Problem 19 - Algebra for College Students

Question

Solve each equation. $$ \frac{-1}{2 x-5}+\frac{2 x-4}{4 x^{2}-25}=\frac{5}{6 x+15} $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{3}{7}\).

Step 1: Identify Common Denominators

Recognize that the equation involves rational expressions with different denominators: \(2x-5\) in the first term, \(4x^2-25\) in the second, and \(6x+15\) in the third. We need a common denominator to combine these fractions.

Step 2: Factorize the Denominators

Factor the denominators where possible. For the second term, \(4x^2-25\) is a difference of squares and factors as \((2x-5)(2x+5)\). For the third term, factor out 3 from \(6x+15\) to get \(3(2x+5)\). Thus, the common denominator is \(3(2x-5)(2x+5)\).

Step 3: Rewrite Each Fraction with the Common Denominator

Rewrite each term of the equation with the common denominator. The first term becomes \(\frac{-3(2x+5)}{3(2x-5)(2x+5)}\), the second term becomes \(\frac{2x-4}{4x^2-25} = \frac{2x-4}{(2x-5)(2x+5)}\), and the third term becomes \(\frac{5}{3(2x+5)}\).

Step 4: Combine Fractions

Combine the fractions on the left into one fraction. This gives the equation: \[\frac{-3(2x+5) + (2x-4)}{3(2x-5)(2x+5)} = \frac{5}{3(2x+5)}\]. Multiply each side by the common denominator to clear the fractions.

Step 5: Solve the Numerator Equation

Set \(-3(2x+5) + (2x-4) = 5(2x-5)\) and solve for \(x\). Distribute and combine like terms: \(-6x - 15 + 2x - 4 = 10x - 25\), leading to \(-4x - 19 = 10x - 25\).

Step 6: Simplify and Solve for x

Continue solving \(-4x - 19 = 10x - 25\) by adding \(4x\) to both sides: \[-19 = 14x - 25\]. Then add 25 to both sides: \[6 = 14x\]. Divide by 14: \(x = \frac{3}{7}\).

Step 7: Check for Extraneous Solutions

Ensure that \(x = \frac{3}{7}\) does not create a zero in any original fractions. Substituting into any denominator (\(2x-5\), \(4x^2-25\), or \(6x+15\)) does not result in zero, so \(x = \frac{3}{7}\) is valid.

Key Concepts

Common Denominators

When solving rational equations, identifying a common denominator is crucial. Think of it like finding a common language for different fractions, so they can be compared or combined effectively. In this exercise, the sound of the task is: find a denominator that can represent all the terms without any remainder. You'll notice that each term in the equation holds a different denominator:

• For the first term, it's \(2x-5\),
• For the second, it's the quadratic \(4x^2-25\),
• And for the third, it's \(6x+15\).

Identifying what common ground these varied terms can share makes it easier when you start to factor and mix them.

Factoring Quadratics

Factoring quadratics, like our \(4x^2-25\), can sometimes look intimidating. However, it often involves recognizing patterns, such as the difference of squares. Consider \(a^2-b^2\), which factors as \((a-b)(a+b)\), a formula quite handy here. Now, notice that \(4x^2-25\) matches this form, breaking down into \((2x-5)(2x+5)\).

The third denominator, \(6x+15\), simplifies by factoring out the greatest common factor, in this case, 3, yielding \(3(2x+5)\). By tackling these individually, we convert a pair of complex terms into simpler factors, allowing for a clearer pathway to the common denominator.

Combining Fractions

Once the denominators are factored and aligned, the next step involves combining the fractions. This means rewriting each fraction so that they share the same denominator, which we determined earlier as \(3(2x-5)(2x+5)\).

Adjust each fraction accordingly:

• The first becomes \(\frac{-3(2x+5)}{3(2x-5)(2x+5)}\),
• The second remains as \(\frac{2x-4}{(2x-5)(2x+5)}\),
• And the third turns into \(\frac{5}{3(2x+5)}\).

After these adjustments, you can combine terms on the left side of the equation while ensuring the integrity of mathematical operations remains intact, carefully adhering to distributive rules to merge these fractions into one expression.

Checking Extraneous Solutions

One critical step in solving rational equations is verifying whether the solution found is extraneous, meaning it doesn't satisfy the original equation. Often, solutions might appear valid mathematically but, when substituted back, create a zero in the denominator—rendering the expression undefined.

For our solution, \(x = \frac{3}{7}\), ensuring it doesn’t zero out any original denominators:

• \(2x-5\),
• \(4x^2-25\),
• \(6x+15\).

Checking each, we find that no denominator becomes zero, thus confirming \(x = \frac{3}{7}\) as a valid solution. This validation process is essential in maintaining mathematical truth throughout computation.