Solving System Of Equations 3y + 11 = 4x And 10x + 2y + 1 = 0 A Step-by-Step Guide

Introduction: Unveiling the World of Linear Equations

In the realm of mathematics, solving systems of linear equations stands as a cornerstone concept with far-reaching applications across various disciplines. From engineering and physics to economics and computer science, the ability to efficiently and accurately solve these systems is paramount. This article delves into the intricacies of solving a specific system of linear equations: $\begin{aligned} 3y + 11 &= 4x \\ 10x + 2y + 1 &= 0 \end{aligned}$. We will explore different methods, providing a step-by-step guide to unravel the solution while highlighting the underlying principles and potential pitfalls. Understanding the system of linear equations is essential for anyone seeking to master mathematical problem-solving.

At its core, a system of linear equations represents a set of two or more equations involving the same variables. The solution to such a system is the set of values for the variables that satisfy all equations simultaneously. Geometrically, each linear equation represents a straight line, and the solution corresponds to the point(s) where these lines intersect. This geometric interpretation provides a valuable visual aid for understanding the nature of solutions: one unique solution (lines intersect at a single point), infinitely many solutions (lines coincide), or no solution (lines are parallel). In the given system, we have two equations with two unknowns (x and y), which is a common scenario in linear algebra. The challenge lies in manipulating these equations to isolate the variables and determine their values. The solution to this specific system will be found using multiple approaches to demonstrate the versatility of linear equation solving.

Before diving into the methods, let's emphasize the importance of accuracy in each step. A small error in arithmetic or algebraic manipulation can lead to an incorrect solution. Therefore, it is crucial to double-check every calculation and maintain a systematic approach. This not only ensures the correctness of the final answer but also enhances the understanding of the underlying concepts. The ability to solve linear equations accurately forms the foundation for tackling more complex mathematical problems. This article aims to not only provide a solution to the given system but also to equip readers with the skills and confidence to solve similar problems independently. We will be covering substitution, elimination, and graphical methods, each offering a unique perspective on the solution process.

Method 1: The Substitution Method

The substitution method is a powerful technique for solving systems of linear equations. The core idea is to solve one equation for one variable and then substitute that expression into the other equation. This process transforms the system into a single equation with one variable, which can be easily solved. Once the value of that variable is found, it can be substituted back into either of the original equations to find the value of the other variable. Let's apply this method to our system: $\begin{aligned} 3y + 11 &= 4x \\ 10x + 2y + 1 &= 0 \end{aligned}$.

Step 1: Solve one equation for one variable.

Let's choose the first equation, $3y + 11 = 4x$, and solve for $x$. To do this, we can divide both sides of the equation by 4 after isolating the $4x$ term. Rearranging the equation gives us $4x = 3y + 11$. Now, dividing by 4, we get $x = \frac{3y + 11}{4}$. This expression now represents $x$ in terms of $y$. This step is crucial as it sets the stage for the substitution process. By isolating $x$, we have created a direct link between the two variables, which will allow us to reduce the system to a single equation. The choice of which variable to isolate often depends on the specific equations. In this case, solving the first equation for $x$ seems straightforward as it avoids fractions in the intermediate steps. However, solving for $y$ first would also be a valid approach, though it might involve slightly more complex fractions.

Step 2: Substitute the expression into the other equation.

Now, we substitute the expression for $x$, which is $x = \frac{3y + 11}{4}$, into the second equation, $10x + 2y + 1 = 0$. Replacing $x$ with its equivalent expression, we get $10(\frac{3y + 11}{4}) + 2y + 1 = 0$. This is a single equation in terms of $y$, which we can solve. This step is the heart of the substitution method, where we replace one variable with its equivalent expression, effectively eliminating that variable from the equation. The resulting equation might look complex at first, but it is a single-variable equation that can be solved using standard algebraic techniques. The key is to carefully distribute and combine like terms to simplify the equation and isolate the variable. The accuracy of this substitution is paramount, as any error here will propagate through the rest of the solution.

Step 3: Solve for the remaining variable.

Simplifying the equation $10(\frac{3y + 11}{4}) + 2y + 1 = 0$, we first multiply both sides by 4 to eliminate the fraction: $10(3y + 11) + 8y + 4 = 0$. Expanding the terms, we get $30y + 110 + 8y + 4 = 0$. Combining like terms, we have $38y + 114 = 0$. Subtracting 114 from both sides gives $38y = -114$. Finally, dividing by 38, we find $y = -3$. This step involves algebraic manipulation to isolate the variable $y$. The process includes distributing constants, combining like terms, and performing inverse operations to solve for the unknown. Each step must be performed carefully to maintain the equality of the equation and avoid errors. The result, $y = -3$, is one part of the solution to the system of equations.

Step 4: Substitute back to find the other variable.

