Solving For Y In The Equation 30 - 28 = 29y - 12y - 2y

This article provides a step-by-step solution for the equation 30 - 28 = 29y - 12y - 2y. We will simplify the equation, combine like terms, and isolate the variable y to find its value. This is a fundamental algebraic problem that demonstrates how to solve linear equations, which are prevalent in various fields of mathematics and science. Understanding how to manipulate and solve such equations is crucial for tackling more complex problems and real-world applications. This process involves several key algebraic principles, including the order of operations, combining like terms, and the properties of equality. By carefully applying these principles, we can systematically solve for the unknown variable and arrive at the correct solution. The steps outlined in this article are designed to be clear and concise, making it easy for readers to follow along and understand the underlying concepts. Furthermore, this solution is not just about finding the value of y; it's about building a foundation in algebraic problem-solving that can be applied to a wide range of mathematical challenges. Whether you are a student learning algebra for the first time or someone looking to refresh your skills, this article provides a comprehensive guide to solving this particular equation and similar ones.

The given equation is:

$$30 - 28 = 29y - 12y - 2y$$

Our goal is to find the value of y that satisfies this equation. This involves simplifying both sides of the equation and isolating y on one side. The process begins by performing the subtraction on the left side and then combining the terms with y on the right side. This consolidation of terms is crucial for simplifying the equation and bringing it closer to a solvable form. Once the terms are combined, the equation will be in a simpler format, making it easier to isolate y. The subsequent steps involve using algebraic manipulations to get y by itself. This typically involves performing the same operation on both sides of the equation to maintain equality while moving terms around. For instance, if y is multiplied by a coefficient, we would divide both sides by that coefficient to isolate y. The ultimate aim is to have y equal to a constant, which will be the solution to the equation. This solution represents the value of y that makes the original equation true. The problem-solving strategy applied here is a fundamental aspect of algebra, applicable to numerous equations of varying complexity. Therefore, mastering this method is essential for anyone looking to excel in mathematics.

Step-by-step Solution

Step 1: Simplify the left side of the equation

First, we simplify the left side of the equation by performing the subtraction:

$$30 - 28 = 2$$

So the equation becomes:

$$2 = 29y - 12y - 2y$$

This initial simplification is a critical step in solving the equation. By reducing the left side to a single numerical value, we make the equation more manageable and easier to work with. This step aligns with the fundamental principle of simplifying mathematical expressions before attempting to solve them. It reduces the complexity of the equation and sets the stage for combining like terms on the right side. The subtraction operation here is straightforward, but it's an essential foundation for the subsequent steps. The result of this subtraction, 2, now represents the constant value that the expression involving y on the right side must equal. This constant value serves as a benchmark against which we can compare the simplified expression on the right side, helping us to eventually isolate y and find its value. In essence, this first step is about making the equation as clean and simple as possible, which is a common strategy in algebraic problem-solving. The simplicity achieved here allows us to focus on the terms involving y, setting up the next steps in the solution process.

Step 2: Combine like terms on the right side

Next, we combine the terms with y on the right side:

$$29y - 12y - 2y = (29 - 12 - 2)y$$

$$= (17 - 2)y$$

$$= 15y$$

So the equation now is:

$$2 = 15y$$

Combining like terms is a fundamental algebraic technique that simplifies equations by grouping terms with the same variable. In this step, we focus on the right side of the equation, where we have three terms involving y. By combining these terms, we reduce the complexity of the equation and make it easier to isolate y. The process involves performing the arithmetic operations on the coefficients of y, which in this case are 29, -12, and -2. We first subtract 12 from 29, resulting in 17, and then subtract 2 from 17, resulting in 15. This means that the three terms 29y, -12y, and -2y combine to give 15y. This simplification is crucial because it reduces the number of terms involving y from three to one, making the equation much more manageable. The equation 2 = 15y is now in a form where we can easily isolate y by performing a single algebraic operation. This step highlights the importance of recognizing and combining like terms in algebraic problem-solving, as it significantly simplifies the equation and brings us closer to the solution. The ability to combine like terms is a core skill in algebra, and it is essential for solving a wide range of equations and mathematical problems.

Step 3: Isolate y

To isolate y, we divide both sides of the equation by 15:

$$\frac{2}{15} = \frac{15y}{15}$$

$$\frac{2}{15} = y$$

So, the solution is:

$$y = \frac{2}{15}$$

Isolating y is the final step in solving for the variable. In this step, we aim to get y by itself on one side of the equation. We achieve this by performing the inverse operation of whatever is being done to y. In our equation, y is being multiplied by 15. To undo this multiplication, we divide both sides of the equation by 15. This is a critical application of the properties of equality, which state that performing the same operation on both sides of an equation maintains the balance and equality. By dividing both sides by 15, we cancel out the 15 on the right side, leaving y isolated. On the left side, we have 2 divided by 15, which gives us the fraction 2/15. This fraction represents the value of y that satisfies the original equation. The solution y = 2/15 is a precise and exact value, and it can be used to verify the solution by substituting it back into the original equation. If the substitution results in both sides of the equation being equal, then the solution is correct. This step of isolating y is a cornerstone of algebraic problem-solving, and it is a technique that is used extensively in various mathematical contexts. The ability to isolate a variable is essential for solving equations and for understanding the relationships between variables in mathematical models.

Final Answer

The solution for the equation $30 - 28 = 29y - 12y - 2y$ is:

$$y = \frac{2}{15}$$

This final answer represents the value of y that makes the original equation true. To arrive at this solution, we followed a series of logical steps, including simplifying both sides of the equation, combining like terms, and isolating the variable y. This process demonstrates the fundamental principles of algebraic problem-solving, which are applicable to a wide range of equations. The solution y = 2/15 can be verified by substituting it back into the original equation and confirming that both sides of the equation are equal. This verification step is an important part of the problem-solving process, as it ensures that the solution is accurate. The fraction 2/15 is the exact value of y, and it represents the proportion that satisfies the given equation. This solution is a clear and concise answer to the problem, and it is the culmination of the algebraic manipulations performed. The process of solving this equation reinforces the importance of algebraic techniques such as simplifying expressions, combining like terms, and using inverse operations to isolate variables. These techniques are essential for success in algebra and higher levels of mathematics. The final answer y = 2/15 provides a concrete solution to the problem, and it demonstrates the power of algebraic methods in solving equations.