• For \(\frac{4}{2x-3}\), multiply both the numerator and the denominator by \(2x+3\).

This establishes a standard denominator for easier manipulation and equation-solving.

Linear Equations

Once all terms are expressed with a common denominator, solving a rational equation typically involves clearing denominators, leaving you with a linear equation.
In our example, after equating the numerators, you get:
\[ 18x - 3 = 14x + 3 \].
The task is to solve for \(x\):

• First, subtract \(14x\) from both sides to isolate terms involving \(x\): \(4x - 3 = 3\).
• Then, add 3 to both sides: \(4x = 6\).
• Lastly, divide by 4 to solve for \(x\): \(x = \frac{6}{4} = \frac{3}{2}\).

Linear equations like this are characterized by expressions where each term is either a constant or a product of a constant and a single variable.

Division by Zero

Division by zero is a fundamental undefined operation in mathematics that you must avoid when solving equations. It occurs when the denominator of a fraction equals zero, nullifying the expression or result.
In rational equations, always validate potential solutions by plugging them back into the original equation.

• Check if the evaluated denominators equal zero.
• For our equation with \((2x+3)(2x-3)\) as the denominator, this evaluates to zero if \(x = \pm \frac{3}{2}\).

Thus, substitute \(x = \frac{3}{2}\) back into the equation. Here both terms of multiplication become zero, indicating division by zero.
Therefore, the initially found solution \(x = \frac{3}{2}\) is invalid, leading to no acceptable solution, as predicted by ensuring no division by zero occurs in the original rational expression.