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Solving algebraic equations involves finding the value of one or more unknown variables. In this problem, our goal is to solve for the variable \(s\) in the given equation. To tackle any equation, follow these basic steps:

• Identify what you're solving for — in this case, \(s\).
• Use algebraic manipulations to isolate the variable. Keep the equation balanced by performing the same operations on both sides.
• Simplify the equation step-by-step until the variable is isolated.

Understanding these fundamental techniques can help you navigate through simple to complex equations with ease. Remember, always ensure each manipulation keeps the equation balanced.

Isolating a variable means rearranging an equation so that the variable ends up on one side by itself. In our problem, we want \(s\) to be alone. Here's how you can achieve it:1. Start by eliminating any fractions or denominators. Multiply both sides by \((r+s)\) to get rid of the fraction in the equation: \[(r+s) h = 3A\] 2. Distribute any factors over terms on one side of the equation. Here, multiply \(h\) through the binomial on the left side: \[rh + sh = 3A\] 3. Move all terms not including your target variable to the other side: Subtract \(rh\) from both sides: \[sh = 3A - rh\] 4. Finally, divide by the coefficient of \(s\), which is \(h\), to leave \(s\) by itself: \[s = \frac{3A - rh}{h}\] Following these steps ensures that you isolate the variable effectively. It's like systematically peeling off layers to find what you're looking for.

Rational equations have variables in the denominator. Solving these equations often involves steps to eliminate the fractions first, making the problem easier to handle.Here's a quick guide:

• Identify the rational part (the fraction). In our exercise, that's \(\frac{3A}{r+s}\).
• Clear fractions by multiplying both sides of the equation by the denominator, \(r+s\), to eliminate the fraction.
• Proceed with solving as you would any linear equation once the rational part is cleared away.

Understanding rational equations allows you to solve problems even when variables divide terms. The key is to first clear out the denominator using multiplication, which simplifies your path to isolating variables.