Solving Systems Of Equations Find The Value Of X/2

In the realm of mathematics, solving systems of equations is a fundamental skill with applications spanning various fields, from engineering to economics. This article delves into the intricacies of solving a specific system of linear equations and determining the value of ${ \frac{x}{2} }$. We will embark on a step-by-step journey, unraveling the underlying concepts and techniques involved in arriving at the solution. Whether you are a student grappling with algebraic equations or a seasoned mathematician seeking a refresher, this guide aims to provide a clear and comprehensive understanding of the process.

The system of equations we will be tackling is:

${ \begin{aligned} \frac{x}{2} + 2y &= 14 \\ x - \frac{y}{2} &= 1 \end{aligned} }$

Our objective is to find the value of ${ \frac{x}{2} }$. To achieve this, we will employ a combination of algebraic manipulation and substitution methods. The journey may seem intricate at first, but with careful attention to detail and a systematic approach, we can conquer this mathematical challenge.

Unveiling the Equations The First Step

Before we dive into the solution, let's take a closer look at the equations themselves. The system comprises two linear equations, each with two variables, x and y. Our primary goal is to find the values of x and y that satisfy both equations simultaneously. Once we determine the value of x, we can easily calculate ${ \frac{x}{2} }$.

The first equation, ${ \frac{x}{2} + 2y = 14 }$, establishes a relationship between x and y. It states that half of x plus twice y equals 14. This equation can be visualized as a straight line on a coordinate plane, where every point on the line represents a solution that satisfies the equation.

The second equation, ${ x - \frac{y}{2} = 1 }$, presents another connection between x and y. It indicates that x minus half of y equals 1. Similar to the first equation, this equation can also be represented as a straight line on a coordinate plane.

The solution to the system of equations lies at the point where these two lines intersect. This intersection point represents the unique pair of (x, y) values that satisfy both equations. Our task is to find this intersection point, or in other words, solve for x and y.

Strategies for Solving Systems of Equations

There are several methods for solving systems of linear equations, each with its own strengths and weaknesses. Some common techniques include:

• Substitution: This method involves solving one equation for one variable and substituting that expression into the other equation. This reduces the system to a single equation with one variable, which can then be solved directly.
• Elimination: This method involves manipulating the equations to eliminate one variable by adding or subtracting the equations. This also results in a single equation with one variable.
• Graphing: This method involves plotting the equations on a coordinate plane and finding the point of intersection. This method is visually intuitive but may not be accurate for complex systems.
• Matrix methods: These methods involve representing the system of equations in matrix form and using matrix operations to solve for the variables. These methods are efficient for large systems of equations.

In this article, we will primarily use the substitution method to solve the given system of equations. This method is well-suited for this particular system and provides a clear and systematic approach.

The Substitution Method A Detailed Walkthrough

Now, let's apply the substitution method to solve the system of equations:

${ \begin{aligned} \frac{x}{2} + 2y &= 14 \\ x - \frac{y}{2} &= 1 \end{aligned} }$

Our first step is to solve one of the equations for one variable. Let's choose the second equation, ${ x - \frac{y}{2} = 1 }$, and solve for x. To do this, we add ${ \frac{y}{2} }$ to both sides of the equation:

${ x = 1 + \frac{y}{2} }$

Now we have an expression for x in terms of y. This is a crucial step in the substitution method. We will now substitute this expression for x into the first equation.

Substituting into the First Equation

The first equation is ${ \frac{x}{2} + 2y = 14 }$. We will replace x with the expression we just found, ${ 1 + \frac{y}{2} }$:

${ \frac{1 + \frac{y}{2}}{2} + 2y = 14 }$

This equation now contains only one variable, y. Our next step is to simplify this equation and solve for y. To do this, we first distribute the division by 2 in the first term:

${ \frac{1}{2} + \frac{y}{4} + 2y = 14 }$

Now, we need to combine the y terms. To do this, we find a common denominator for the y terms, which is 4. We rewrite 2y as ${ \frac{8y}{4} }$:

${ \frac{1}{2} + \frac{y}{4} + \frac{8y}{4} = 14 }$

Combining the y terms, we get:

${ \frac{1}{2} + \frac{9y}{4} = 14 }$

Isolating y

To isolate the y term, we subtract ${ \frac{1}{2} }$ from both sides of the equation:

${ \frac{9y}{4} = 14 - \frac{1}{2} }$

We need to find a common denominator to subtract the fractions on the right side. We rewrite 14 as ${ \frac{28}{2} }$:

${ \frac{9y}{4} = \frac{28}{2} - \frac{1}{2} }$

Subtracting the fractions, we get:

${ \frac{9y}{4} = \frac{27}{2} }$

Now, to solve for y, we multiply both sides of the equation by ${ \frac{4}{9} }$:

${ y = \frac{27}{2} \cdot \frac{4}{9} }$

Simplifying the expression, we get:

${ y = 6 }$

We have now found the value of y. The next step is to substitute this value back into one of the original equations to solve for x.

