Solving Systems Of Equations A Detailed Solution For 3x-5y=21 And X-4y=3
Introduction to Systems of Linear Equations
In the realm of mathematics, solving systems of linear equations is a fundamental skill with wide-ranging applications in various fields, including engineering, economics, and computer science. A system of linear equations is a set of two or more linear equations containing the same variables. The solution to a system of linear equations is the set of values for the variables that satisfy all equations simultaneously. This article will delve into the step-by-step process of solving a specific system of linear equations: \begin{array}{l} 3x-5y=21 \ x-4y=3 \end{array}. We will explore different methods, including substitution, elimination, and matrix methods, providing a comprehensive understanding of how to tackle such problems.
Understanding the Importance of Linear Equations
Linear equations, characterized by their straight-line graphs, are the building blocks of many mathematical models. They represent relationships where the change in one variable is directly proportional to the change in another. The ability to solve systems of linear equations is crucial because it allows us to find the points where these relationships intersect, representing solutions that satisfy multiple conditions simultaneously. For instance, in economics, these systems can model supply and demand curves, and the solution represents the market equilibrium point. In engineering, they can be used to analyze electrical circuits or structural systems. Mastering these techniques not only strengthens mathematical proficiency but also unlocks the potential to solve real-world problems.
The Specific System: A Closer Look
The system of equations we will be focusing on is:
\begin{array}{l} 3x - 5y = 21 \ x - 4y = 3 \end{array}
This system consists of two linear equations with two unknowns, x and y. Our goal is to find the values of x and y that make both equations true. This can be visualized graphically as finding the point of intersection of the two lines represented by these equations. There are several methods we can use to achieve this, each with its own advantages and disadvantages. We will explore the most common methods, providing clear explanations and examples to ensure a thorough understanding of the solution process.
Method 1: Substitution
The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This reduces the system to a single equation with one variable, which can then be easily solved. Let's apply this method to our system:
\begin{array}{l} 3x - 5y = 21 \ x - 4y = 3 \end{array}
Step 1: Solve one equation for one variable
It's often easiest to choose an equation where one of the variables has a coefficient of 1 or -1. In this case, the second equation, x - 4y = 3, is ideal. We can solve for x by adding 4y to both sides:
x = 4y + 3
Step 2: Substitute into the other equation
Now we substitute this expression for x into the first equation:
3(4y + 3) - 5y = 21
Step 3: Solve for the remaining variable
Distribute the 3 and simplify:
12y + 9 - 5y = 21
Combine like terms:
7y + 9 = 21
Subtract 9 from both sides:
7y = 12
Divide by 7:
y = 12/7
Step 4: Substitute back to find the other variable
Now that we have the value of y, we can substitute it back into either of the original equations or the expression we found for x. Using x = 4y + 3 is the easiest option:
x = 4(12/7) + 3
x = 48/7 + 3
To add these, we need a common denominator. Convert 3 to 21/7:
x = 48/7 + 21/7
x = 69/7
Therefore, the solution to the system of equations using the substitution method is x = 69/7 and y = 12/7.
The Substitution Method is particularly useful when one of the equations can be easily solved for one variable in terms of the other. It provides a direct way to reduce the system to a single equation, making it a powerful tool in solving linear systems.
Method 2: Elimination
The elimination method, also known as the addition method, involves manipulating the equations so that the coefficients of one variable 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{array}{l} 3x - 5y = 21 \ x - 4y = 3 \end{array}
Step 1: Multiply equations to get opposite coefficients
We want to eliminate either x or y. Let's eliminate x. To do this, we can multiply the second equation by -3:
-3(x - 4y) = -3(3)
-3x + 12y = -9
Now our system looks like this:
\begin{array}{l} 3x - 5y = 21 \ -3x + 12y = -9 \end{array}
Notice that the coefficients of x are now 3 and -3, which are opposites.
Step 2: Add the equations
Add the two equations together:
(3x - 5y) + (-3x + 12y) = 21 + (-9)
3x and -3x cancel out, leaving:
7y = 12
Step 3: Solve for the remaining variable
Divide both sides by 7:
y = 12/7
Step 4: Substitute back to find the other variable
Now we substitute y = 12/7 back into one of the original equations. Let's use the second equation, x - 4y = 3:
x - 4(12/7) = 3
x - 48/7 = 3
Add 48/7 to both sides:
x = 3 + 48/7
Convert 3 to 21/7:
x = 21/7 + 48/7
x = 69/7
Therefore, the solution to the system of equations using the elimination method is x = 69/7 and y = 12/7, which is the same solution we found using substitution. This consistency across different methods reinforces the accuracy of our solution.
The Elimination Method shines when the coefficients of one variable are easily made opposites through multiplication. It's a systematic approach that eliminates one variable at a time, simplifying the system to a manageable equation.
Method 3: Matrix Methods (Optional)
For larger systems of equations, matrix methods provide a more efficient and organized approach. While we won't delve into the full complexity of matrix operations, we can briefly illustrate how our system can be represented in matrix form and solved using techniques like Gaussian elimination or matrix inversion.
Our system can be written as:
\begin{bmatrix} 3 & -5 \ 1 & -4 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 21 \ 3 \end{bmatrix}
This is in the form Ax = b, where A is the coefficient matrix, x is the variable matrix, and b is the constant matrix. Solving for x involves finding the inverse of A (if it exists) and multiplying it by b: x = A⁻¹b. While the manual computation of matrix inverses can be tedious for larger matrices, it's a standard operation in computational software and calculators. Matrix methods are essential for solving systems with many variables, as they provide a structured and scalable approach.
Verifying the Solution
It's always a good practice to verify the solution by plugging the values of x and y back into the original equations to ensure they are satisfied. Let's check our solution, x = 69/7 and y = 12/7:
Equation 1: 3x - 5y = 21
3(69/7) - 5(12/7) = 207/7 - 60/7 = 147/7 = 21 (Correct)
Equation 2: x - 4y = 3
69/7 - 4(12/7) = 69/7 - 48/7 = 21/7 = 3 (Correct)
Since the solution satisfies both equations, we can confidently conclude that x = 69/7 and y = 12/7 is the correct solution to the system of equations. Verification is a crucial step in the problem-solving process, ensuring accuracy and building confidence in the results.
Conclusion
In this article, we've explored the process of solving the system of linear equations \begin{array}{l} 3x-5y=21 \ x-4y=3 \end{array} using various methods, including substitution and elimination. We also touched upon the concept of matrix methods for larger systems. Each method offers a unique approach, and the choice of method often depends on the specific system and personal preference. The key takeaway is the understanding of the underlying principles and the ability to apply these methods systematically to arrive at the correct solution. The solution we found, x = 69/7 and y = 12/7, highlights the power of these techniques in solving mathematical problems with practical implications. Mastering these skills is essential for anyone pursuing studies or careers in fields that require mathematical proficiency.
