Solving Systems Of Equations 4x + 3y = 9 And 2x + 5y = 15
Introduction to Systems of Linear Equations
In the realm of mathematics, particularly in algebra, a system of linear equations represents a collection of two or more linear equations involving the same set of variables. Solving such a system entails finding values for the variables that satisfy all equations simultaneously. This concept is fundamental in various fields, including engineering, economics, and computer science, where real-world problems often translate into systems of equations. The system presented here, consisting of the equations 4x + 3y = 9 and 2x + 5y = 15, provides a classic example of a two-variable linear system. Our goal is to explore different methods to determine the values of x and y that make both equations true.
Understanding the nature of these systems is crucial. A system of linear equations can have one solution (a unique pair of x and y values), no solution (the lines are parallel and never intersect), or infinitely many solutions (the equations represent the same line). The graphical interpretation of a system of two linear equations is the intersection of two lines on a coordinate plane. The point of intersection, if it exists, represents the solution to the system. However, graphical methods, while intuitive, may not always provide precise solutions, especially when dealing with non-integer values. Therefore, algebraic methods like substitution and elimination offer more accurate and efficient ways to solve these systems.
Solving systems of linear equations is not just an abstract mathematical exercise; it has practical applications in numerous real-world scenarios. For instance, in economics, these systems can model supply and demand curves to find equilibrium prices and quantities. In engineering, they can be used to analyze circuits and structural frameworks. Furthermore, in computer graphics and data analysis, solving systems of linear equations is essential for tasks like image processing and regression analysis. Therefore, mastering the techniques for solving these systems is a valuable skill across various disciplines.
Methods to Solve the System
There are primarily two algebraic methods used to solve systems of linear equations: substitution and elimination. Each method has its advantages, and the choice of method often depends on the specific structure of the equations. We will delve into both methods, demonstrating how they can be applied to the system 4x + 3y = 9 and 2x + 5y = 15.
1. Substitution Method
The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This process reduces the system to a single equation with one variable, which can then be easily solved. Once the value of one variable is found, it can be substituted back into either of the original equations to find the value of the other variable. This method is particularly effective when one of the equations is easily solved for one variable in terms of the other.
Let's apply the substitution method to our system. We can start by solving the second equation, 2x + 5y = 15, for x. Subtracting 5y from both sides gives us 2x = 15 - 5y. Dividing both sides by 2, we get x = (15 - 5y) / 2. Now we substitute this expression for x into the first equation, 4x + 3y = 9. This gives us 4((15 - 5y) / 2) + 3y = 9. Simplifying this equation will lead us to a single equation in terms of y, which we can solve.
After solving for y, we substitute the value we found back into the expression x = (15 - 5y) / 2 to find the value of x. This process effectively eliminates one variable at a time, allowing us to systematically solve for both x and y. The substitution method is a powerful technique that underscores the importance of algebraic manipulation and the ability to transform equations while preserving their solutions.
2. Elimination Method
The elimination method, also known as the addition method, involves manipulating the equations so that the coefficients of one of the variables are opposites. When the equations are added together, that variable is eliminated, leaving a single equation in one variable. This method is particularly useful when the coefficients of one variable are multiples of each other or can be easily made so. The elimination method relies on the principle that adding equal quantities to both sides of an equation does not change its solution.
To apply the elimination method to our system, 4x + 3y = 9 and 2x + 5y = 15, we can multiply the second equation by -2. This will give us a new equation, -4x - 10y = -30. Now, we can add this modified equation to the first equation, 4x + 3y = 9. Notice that the x terms will cancel out (4x and -4x), leaving us with an equation in terms of y only. This resulting equation can be easily solved for y.
Once we have the value of y, we can substitute it back into either of the original equations to solve for x. The elimination method provides a structured approach to solving systems of equations by systematically eliminating variables. It highlights the importance of carefully choosing multipliers to simplify the system and arrive at a solution efficiently. This method is especially advantageous when dealing with systems where the coefficients are such that direct substitution might lead to more complex fractions or calculations.
Solving by Substitution: A Detailed Walkthrough
Let's dive deeper into solving the system 4x + 3y = 9 and 2x + 5y = 15 using the substitution method. As previously outlined, the first step is to solve one equation for one variable. We'll choose the second equation, 2x + 5y = 15, and solve for x because it has smaller coefficients, which can simplify the calculations.
Subtracting 5y from both sides of 2x + 5y = 15 gives us 2x = 15 - 5y. Next, we divide both sides by 2 to isolate x: x = (15 - 5y) / 2. This expression represents x in terms of y. Now, we substitute this expression into the first equation, 4x + 3y = 9. Replacing x with (15 - 5y) / 2 gives us the equation 4((15 - 5y) / 2) + 3y = 9.
