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When solving equations involving fractions, one of the key steps is to identify a common denominator. This concept refers to a shared multiple of the denominators present in the equation. Understanding how to find a common denominator will allow you to combine fractions or, as we often aim for in algebra, clear the fractions completely for easier manipulations.

In our example, \( \frac{2x-3}{x+1} = \frac{2}{5} \), we have denominators \(x + 1\) and \(5\). Therefore, we look for the least common denominator (LCD) to combine these efficiently without altering the values of the fractions. The LCD for the denominators \(x+1\) and \(5\) is their product, \(5(x+1)\), because \(x+1\) and \(5\) share no common factors other than 1. In practice, the LCD makes it possible to streamline the equation, setting the stage for the solution process.

The process of clearing fractions simplifies equations by eliminating the denominator, making algebraic manipulation more straightforward. To accomplish this, you multiply every term in the equation by the common denominator we identified earlier.

For the equation \( \frac{2x-3}{x+1} = \frac{2}{5} \), we multiply both sides by the LCD, \(5(x+1)\). This step effectively removes the fractions, resulting in an equation with integers and variables that are easier to handle: \[ (2x-3)(5) = (2)(x+1) \]. Here, \(x+1\) cancels out on the left side, and \(5\) cancels out on the right side, elegantly simplifying our task.

With the fractions cleared, the next phase is equation simplification. This involves expanding expressions, combining like terms, and arranging the equation in a more solvable form.

We expand the terms by distributing multiplication over addition and subtraction, then combine like terms by adding or subtracting variables and constants separately: \[ 10x - 15 = 2x + 2 \]. To further simplify, subtract \(2x\) from both sides to get all the variable terms on one side, and add \(15\) to both sides to move the constant terms to the opposite side. These operations condense the equation to \(8x = 17\), stripping away complexities and bringing us closer to isolating the variable.

The last step in solving rational equations is isolating the variable. This step is crucial as it gives the solution to the equation by ensuring the variable stands alone on one side of the equation.

In our working example, we have the simplified form \(8x = 17\). To isolate \(x\), we divide both sides by \(8\), giving us \(x = \frac{17}{8}\). This solution signifies that for the equation to hold true, \(x\) must be equal to \(\frac{17}{8}\), illustrating the power of isolating the variable in revealing the answer to an algebraic problem. Such a methodical approach is universally applicable across a wide array of algebraic equations.