Solving For S In The Equation 3r = 10 + 5s When R = 10

Introduction

In this article, we will delve into solving a fundamental algebraic equation. Our primary goal is to determine the value of the variable 's' within the equation 3r = 10 + 5s, given that the value of 'r' is 10. This exercise is a classic example of how to manipulate equations and isolate variables, a crucial skill in mathematics and various scientific disciplines. We will explore the step-by-step process of substituting the known value, simplifying the equation, and ultimately solving for the unknown variable. This process not only provides a solution to the specific problem but also reinforces the fundamental principles of algebraic manipulation. Understanding these principles is essential for tackling more complex mathematical challenges in the future.

Understanding the Problem

To effectively solve for the value of 's', we must first thoroughly understand the given equation and the information provided. The equation we are working with is 3r = 10 + 5s. This is a linear equation, which means it represents a straight line when graphed. Linear equations are fundamental in mathematics and are used to model various real-world scenarios. The equation contains two variables, 'r' and 's', and our task is to find the value of 's' when 'r' is equal to 10. The fact that we are given a specific value for 'r' allows us to transform the equation into one with a single unknown, making it solvable. The process of solving this equation involves substituting the given value of 'r', simplifying the equation, and then using algebraic manipulation to isolate 's' on one side of the equation. This approach is a cornerstone of algebraic problem-solving and is applicable to a wide range of mathematical problems.

Step-by-Step Solution

1. Substituting the Value of 'r'

The first crucial step in solving for 's' is to substitute the given value of 'r', which is 10, into the equation 3r = 10 + 5s. This substitution allows us to eliminate one variable and work with an equation that only involves 's'. When we replace 'r' with 10, the equation transforms into 3 * 10 = 10 + 5s. This step is a direct application of the principle of substitution, a fundamental technique in algebra where a variable is replaced with its known value. The resulting equation is a simplified form that is easier to manipulate and solve. This substitution is a critical step as it sets the stage for the subsequent steps in isolating and finding the value of 's'.

2. Simplifying the Equation

Following the substitution, the equation now reads 3 * 10 = 10 + 5s. Our next step is to simplify both sides of the equation. On the left side, we perform the multiplication: 3 multiplied by 10 equals 30. So, the equation becomes 30 = 10 + 5s. This simplification is a crucial step in making the equation more manageable. It reduces the complexity and allows us to focus on isolating the term containing 's'. Simplifying equations is a fundamental skill in algebra, as it often reveals the underlying structure of the problem and makes it easier to apply further algebraic manipulations. In this case, the simplification has brought us closer to isolating 's' and ultimately finding its value.

3. Isolating the Term with 's'

Now that we have the simplified equation 30 = 10 + 5s, our objective is to isolate the term containing 's', which is 5s. To do this, we need to eliminate the '+ 10' on the right side of the equation. We achieve this by subtracting 10 from both sides of the equation. This operation maintains the equality because whatever is done on one side must be done on the other. Subtracting 10 from both sides gives us 30 - 10 = 10 + 5s - 10. This simplifies to 20 = 5s. Isolating the term with the variable is a key step in solving for the variable, as it brings us closer to having the variable alone on one side of the equation.

4. Solving for 's'

After isolating the term 5s, we have the equation 20 = 5s. To finally solve for 's', we need to isolate 's' completely. This is achieved by dividing both sides of the equation by the coefficient of 's', which is 5. Dividing both sides by 5 gives us 20 / 5 = 5s / 5. This simplifies to 4 = s. Therefore, the value of 's' is 4. This final step demonstrates the principle of inverse operations, where we use the opposite operation (division in this case) to undo the multiplication and isolate the variable. The result, s = 4, is the solution to the problem and represents the value of 's' that satisfies the original equation when r = 10.

Final Answer

By following these steps, we have successfully determined the value of 's' in the equation 3r = 10 + 5s, given that r = 10. Our step-by-step solution involved substituting the value of 'r', simplifying the equation, isolating the term with 's', and finally solving for 's'. The final answer is s = 4. This process illustrates the fundamental principles of algebraic manipulation and problem-solving. Understanding and applying these principles is crucial for success in mathematics and related fields. The ability to solve equations like this one forms the foundation for tackling more complex mathematical problems and real-world applications.