Question

Solve each equation, if possible. $$ \frac{4}{x-2}=\frac{-3}{x+5}+\frac{7}{(x+5)(x-2)} $$

Short answer

The solution is \(x = -1\).

Step-by-step solution

Identify the Least Common Denominator (LCD)

To combine the fractions, determine the least common denominator (LCD) of the denominators. The denominators are: \(x-2\), \(x+5\), and \((x+5)(x-2)\). The LCD is \((x+5)(x-2)\).

Rewrite the Fractions with the LCD

Rewrite each fraction with the common denominator \((x+5)(x-2)\). \[ \frac{4}{x-2} = \frac{4(x+5)}{(x-2)(x+5)} \ \frac{-3}{x+5} = \frac{-3(x-2)}{(x+5)(x-2)} \ \frac{7}{(x+5)(x-2)} \text{ remains the same.} \]

Set Up the Equation

Combine the fractions into one equation with the common denominator: \[ \frac{4(x+5)}{(x-2)(x+5)} = \frac{-3(x-2) + 7}{(x+5)(x-2)} \]

Eliminate the Denominator

Since the denominators on both sides are the same, we can eliminate them, yielding: \[ 4(x+5) = -3(x-2) + 7 \]

Simplify Both Sides

Expand and simplify both sides of the equation: \[ 4x + 20 = -3x + 6 + 7 \]

Combine Like Terms

Combine like terms: \[ 4x + 20 = -3x + 13 \]

Solve for x

Isolate \(x\) by adding \(3x\) to both sides and subtracting 20: \[ 4x + 3x = 13 - 20 \ 7x = -7 \ x = -1 \]

Verify the Solution

Verify by substituting \(x = -1\) into the original equation: \[ \frac{4}{-1-2} = \frac{-3}{-1+5} + \frac{7}{(-1+5)(-1-2)} \] \[ \frac{4}{-3} = \frac{-3}{4} + \frac{7}{4 \times -3} \ -\frac{4}{3} = -\frac{3}{4} - \frac{7}{12} \ -\frac{4}{3} = -\frac{9}{12} - \frac{7}{12} \ -\frac{4}{3} = -\frac{16}{12} \ -\frac{4}{3} = -\frac{4}{3} \text{ (True)} \] Therefore, \(x = -1\) is the correct solution.

Key concepts

Least Common Denominator

To solve rational equations, it is crucial to identify the Least Common Denominator (LCD). The LCD is the smallest expression that each of the denominators in the equation can divide into evenly. For example, if you have the denominators \(x-2\), \(x+5\), and \((x+5)(x-2)\), the LCD would be \((x+5)(x-2)\) since it is the most comprehensive denominator shared by all terms.
This step is important because it allows you to combine and compare fractions by giving them a common denominator. Once this common ground is established, you can rewrite each fraction in terms of this LCD, which simplifies the equation significantly.
Identifying and using the LCD makes it easier to align the terms of the equation and proceed with the solution efficiently.

Simplifying Rational Expressions

Simplifying rational expressions is an essential part of solving rational equations. This process involves rewriting fractions so that they have a common denominator. In our example, we start with the fractions: \(\frac{4}{x-2}\), \(\frac{-3}{x+5}\), and \(\frac{7}{(x+5)(x-2)}\). Each must be rewritten to have the least common denominator \((x+5)(x-2)\).

• For \(\frac{4}{x-2}\), multiply the numerator and the denominator by \(x+5\): \(\frac{4(x+5)}{(x-2)(x+5)}\).
• For \(\frac{-3}{x+5}\), multiply the numerator and the denominator by \(x-2\): \(\frac{-3(x-2)}{(x+5)(x-2)}\).
• \(\frac{7}{(x+5)(x-2)}\) remains the same.

By rewriting the fractions in this way, you align them under a common denominator, allowing you to combine them into a single equation. With the denominators normalized, you can eliminate them from both sides of the equation, significantly simplifying the solving process.

Verifying Solutions

Verifying solutions is the final important step in solving rational equations. This involves substituting the solution back into the original equation to confirm that it holds true. For instance, when we found that \(x = -1\) in the given problem, we substituted \(-1\) back into the original equation: \(\frac{4}{-1-2} = \frac{-3}{-1+5} + \frac{7}{(-1+5)(-1-2)}\).
By evaluating each term:

• \(\frac{4}{-3}\)
• \(\frac{-3}{4}\)
• \(\frac{7}{4 \times -3}\)

we confirmed that both sides of the equation are equal: \(-\frac{4}{3} = -\frac{4}{3}\). This tells us that our solution is correct.
Verifying ensures that no mistakes were made during the simplification and solving process, and it also checks for extraneous solutions that don't actually satisfy the original equation. Always verify your solutions to ensure they are accurate and valid in the context of the problem.