Now that we have found $y = -3$, we substitute this value back into either of the original equations or the expression we derived for $x$. Let's use the expression $x = \frac{3y + 11}{4}$. Substituting $y = -3$, we get $x = \frac{3(-3) + 11}{4} = \frac{-9 + 11}{4} = \frac{2}{4} = \frac{1}{2}$. Therefore, $x = \frac{1}{2}$. This final step completes the substitution method by finding the value of the second variable. Once one variable is known, substituting its value back into a previous equation allows us to determine the value of the remaining variable. The choice of which equation to use for this substitution is often a matter of convenience; however, using the expression we derived for $x$ directly simplifies the calculation in this case.

Solution:

The solution to the system of equations is $x = \frac{1}{2}$ and $y = -3$. We can write this as an ordered pair: $(\frac{1}{2}, -3)$.

Method 2: The Elimination Method

The elimination method, also known as the addition method, is another powerful technique for solving systems of linear equations. The key idea behind this method is to manipulate the equations so that the coefficients of one of the variables are opposites. Then, by adding the equations together, that variable is eliminated, leaving a single equation with one variable. Let's apply this method to our system: $\begin{aligned} 3y + 11 &= 4x \\ 10x + 2y + 1 &= 0 \end{aligned}$.

Step 1: Rewrite the equations in standard form.

First, we rewrite both equations in the standard form $Ax + By = C$. For the first equation, $3y + 11 = 4x$, we subtract $4x$ from both sides to get $-4x + 3y = -11$. For the second equation, $10x + 2y + 1 = 0$, we subtract 1 from both sides to get $10x + 2y = -1$. So, our system now looks like this: $\begin{aligned} -4x + 3y &= -11 \\ 10x + 2y &= -1 \end{aligned}$. This step is crucial for the elimination method because it aligns the variables, making it easier to identify the coefficients and manipulate the equations. The standard form facilitates the process of multiplying equations by constants to create opposing coefficients. Ensuring that both equations are in this form is a critical prerequisite for the next steps.

Step 2: Multiply the equations to make the coefficients of one variable opposites.

To eliminate the $y$ variable, we can multiply the first equation by 2 and the second equation by -3. This will make the coefficients of $y$ be 6 and -6, respectively. Multiplying the first equation, $-4x + 3y = -11$, by 2 gives us $-8x + 6y = -22$. Multiplying the second equation, $10x + 2y = -1$, by -3 gives us $-30x - 6y = 3$. Now, our system looks like this: $\begin{aligned} -8x + 6y &= -22 \\ -30x - 6y &= 3 \end{aligned}$. This step is the heart of the elimination method. The goal is to strategically multiply each equation by a constant so that the coefficients of one of the variables become additive inverses (opposites). This allows for the cancellation of that variable when the equations are added together. The choice of which variable to eliminate and the constants to multiply by depends on the specific equations in the system. Here, we chose to eliminate $y$ because the coefficients were relatively small and easily manipulated.

Step 3: Add the equations to eliminate one variable.

Adding the two equations, $\begin{aligned} -8x + 6y &= -22 \\ -30x - 6y &= 3 \end{aligned}$, we get $(-8x - 30x) + (6y - 6y) = -22 + 3$, which simplifies to $-38x = -19$. This step is the culmination of the elimination process. By adding the equations together, the $y$ terms cancel out, leaving us with a single equation in terms of $x$. This significantly simplifies the problem, as we can now easily solve for $x$. The elimination of one variable is the key advantage of this method, as it reduces the system to a solvable single-variable equation.

Step 4: Solve for the remaining variable.

Dividing both sides of $-38x = -19$ by -38, we get $x = \frac{-19}{-38} = \frac{1}{2}$. This step involves solving the single-variable equation obtained after elimination. In this case, it is a simple division operation to isolate $x$. The result, $x = \frac{1}{2}$, is one part of the solution to the system of equations.

Step 5: Substitute back to find the other variable.

Now that we have found $x = \frac{1}{2}$, we substitute this value back into either of the original equations in standard form. Let's use the equation $10x + 2y = -1$. Substituting $x = \frac{1}{2}$, we get $10(\frac{1}{2}) + 2y = -1$, which simplifies to $5 + 2y = -1$. Subtracting 5 from both sides gives $2y = -6$. Finally, dividing by 2, we find $y = -3$. This final step completes the elimination method by finding the value of the second variable. Similar to the substitution method, once one variable is known, substituting its value back into a previous equation allows us to determine the value of the remaining variable. The choice of which equation to use for this substitution is often a matter of convenience.

Solution:

The solution to the system of equations is $x = \frac{1}{2}$ and $y = -3$. We can write this as an ordered pair: $(\frac{1}{2}, -3)$.

