Solving For Y In The Equation 2(3 Y+5)=3(5 Y+1/3)

Jul 11, 2025 by qnaftunila 50 views

Solve for y 2(3 y+5)=3(5 y+1/3). The solution is y=

Unveiling the Solution to the Linear Equation: 2(3y + 5) = 3(5y + 1/3)

In this comprehensive guide, we embark on a step-by-step journey to unravel the solution for the linear equation 2(3y + 5) = 3(5y + 1/3). Linear equations, the cornerstone of algebra, play a vital role in various mathematical and real-world applications. Mastering the art of solving these equations equips us with the power to decipher unknown variables and gain valuable insights into the relationships between them. Through a meticulous and detailed approach, we will navigate the intricacies of this equation, leaving no stone unturned until we arrive at the definitive solution for y. Join us as we delve into the realm of algebraic manipulation and equation solving.

Dissecting the Equation: A Step-by-Step Approach

Our quest to solve the equation 2(3y + 5) = 3(5y + 1/3) begins with a strategic plan. We will employ the fundamental principles of algebra, meticulously applying each step to maintain the equation's balance and integrity. Our arsenal includes the distributive property, the art of combining like terms, and the strategic use of inverse operations. With these tools at our disposal, we will systematically isolate the variable y on one side of the equation, ultimately unveiling its true value.

Unleashing the Distributive Property: Our initial move involves invoking the distributive property, a powerful tool that allows us to simplify expressions enclosed within parentheses. By multiplying the constants outside the parentheses with each term inside, we break free from the confines of grouping symbols. Applying this property to our equation, we obtain:
```
2 * 3y + 2 * 5 = 3 * 5y + 3 * (1/3)
```
Simplifying the multiplication, the equation transforms into:
```
6y + 10 = 15y + 1
```
The distributive property has paved the way for further simplification, setting the stage for the next phase of our solution.
Gathering Like Terms: A Strategic Consolidation: Now that we have liberated the terms from their parentheses, our next objective is to consolidate like terms. This involves strategically grouping terms that share the same variable (y in this case) and constant terms. To achieve this, we will employ the art of adding or subtracting terms from both sides of the equation, ensuring that the balance remains undisturbed.

Let's begin by subtracting 6y from both sides of the equation:
```
6y + 10 - 6y = 15y + 1 - 6y
```
This maneuver strategically eliminates the y term from the left side, leaving us with:
```
10 = 9y + 1
```
Next, we will subtract 1 from both sides to isolate the y term further:
```
10 - 1 = 9y + 1 - 1
```
This yields:
```
9 = 9y
```
We have successfully gathered the like terms, bringing us closer to the solution.
Isolating the Variable: The Final Act: With the like terms consolidated, our final mission is to isolate the variable y completely. This involves neutralizing the coefficient that clings to y. In this case, y is multiplied by 9. To undo this multiplication, we will employ the inverse operation: division.

Dividing both sides of the equation by 9, we get:
```
9 / 9 = 9y / 9
```
This simplifies to:
```
1 = y
```
Or, more conventionally:
```
y = 1
```
Eureka! We have successfully isolated y and discovered its value. The solution to the equation 2(3y + 5) = 3(5y + 1/3) is y = 1.

Verifying the Solution: A Crucial Step

In the realm of mathematics, accuracy reigns supreme. To ensure the validity of our solution, we must embark on a crucial step: verification. This involves substituting the value we obtained for y (which is 1) back into the original equation. If the left-hand side (LHS) of the equation equals the right-hand side (RHS) after the substitution, our solution stands vindicated.

Let's substitute y = 1 into the original equation:

2(3 * 1 + 5) = 3(5 * 1 + 1/3)

Simplifying the expressions within the parentheses, we get:

2(3 + 5) = 3(5 + 1/3)

2(8) = 3(16/3)

Performing the multiplication, we have:

16 = 16

The LHS (16) indeed equals the RHS (16). This resounding confirmation solidifies our solution. We have verified that y = 1 is indeed the correct solution to the equation 2(3y + 5) = 3(5y + 1/3).

Mastering Linear Equations: A Gateway to Mathematical Proficiency

Through this detailed exploration, we have successfully navigated the intricacies of solving the linear equation 2(3y + 5) = 3(5y + 1/3). By meticulously applying the distributive property, consolidating like terms, and employing inverse operations, we have unveiled the value of y. The journey has underscored the importance of each step, emphasizing the need for precision and a systematic approach.

Linear equations serve as the bedrock of algebra and find widespread applications in various fields, from science and engineering to economics and finance. By mastering the techniques for solving linear equations, we empower ourselves to tackle complex problems and make informed decisions. The ability to manipulate equations, isolate variables, and interpret solutions is a valuable asset in both academic pursuits and real-world scenarios.

As we conclude this exploration, let us carry forward the lessons learned. The principles and techniques we have employed will serve as a guiding light as we encounter more intricate mathematical challenges. With a solid foundation in linear equations, we are well-equipped to embark on further mathematical explorations and unlock the boundless potential that lies within the realm of numbers and equations.

Therefore, the final solution to the equation 2(3y + 5) = 3(5y + 1/3) is y = 1.