Solving Systems Of Equations A Comprehensive Guide

Solving systems of equations is a fundamental concept in mathematics with wide-ranging applications in various fields, including science, engineering, economics, and computer science. A system of equations is a set of two or more equations that share the same variables. The solution to a system of equations is the set of values for the variables that satisfy all equations simultaneously. In this article, we will delve into the methods for solving systems of equations, focusing on a specific example and providing a detailed step-by-step solution. Understanding how to solve these systems is crucial for anyone working with mathematical models and real-world problems. The ability to find solutions efficiently and accurately is a cornerstone of mathematical proficiency. This guide aims to equip you with the knowledge and skills necessary to tackle such problems with confidence.

Before diving into the solution, let's clarify what a system of equations is and why it's important. A system of equations consists of two or more equations with the same set of variables. The goal is to find the values of these variables that make all the equations true simultaneously. This concept is not just an abstract mathematical exercise; it models many real-world situations. For instance, in economics, it might represent supply and demand curves, where the solution represents the market equilibrium. In physics, it could describe the motion of objects under multiple forces. Therefore, understanding how to solve systems of equations is essential for making predictions and decisions in numerous contexts. The methods we will explore, such as substitution and elimination, are versatile tools applicable across various mathematical and scientific domains. Furthermore, these techniques build the foundation for more advanced mathematical concepts, such as linear algebra and calculus. By mastering the basics, you are preparing yourself for more complex problem-solving scenarios in the future.

In this article, we will focus on the following system of equations:

4x + 5y = 7
3x - 2y = -12

This is a system of two linear equations with two variables, x and y. Our objective is to find the values of x and y that satisfy both equations simultaneously. This type of system is common in introductory algebra courses and serves as a building block for more complex systems. The linearity of the equations means that they represent straight lines when graphed, and the solution to the system corresponds to the point where these lines intersect. However, we will focus on algebraic methods to find the solution, which are more precise and applicable to systems with more than two variables. The techniques we use here, such as elimination or substitution, can be extended to solve larger systems. The key is to manipulate the equations in a way that simplifies the system, allowing us to isolate one variable at a time. This systematic approach ensures we arrive at the correct solution efficiently and accurately. Mastering this process is a valuable skill for any student or professional dealing with mathematical problems.

The elimination method is a powerful technique for solving systems of equations. The core idea is to manipulate the equations so that, when added or subtracted, one of the variables is eliminated. This leaves us with a single equation in one variable, which we can easily solve. Then, we substitute this value back into one of the original equations to find the value of the other variable. This method is particularly effective when the coefficients of one variable are multiples of each other or can be easily made so. In our example, we will multiply the equations by suitable constants to make the coefficients of either x or y opposites, allowing them to cancel out when the equations are added. This systematic approach minimizes the chances of error and provides a clear path to the solution. The elimination method is a cornerstone of algebraic problem-solving and is widely used in various mathematical and scientific contexts. It's a versatile technique that can handle a wide range of systems of equations, making it an essential tool for anyone working with mathematical models.

Step 1: Multiply the Equations

To eliminate one of the variables, we need to multiply each equation by a number that produces opposite coefficients for either x or y. Let's choose to eliminate x. The coefficients of x are 4 and 3. The least common multiple of 4 and 3 is 12. So, we want to make the coefficients of x equal to 12 and -12. To achieve this, we multiply the first equation by 3 and the second equation by -4:

3 * (4x + 5y) = 3 * 7
-4 * (3x - 2y) = -4 * (-12)

This gives us:

12x + 15y = 21
-12x + 8y = 48

This step is crucial in the elimination method. By carefully choosing the multipliers, we set up the equations so that adding them together will eliminate one variable. The multipliers are selected to ensure that the coefficients of the chosen variable are opposites. This allows for a clean cancellation when the equations are combined. The process may seem a bit tedious, but it significantly simplifies the system and allows us to solve for the remaining variable. This step demonstrates the power of algebraic manipulation in making complex problems manageable. The strategic multiplication of equations is a fundamental skill in solving systems of equations and is applicable to a wide range of mathematical problems.