Finding x and Calculating x/2

We have determined that y = 6. Now we need to find the value of x. We can substitute this value into either of the original equations. Let's use the equation we solved for x earlier:

${ x = 1 + \frac{y}{2} }$

Substituting y = 6, we get:

${ x = 1 + \frac{6}{2} }$

Simplifying the expression, we get:

${ x = 1 + 3 }$

${ x = 4 }$

Now that we have found the value of x, which is 4, we can calculate ${ \frac{x}{2} }$. Dividing x by 2, we get:

${ \frac{x}{2} = \frac{4}{2} }$

${ \frac{x}{2} = 2 }$

Therefore, the value of ${ \frac{x}{2} }$ is 2.

Verification

To ensure that our solution is correct, we can substitute the values of x and y back into the original equations and verify that they hold true.

For the first equation, ${ \frac{x}{2} + 2y = 14 }$, we substitute x = 4 and y = 6:

${ \frac{4}{2} + 2(6) = 14 }$

${ 2 + 12 = 14 }$

${ 14 = 14 }$

The first equation holds true.

For the second equation, ${ x - \frac{y}{2} = 1 }$, we substitute x = 4 and y = 6:

${ 4 - \frac{6}{2} = 1 }$

${ 4 - 3 = 1 }$

${ 1 = 1 }$

The second equation also holds true. This confirms that our solution, x = 4 and y = 6, is correct.

Conclusion The Value of x/2

In this comprehensive guide, we have successfully solved the system of equations:

${ \begin{aligned} \frac{x}{2} + 2y &= 14 \\ x - \frac{y}{2} &= 1 \end{aligned} }$

We employed the substitution method, a powerful technique for solving systems of linear equations. Through a series of algebraic manipulations and substitutions, we determined the values of x and y that satisfy both equations simultaneously.

Our primary objective was to find the value of ${ \frac{x}{2} }$. After solving for x, we calculated ${ \frac{x}{2} }$ and arrived at the solution:

${ \frac{x}{2} = 2 }$

Therefore, the value of ${ \frac{x}{2} }$ is 2. This corresponds to option D in the given choices.

Key Takeaways

• Solving systems of equations is a fundamental skill in mathematics with wide-ranging applications.
• The substitution method is a versatile technique for solving systems of linear equations.
• Careful attention to detail and a systematic approach are crucial for success in solving mathematical problems.
• Verification is an essential step to ensure the accuracy of the solution.

By mastering the concepts and techniques presented in this article, you will be well-equipped to tackle a wide range of systems of equations and related problems. Whether you are a student, educator, or simply a math enthusiast, the ability to solve systems of equations is a valuable asset.

Practice Problems to Hone Your Skills

To further solidify your understanding and enhance your problem-solving abilities, consider working through the following practice problems:

1. Solve the system of equations:

   ${ \begin{aligned} 2x + y &= 7 \\ x - y &= 2 \end{aligned} }$

   Find the values of x and y.

2. Solve the system of equations:

   ${ \begin{aligned} 3x - 2y &= 5 \\ x + y &= 4 \end{aligned} }$

   What is the value of x + y?

3. Solve the system of equations:

   ${ \begin{aligned} \frac{x}{3} + y &= 5 \\ x - 2y &= 2 \end{aligned} }$

   Determine the value of ${ \frac{y}{2} }$.

By tackling these practice problems, you will gain confidence in your ability to solve systems of equations and apply the substitution method effectively. Remember, practice makes perfect!

Further Exploration

If you are interested in delving deeper into the world of systems of equations, there are numerous resources available to expand your knowledge and skills. Consider exploring the following topics:

• Elimination method: Learn how to solve systems of equations by eliminating variables through addition or subtraction.
• Matrix methods: Discover how to represent systems of equations in matrix form and use matrix operations to solve them.
• Applications of systems of equations: Explore real-world applications of systems of equations in various fields, such as engineering, economics, and computer science.
• Non-linear systems of equations: Investigate systems of equations that involve non-linear equations, such as quadratic or exponential equations.

By expanding your knowledge in these areas, you will gain a more comprehensive understanding of systems of equations and their significance in mathematics and beyond.

This guide has provided a detailed explanation of how to solve a specific system of equations and find the value of ${ \frac{x}{2} }$. By following the steps outlined and practicing regularly, you can master this essential mathematical skill and unlock new levels of problem-solving proficiency. Remember, mathematics is a journey of exploration and discovery, so embrace the challenges and enjoy the process of learning!