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When tackling absolute value equations, it is essential to understand that you are dealing with expressions that can yield two different scenarios. The absolute value, denoted by vertical bars like \(|a|\), represents the distance of a number from zero on the number line. This means it can encompass both positive and negative solutions.

In the exercise where we have \(|2x - 1| = 5\), the critical step is to recognize that this equation implies two separate possibilities. Firstly, the expression inside the absolute value could equal 5, which gives us the equation \(2x - 1 = 5\). Secondly, it could also equal -5, resulting in \(2x - 1 = -5\). These are the two equations you need to solve.

To solve each equation, isolate the variable. For \(2x - 1 = 5\), add 1 to both sides to get \(2x = 6\), then divide by 2 to find \(x = 3\). Similarly, for \(2x - 1 = -5\), add 1 to both sides yielding \(2x = -4\) and then divide by 2 to get \(x = -2\). Thus, our potential solutions are \(x = 3\) and \(x = -2\).

After finding potential solutions to the absolute value equation, it is crucial to verify their validity by plugging them back into the original equation. This ensures that the solutions satisfy the condition set by the absolute value.

For instance, re-substitute \(x = 3\) into the original equation \(|2x - 1| = 5\). This gives \(|2 \times 3 - 1| = 5\), simplifying to \(|6 - 1| = 5\) or \(|5| = 5\), which holds true.

Do the same with \(x = -2\). Substituting it into the original equation gives \(|2 \times -2 - 1| = 5\), simplifying to \(|-4 - 1| = 5\) or \(|-5| = 5\), which is also valid. So both solutions check out.

Algebraic manipulation is a fundamental aspect of solving equations, particularly when dealing with equations that include absolute values. Manipulating the equations correctly requires adhering to mathematical rules and performing operations systematically to isolate the variable.

In this exercise, you saw that each absolute value equation was transformed into two separate algebraic equations. The steps involved operations such as adding or subtracting the same number from both sides, followed by dividing or multiplying both sides to finally solve for the variable. For instance, when solving \(2x - 1 = 5\), adding 1 to both sides and then dividing by 2 effectively isolated \(x\).

This process of systematically rearranging equations by undoing operations in reverse order — starting from the most outside operation and working inward — is fundamental in algebra. Mastery of these techniques enables students to solve more complex equations with confidence.