Solving Systems Of Equations By Substitution A Comprehensive Guide
In the realm of mathematics, solving systems of equations is a fundamental skill. This article delves into the method of substitution, a powerful technique for finding solutions to systems of equations, particularly those involving three variables. We will explore this method through a detailed example, providing a step-by-step guide to ensure clarity and understanding. This technique, substitution, is a cornerstone of algebra, enabling us to unravel the relationships between multiple variables and arrive at precise solutions. The beauty of substitution lies in its ability to transform a complex problem into a series of simpler steps, making it accessible even to those new to the world of simultaneous equations. By mastering this method, you will gain a valuable tool for tackling a wide range of mathematical challenges.
Understanding the Substitution Method
The substitution method involves solving one equation for one variable and then substituting that expression into another equation. This process reduces the number of variables in the system, making it easier to solve. In essence, we are replacing one variable with an equivalent expression, allowing us to isolate and determine the value of the remaining variables. This method is particularly effective when one or more equations can be easily solved for a single variable. For example, if an equation is already in the form x = ... or y = ..., substitution becomes a natural and efficient choice. However, even if the equations are not initially in this form, we can often manipulate them to isolate a variable and pave the way for substitution. The key is to identify the equation and variable that will lead to the simplest subsequent steps. This strategic approach can significantly streamline the solving process and minimize the chances of errors.
When to Use Substitution
Substitution shines when dealing with systems where one variable can be easily isolated. This often occurs when an equation has a variable with a coefficient of 1 or -1. In such cases, solving for that variable involves minimal algebraic manipulation, making substitution a highly efficient strategy. However, substitution is not limited to these scenarios. It can be applied to any system of equations, although it might become more cumbersome if the equations are complex or require extensive manipulation to isolate a variable. In those situations, other methods, such as elimination, might be more advantageous. Ultimately, the choice of method depends on the specific characteristics of the system and the solver's personal preference and skill. A proficient mathematician is adept at recognizing the most efficient approach for each problem, and often, a combination of methods can be the key to unlocking the solution.
Example: Solving a System of Equations
Let's consider the following system of equations:
\left\{
\begin{array}{l}
x + 2z = 9 \\
5x + y + 7z = 35 \\
2x + 6y + z = 18
\end{array}
\right.
Our goal is to find the values of x, y, and z that satisfy all three equations simultaneously. We will achieve this by systematically applying the substitution method.
Step 1: Solve for One Variable
Looking at the system, the first equation, x + 2z = 9, appears to be the simplest to manipulate. We can easily solve for x:
x = 9 - 2z
This isolates x in terms of z, providing us with the expression we will use for substitution. The strategic choice of which variable to solve for is crucial in making the process smoother and less prone to errors. In this case, choosing the first equation and solving for x was a deliberate decision to minimize the complexity of the subsequent steps. The expression x = 9 - 2z now becomes our key to unlocking the solution to the entire system.
Step 2: Substitute into Other Equations
Now, we substitute the expression for x (9 - 2z) into the second and third equations:
- Equation 2: 5(9 - 2z) + y + 7z = 35
- Equation 3: 2(9 - 2z) + 6y + z = 18
This substitution eliminates x from these equations, leaving us with a system of two equations in two variables (y and z). The act of substitution is a pivotal step in reducing the complexity of the system. By replacing x with its equivalent expression in terms of z, we effectively decouple the equations, allowing us to focus on the remaining variables. This strategic simplification is at the heart of the substitution method's power and efficiency.
Step 3: Simplify the Equations
Let's simplify the equations we obtained in the previous step:
- Equation 2 (Simplified): 45 - 10z + y + 7z = 35 => y - 3z = -10
- Equation 3 (Simplified): 18 - 4z + 6y + z = 18 => 6y - 3z = 0
We now have a simpler system:
\left\{
\begin{array}{l}
y - 3z = -10 \\
6y - 3z = 0
\end{array}
\right.
These simplified equations are much easier to work with. The process of simplification is crucial in ensuring that we can effectively solve the system. By combining like terms and rearranging the equations, we reduce the risk of errors and make the next steps more manageable. The system now presents a clearer path to finding the values of y and z, bringing us closer to the complete solution.
Step 4: Solve for Another Variable
We can solve the first equation (y - 3z = -10) for y:
y = 3z - 10
This isolates y in terms of z, setting us up for another substitution. The decision to solve the first equation for y is based on its simplicity and the ease with which y can be isolated. This strategic choice further streamlines the solution process. By expressing y in terms of z, we are one step closer to having a single equation with a single unknown, which can be readily solved.
Step 5: Substitute Again
Substitute this expression for y into the second simplified equation (6y - 3z = 0):
6(3z - 10) - 3z = 0
This substitution eliminates y, leaving us with an equation in just z. The power of substitution is once again evident here. By replacing y with its equivalent expression, we have successfully reduced the system to a single equation with a single variable. This is a major milestone in the solving process, as it allows us to isolate and determine the value of z directly.
Step 6: Solve for z
Simplify and solve for z:
18z - 60 - 3z = 0
15z = 60
z = 4
We have found the value of z! This is a significant breakthrough, as it provides a foundation for finding the values of the other variables. The algebraic manipulations involved in solving for z are straightforward, demonstrating the effectiveness of the substitution method in simplifying the problem. With z now known, we can backtrack through our previous substitutions to determine the values of x and y.
Step 7: Back-Substitute to Find y
Substitute the value of z (4) back into the equation y = 3z - 10:
y = 3(4) - 10
y = 12 - 10
y = 2
We have found the value of y! The process of back-substitution is a crucial step in completing the solution. By plugging the value of z back into the equation where y was expressed in terms of z, we can easily calculate the value of y. This step highlights the interconnectedness of the variables in the system and how the solution for one variable leads to the solution for others.
Step 8: Back-Substitute to Find x
Substitute the value of z (4) back into the equation x = 9 - 2z:
x = 9 - 2(4)
x = 9 - 8
x = 1
We have found the value of x! This final back-substitution completes the solution process. By plugging the value of z into the equation where x was expressed in terms of z, we determine the value of x. This step demonstrates the elegance of the substitution method, where each variable is systematically solved by leveraging the relationships established in the previous steps.
Solution
The solution to the system of equations is:
(x, y, z) = (1, 2, 4)
This means that the values x = 1, y = 2, and z = 4 satisfy all three equations in the original system. The solution represents the point where all three planes defined by the equations intersect in three-dimensional space. Verifying the solution by plugging these values back into the original equations is a crucial step to ensure accuracy and confirm that no errors were made during the solving process. This final check provides confidence in the correctness of the solution and completes the problem-solving journey.
Conclusion
The substitution method is a versatile tool for solving systems of equations. By systematically solving for variables and substituting expressions, we can reduce complex problems into simpler ones. This example demonstrates the power and efficiency of substitution, providing a clear path to finding solutions for systems with multiple variables. Mastering this method is essential for anyone delving into the world of algebra and beyond. The ability to solve systems of equations is a fundamental skill that has wide-ranging applications in various fields, including physics, engineering, economics, and computer science. By understanding and practicing the substitution method, you equip yourself with a valuable problem-solving tool that will serve you well in your mathematical endeavors.
This comprehensive guide should empower you to tackle similar problems with confidence. Remember, practice is key to mastering any mathematical technique. So, delve into more examples, and you'll soon find yourself effortlessly navigating the world of systems of equations.