Step 2: Add the Equations

Now, we add the two modified equations together:

(12x + 15y) + (-12x + 8y) = 21 + 48

Combining like terms, we get:

23y = 69

Notice that the x terms have canceled out, which is the goal of the elimination method. This simplification is a direct result of the strategic multiplication we performed in the previous step. By adding the equations, we have reduced the system to a single equation with one variable, making it much easier to solve. This step highlights the elegance of the elimination method – by carefully manipulating the equations, we can isolate the variables and solve for them one at a time. The cancellation of terms is a powerful technique in algebra and is used in many different contexts. This step is a critical juncture in the solution process, as it bridges the gap between a complex system of equations and a simple, solvable equation. The ability to see these cancellations and strategically set them up is a key skill in mathematical problem-solving.

Step 3: Solve for y

To solve for y, we divide both sides of the equation by 23:

y = 69 / 23
y = 3

Now we have found the value of y. This step is a straightforward application of algebraic principles. By dividing both sides of the equation by the coefficient of y, we isolate the variable and determine its value. This step is a testament to the power of basic algebraic operations in solving complex problems. The simplicity of this step belies the effort that went into setting it up in the previous steps. The value of y = 3 is a crucial piece of the solution, and we will use it in the next step to find the value of x. This step demonstrates the importance of precision in mathematical calculations. A small error in this step could propagate through the rest of the solution, leading to an incorrect answer. Therefore, it's essential to double-check your work and ensure that each step is performed accurately.

Step 4: Substitute y into One of the Original Equations

Now that we have the value of y, we can substitute it into either of the original equations to solve for x. Let's use the first equation:

4x + 5y = 7

Substitute y = 3:

4x + 5(3) = 7
4x + 15 = 7

This step is a classic example of how substitution can simplify a problem. By replacing the variable y with its known value, we transform the equation into one with only one unknown, x. This makes the equation much easier to solve. The choice of which equation to substitute into is often a matter of convenience; either equation will work, but one might lead to simpler arithmetic. In this case, we chose the first equation, but the second equation would have worked just as well. This step is a critical link in the chain of solving the system of equations. It demonstrates the interplay between the two equations and how knowing the value of one variable allows us to find the value of the other. The ability to strategically substitute values is a fundamental skill in algebra and is used in various mathematical contexts.

Step 5: Solve for x

Subtract 15 from both sides:

4x = 7 - 15
4x = -8

Divide both sides by 4:

x = -8 / 4
x = -2

Now we have found the value of x. This step is another straightforward application of algebraic principles. By isolating the term with x and then dividing by its coefficient, we determine the value of x. This step is the culmination of the elimination method, where we have systematically reduced the system of equations to a single value for x. The value of x = -2, along with the value of y = 3 that we found earlier, constitutes the solution to the system of equations. This step underscores the importance of meticulous execution in mathematical problem-solving. Each algebraic manipulation must be performed correctly to arrive at the correct answer. The ability to confidently solve for x is a testament to the effectiveness of the elimination method and the power of algebraic techniques.

The substitution method is another powerful technique for solving systems of equations. In this method, we solve one equation for one variable and then substitute that expression into the other equation. This results in a single equation with one variable, which we can solve. Then, we substitute this value back into one of the original equations to find the value of the other variable. The substitution method is particularly useful when one of the equations can be easily solved for one variable in terms of the other. In our example, we will demonstrate this method by solving one of the equations for x or y and then substituting that expression into the other equation. This approach provides an alternative way to tackle systems of equations and is a valuable tool in any mathematician's arsenal. The substitution method is a cornerstone of algebraic problem-solving and is widely used in various mathematical and scientific contexts. It's a versatile technique that can handle a wide range of systems of equations, making it an essential skill for anyone working with mathematical models.

Step 1: Solve One Equation for One Variable

Let's solve the first equation for x:

4x + 5y = 7

Subtract 5y from both sides:

4x = 7 - 5y

Divide both sides by 4:

x = (7 - 5y) / 4

This step is crucial in the substitution method. By isolating one variable in terms of the other, we create an expression that we can substitute into the other equation. The choice of which equation and which variable to solve for is often a matter of convenience; we chose the first equation and solved for x, but other choices might have been equally valid. The key is to select the option that leads to the simplest algebraic manipulations. This step demonstrates the flexibility of algebraic techniques in manipulating equations to suit our needs. The expression x = (7 - 5y) / 4 now represents x in terms of y, which we will use in the next step. This step is a fundamental building block in the substitution method, and it sets the stage for simplifying the system of equations.

