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When you encounter an equation like \(|3y-1| + 10 = 25\), one of the key objectives is to determine the solution set. A solution set comprises all values that satisfy the equation. For absolute value equations, we often find more than one solution.

This is because the expression inside the absolute value can yield two different scenarios that make it true. You can think of the absolute value function as measuring the distance from zero on a number line. For example, both positive and negative numbers can have the same absolute value since they are equidistant from zero.

In this instance, the solutions - \(y = \frac{16}{3}\) and \(y = -\frac{14}{3}\) - form our solution set. It's crucial to write down all possible solutions when solving such equations to ensure completeness of the solution set.

Solving equations is the process of finding the unknown variable that makes the equation true. Let's focus on the particular equation \(|3y-1| + 10 = 25\).

You start by isolating the absolute value by subtracting 10 from both sides. This step is essential because it simplifies the equation to \(|3y-1| = 15\).

Next, recognize that absolute value equations yield two scenarios. You essentially split the equation into two separate equations:

• \(3y - 1 = 15\)
• \(3y - 1 = -15\)

These two new equations need to be solved independently. Each scenario corresponds to one possible reality that satisfies the absolute value condition.

Algebraic manipulation involves rearranging and combining terms within an equation to solve it. For the problem \(|3y-1| + 10 = 25\), performing algebraic manipulation helps to find the solutions.

First, isolate terms involving \(y\) by moving constant terms to the other side of the equation through subtraction. For example, by subtracting 10 from both sides, you simplify the equation to \(|3y-1| = 15\).

Next, resolve the absolute value to form two linear equations. For each linear equation, isolate \(y\) by adding 1 and then dividing everything by 3. Follow these steps for both possible equations:

• For \(3y - 1 = 15\), add 1 to get \(3y = 16\). Then divide by 3 to find \(y = \frac{16}{3}\).
• For \(3y - 1 = -15\), add 1 to get \(3y = -14\). Then divide by 3 to find \(y = -\frac{14}{3}\).

These procedures highlight how algebraic manipulation simplifies even complex equations into manageable forms.