Eliminating Y Terms And Solving For X A Step By Step Guide

In the realm of algebra, solving systems of equations is a fundamental skill. Often, we encounter scenarios where we have two or more equations with multiple variables, and our goal is to find the values of these variables that satisfy all equations simultaneously. One common technique for solving such systems is the elimination method, which involves strategically manipulating the equations to eliminate one variable, thereby simplifying the problem. This article delves into a specific instance of this method, focusing on eliminating the $y$ terms in a system of two linear equations to solve for $x$. We'll break down the steps involved, highlighting the key constants that need to be multiplied to achieve this elimination in the most efficient manner.

Understanding the Elimination Method

Before diving into the specific problem, let's briefly recap the essence of the elimination method. The core idea is to manipulate the equations in a way that when they are added together, one of the variables cancels out. This is achieved by ensuring that the coefficients of the variable we want to eliminate are opposites of each other in the two equations. For example, if we have the terms $+3y$ in one equation and $-3y$ in another, adding these equations will eliminate the $y$ variable, leaving us with an equation in terms of $x$ only.

Problem Statement: Eliminating $y$ to Solve for $x$

Consider the following system of linear equations:

First equation: $4x - 3y = 34$

Second equation: $3x + 2y = 17$

Our objective is to determine the constants by which we should multiply each equation before adding them together so that the $y$ terms are eliminated, allowing us to solve for $x$ efficiently. This involves strategic manipulation to ensure the coefficients of $y$ become additive inverses.

Strategic Multiplication for Elimination

To eliminate the $y$ terms, we need to find multipliers for each equation that will make the coefficients of $y$ opposites. In the given system, the coefficients of $y$ are -3 in the first equation and 2 in the second equation. The least common multiple (LCM) of 3 and 2 is 6. Therefore, we want to transform the $y$ coefficients to -6 and +6 (or vice versa).

To achieve this, we can multiply the first equation by 2 and the second equation by 3. This approach ensures that the $y$ terms will cancel each other out when the equations are added. Specifically, multiplying the first equation by 2 results in a $y$ coefficient of -6, and multiplying the second equation by 3 results in a $y$ coefficient of +6. This strategic multiplication sets the stage for the elimination of the $y$ variable and allows us to focus on solving for $x$.

Step-by-Step Multiplication

Let's illustrate this multiplication step-by-step:

1. Multiply the first equation ($4x - 3y = 34$) by 2: $2 * (4x - 3y) = 2 * 34$ $8x - 6y = 68$
2. Multiply the second equation ($3x + 2y = 17$) by 3: $3 * (3x + 2y) = 3 * 17$ $9x + 6y = 51$

Now, we have two new equations:

• $8x - 6y = 68$
• $9x + 6y = 51$

Notice that the coefficients of $y$ are now -6 and +6, which are additive inverses.

Adding the Equations to Eliminate $y$

With the coefficients of $y$ as opposites, we can now add the two modified equations together. This will eliminate the $y$ term, leaving us with an equation in terms of $x$ only. By carefully choosing the multipliers in the previous step, we've set up a scenario where the $y$ variables gracefully cancel out, simplifying our path to finding the value of $x$. This strategic maneuver is the heart of the elimination method, transforming a system of two variables into a solvable single-variable equation.

Performing the Addition

Adding the equations $8x - 6y = 68$ and $9x + 6y = 51$ involves combining like terms on both sides of the equations. This process is straightforward and crucial in isolating $x$. The $y$ terms, with their opposite signs, will vanish, paving the way for a direct calculation of $x$. This step is where the elegance of the elimination method shines, as it reduces the complexity of the problem to a simple algebraic equation.

Let's perform the addition:

$(8x - 6y) + (9x + 6y) = 68 + 51$

Combine the $x$ terms: $8x + 9x = 17x$

The $y$ terms cancel out: $-6y + 6y = 0$

Combine the constants: $68 + 51 = 119$

So, the resulting equation is:

$17x = 119$

Solving for $x$

Now that we have eliminated $y$, we have a simple equation in one variable, $x$. To solve for $x$, we need to isolate it on one side of the equation. This is achieved by dividing both sides of the equation by the coefficient of $x$, which in this case is 17. This step is a fundamental algebraic operation, and it directly leads us to the numerical value of $x$ that satisfies the system of equations. Isolating $x$ in this manner is a clear demonstration of how the elimination method simplifies the problem, transforming it from a two-variable system to a single, easily solvable equation.

Isolating $x$

To isolate $x$, divide both sides of the equation $17x = 119$ by 17:

rac{17x}{17} = rac{119}{17}

This simplifies to:

$x = 7$

Therefore, the value of $x$ that satisfies the system of equations is 7. This result is a direct consequence of our strategic elimination of the $y$ variable, which allowed us to isolate and solve for $x$ with ease.

Conclusion: The Constants for Elimination

In conclusion, to eliminate the $y$ terms and solve for $x$ in the given system of equations, we should multiply the first equation ($4x - 3y = 34$) by 2 and the second equation ($3x + 2y = 17$) by 3. This strategic multiplication allows us to create opposite coefficients for the $y$ terms, enabling their elimination when the equations are added together. This, in turn, simplifies the problem, allowing us to solve for $x$ efficiently. The key to mastering the elimination method lies in recognizing the appropriate multipliers that will lead to the cancellation of a chosen variable, thereby making the system solvable. Understanding this principle is crucial for tackling more complex systems of equations in algebra and beyond.

This step-by-step approach not only solves the problem at hand but also reinforces the underlying principles of the elimination method, a valuable tool in the algebraic arsenal. By mastering this technique, students and practitioners can confidently tackle a wide range of problems involving systems of equations.