Step 2: Substitute into the Other Equation

Now, substitute this expression for x into the second equation:

3x - 2y = -12

Replace x with (7 - 5y) / 4:

3 * ((7 - 5y) / 4) - 2y = -12

This step is the heart of the substitution method. By replacing x with its equivalent expression in terms of y, we eliminate x from the second equation, leaving us with an equation that involves only y. This simplification is the key to solving the system. The resulting equation might look more complex at first, but it is a significant step forward because it involves only one unknown. This step demonstrates the power of substitution in transforming a system of equations into a more manageable form. The expression 3 * ((7 - 5y) / 4) - 2y = -12 is a pivotal equation in the solution process, and we will proceed to solve it for y in the next step. This step is a critical juncture in the solution process, as it bridges the gap between a complex system of equations and a single, solvable equation.

Step 3: Solve for y

Multiply both sides by 4 to eliminate the fraction:

3(7 - 5y) - 8y = -48

Distribute the 3:

21 - 15y - 8y = -48

Combine like terms:

21 - 23y = -48

Subtract 21 from both sides:

-23y = -69

Divide both sides by -23:

y = 3

Now we have found the value of y. This step involves a series of algebraic manipulations to isolate y. Each step is a careful application of algebraic principles, such as distribution, combining like terms, and isolating the variable. The process may seem lengthy, but it systematically leads us to the value of y. The value of y = 3 is a crucial piece of the solution, and we will use it in the next step to find the value of x. This step demonstrates the importance of precision in mathematical calculations. A small error in any of these steps could propagate through the rest of the solution, leading to an incorrect answer. Therefore, it's essential to double-check your work and ensure that each step is performed accurately.

Step 4: Substitute y into the Expression for x

Now that we have the value of y, we can substitute it back into the expression we found for x:

x = (7 - 5y) / 4

Substitute y = 3:

x = (7 - 5(3)) / 4
x = (7 - 15) / 4
x = -8 / 4
x = -2

This step is a straightforward application of substitution. By replacing y with its known value, we can easily calculate the value of x. This step completes the solution process, giving us the values of both x and y that satisfy the system of equations. The value of x = -2, along with the value of y = 3 that we found earlier, constitutes the solution to the system of equations. This step highlights the elegance of the substitution method, where we systematically reduce the system of equations to a single value for each variable. The ability to confidently substitute values is a fundamental skill in algebra and is used in various mathematical contexts.

Therefore, the solution to the system of equations is:

x = -2, y = 3

We can write this as an ordered pair: (-2, 3). This means that the point (-2, 3) is the intersection of the two lines represented by the equations in the system. This solution satisfies both equations simultaneously. To verify this, we can substitute these values back into the original equations and check if they hold true. This step is a crucial part of the problem-solving process. Verifying the solution ensures that we have not made any errors in our calculations. It also reinforces our understanding of what the solution represents – the values of the variables that make all equations in the system true. The ability to find and verify solutions to systems of equations is a fundamental skill in mathematics and has wide-ranging applications in various fields.

To verify the solution, we substitute x = -2 and y = 3 into the original equations:

For the first equation:

4x + 5y = 7
4(-2) + 5(3) = -8 + 15 = 7

For the second equation:

3x - 2y = -12
3(-2) - 2(3) = -6 - 6 = -12

Both equations are satisfied, so our solution is correct. This verification step is a crucial part of the problem-solving process. It provides a check on our work and ensures that we have not made any errors in our calculations. By substituting the solution back into the original equations, we confirm that the values of x and y satisfy both equations simultaneously. This step reinforces our understanding of what the solution represents – the values of the variables that make all equations in the system true. The ability to verify solutions is a valuable skill in mathematics and helps build confidence in our problem-solving abilities.

In this article, we have explored two methods for solving systems of equations: elimination and substitution. Both methods are powerful tools for finding the values of variables that satisfy multiple equations simultaneously. We have demonstrated these methods with a specific example, providing a detailed step-by-step solution and verification. Understanding how to solve systems of equations is a fundamental skill in mathematics and has wide-ranging applications in various fields. By mastering these techniques, you will be well-equipped to tackle a wide range of mathematical problems. The ability to solve systems of equations is a cornerstone of mathematical proficiency. It's not just about finding the right answer; it's about understanding the underlying concepts and developing a systematic approach to problem-solving. The skills you gain in solving systems of equations will serve you well in more advanced mathematical studies and in real-world applications.