Method 3: Graphical Method

The graphical method provides a visual approach to solving systems of linear equations. The idea is to graph each equation on the coordinate plane. The solution to the system corresponds to the point(s) where the lines intersect. This method is particularly useful for visualizing the nature of the solutions: a single intersection point indicates a unique solution, coinciding lines indicate infinitely many solutions, and parallel lines indicate no solution. Let's apply this method to our system: $\begin{aligned} 3y + 11 &= 4x \\ 10x + 2y + 1 &= 0 \end{aligned}$.

Step 1: Rewrite the equations in slope-intercept form.

To graph the lines easily, we rewrite each equation in slope-intercept form, which is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. For the first equation, $3y + 11 = 4x$, we subtract 11 from both sides to get $3y = 4x - 11$. Then, dividing by 3, we get $y = \frac{4}{3}x - \frac{11}{3}$. For the second equation, $10x + 2y + 1 = 0$, we subtract $10x$ and 1 from both sides to get $2y = -10x - 1$. Then, dividing by 2, we get $y = -5x - \frac{1}{2}$. So, our equations in slope-intercept form are: $\begin{aligned} y &= \frac{4}{3}x - \frac{11}{3} \\ y &= -5x - \frac{1}{2} \end{aligned}$. Rewriting the equations in slope-intercept form is a critical step for the graphical method. This form makes it easy to identify the slope and y-intercept of each line, which are the two key parameters needed to plot the lines accurately. The slope indicates the steepness and direction of the line, while the y-intercept is the point where the line crosses the y-axis. Having the equations in this form allows for a straightforward visual representation of the system.

Step 2: Graph the lines.

Now, we graph the two lines on the coordinate plane. For the first line, $y = \frac{4}{3}x - \frac{11}{3}$, the slope is $\frac{4}{3}$ and the y-intercept is $-\frac{11}{3}$ (approximately -3.67). For the second line, $y = -5x - \frac{1}{2}$, the slope is -5 and the y-intercept is $-\frac{1}{2}$ (-0.5). Plotting these lines on a graph, we can see that they intersect at a single point. Graphing the lines accurately is essential for the graphical method to yield the correct solution. This involves plotting at least two points on each line and then drawing a straight line through those points. The y-intercept is a convenient point to start with, and then using the slope, we can find another point. For example, if the slope is $\frac{4}{3}$, starting from the y-intercept, we can move 3 units to the right and 4 units up to find another point on the line. The intersection point of the lines visually represents the solution to the system of equations.

Step 3: Identify the point of intersection.

By observing the graph, we can see that the lines intersect at the point $(\frac{1}{2}, -3)$. This point represents the solution to the system of equations. The point of intersection is the key to the graphical method. This point's coordinates satisfy both equations simultaneously, making it the solution to the system. Reading the coordinates of the intersection point from the graph provides the values of $x$ and $y$ that solve the system.

Solution:

The solution to the system of equations, as determined graphically, is $x = \frac{1}{2}$ and $y = -3$. This confirms the solutions obtained using the substitution and elimination methods.

Conclusion: Mastering the Art of Solving Linear Equations

In this comprehensive guide, we've explored three distinct methods for solving systems of linear equations: the substitution method, the elimination method, and the graphical method. Each method offers a unique approach and provides valuable insights into the nature of linear systems. The specific system we tackled, $\begin{aligned} 3y + 11 &= 4x \\ 10x + 2y + 1 &= 0 \end{aligned}$, served as a practical example to illustrate the application of these techniques. The fact that all three methods converged on the same solution, $x = \frac{1}{2}$ and $y = -3$, reinforces the correctness and consistency of these approaches. Mastering these methods is crucial for success in various mathematical and scientific fields.

The substitution method is particularly effective when one of the equations can be easily solved for one variable in terms of the other. This method reduces the system to a single equation with one variable, making it straightforward to solve. The elimination method, on the other hand, shines when the coefficients of one variable can be easily made opposites by multiplying the equations by appropriate constants. This method eliminates one variable, simplifying the system. The graphical method provides a visual representation of the system and the solution, making it an intuitive approach for understanding the concept. While it might not always yield precise solutions for systems with non-integer solutions, it offers a valuable visual confirmation of the algebraic solutions.

The ability to solve systems of linear equations is not just a mathematical skill; it is a fundamental tool for problem-solving in diverse fields. From determining the intersection of supply and demand curves in economics to analyzing the forces acting on a structure in engineering, linear equations play a crucial role. Furthermore, the concepts and techniques learned in solving these systems form the foundation for more advanced topics in linear algebra, such as matrices and vectors. Therefore, a solid understanding of these methods is an investment in your mathematical and analytical capabilities. We encourage you to practice these methods with various systems of equations to solidify your understanding and develop your problem-solving skills. The journey to mastery of linear equations is a rewarding one, opening doors to a wide range of applications and further mathematical explorations.