The next step is to simplify this equation. We can start by canceling the 4 in the numerator with the 2 in the denominator: 2(15 - 5y) + 3y = 9. Distributing the 2, we get 30 - 10y + 3y = 9. Combining like terms, we have 30 - 7y = 9. To isolate the y term, we subtract 30 from both sides: -7y = 9 - 30, which simplifies to -7y = -21. Finally, we divide both sides by -7 to solve for y: y = -21 / -7, so y = 3.
Now that we have the value of y, we can substitute it back into the expression for x that we found earlier: x = (15 - 5y) / 2. Plugging in y = 3, we get x = (15 - 5(3)) / 2. Simplifying, we have x = (15 - 15) / 2, which means x = 0 / 2, so x = 0. Therefore, the solution to the system of equations using the substitution method is x = 0 and y = 3. This means that the point (0, 3) is the intersection of the two lines represented by the equations 4x + 3y = 9 and 2x + 5y = 15 on the coordinate plane.
Solving by Elimination: A Detailed Walkthrough
Now, let's apply the elimination method to the same system of equations, 4x + 3y = 9 and 2x + 5y = 15. The key to the elimination method is to manipulate the equations so that the coefficients of one of the variables are opposites. In this case, we can eliminate x by multiplying the second equation by -2. This will make the coefficient of x in the second equation -4, which is the opposite of the coefficient of x in the first equation.
Multiplying the second equation, 2x + 5y = 15, by -2, we get -4x - 10y = -30. Now, we have the two equations:
- 4x + 3y = 9
- -4x - 10y = -30
We add these two equations together. The x terms cancel out (4x and -4x), leaving us with 3y - 10y = 9 - 30. Combining like terms, we get -7y = -21. Dividing both sides by -7, we find y = -21 / -7, so y = 3.
Now that we have the value of y, we can substitute it back into either of the original equations to solve for x. Let's use the first equation, 4x + 3y = 9. Substituting y = 3, we get 4x + 3(3) = 9. Simplifying, we have 4x + 9 = 9. Subtracting 9 from both sides gives us 4x = 0. Dividing both sides by 4, we find x = 0. Therefore, the solution to the system of equations using the elimination method is x = 0 and y = 3, which matches the solution we found using the substitution method.
The elimination method provides a systematic way to eliminate one variable and solve for the other. It's a powerful technique that demonstrates the flexibility and versatility of algebraic manipulation. By carefully choosing multipliers, we can simplify the system and efficiently arrive at the solution.
Verification of the Solution
To ensure the accuracy of our solution, it's crucial to verify that the values we found for x and y satisfy both original equations. We found that x = 0 and y = 3. Let's substitute these values into the equations 4x + 3y = 9 and 2x + 5y = 15.
First, consider the equation 4x + 3y = 9. Substituting x = 0 and y = 3, we get 4(0) + 3(3) = 9. Simplifying, we have 0 + 9 = 9, which is true. This confirms that our solution satisfies the first equation.
Next, let's check the second equation, 2x + 5y = 15. Substituting x = 0 and y = 3, we get 2(0) + 5(3) = 15. Simplifying, we have 0 + 15 = 15, which is also true. This confirms that our solution satisfies the second equation as well.
Since the values x = 0 and y = 3 satisfy both equations, we can confidently conclude that this is the correct solution to the system of equations. Verification is a critical step in the problem-solving process, especially in mathematics, as it helps to catch any errors made during the solution process. By verifying our solution, we ensure that we have a clear and accurate understanding of the problem and its solution.
Conclusion
In conclusion, we have successfully solved the system of linear equations 4x + 3y = 9 and 2x + 5y = 15 using both the substitution and elimination methods. Both methods yielded the same solution: x = 0 and y = 3. We also verified that this solution satisfies both original equations, ensuring its accuracy. This exercise highlights the power and versatility of algebraic techniques in solving systems of equations.
Understanding and mastering these methods is crucial for various applications in mathematics and other fields. Systems of linear equations are fundamental tools for modeling and solving real-world problems. Whether in engineering, economics, or computer science, the ability to solve these systems efficiently and accurately is invaluable. The substitution method and the elimination method provide two distinct approaches, each with its own advantages, allowing us to tackle a wide range of problems.
The system 4x + 3y = 9 and 2x + 5y = 15 served as a clear example to illustrate these methods. By breaking down the steps and providing detailed explanations, we have shown how to systematically approach and solve such systems. The verification step further reinforces the importance of checking our work and ensuring the validity of our solutions. As we continue to explore more complex mathematical concepts, the foundational skills developed in solving systems of linear equations will undoubtedly prove to be essential building blocks for future learning and problem-solving endeavors.
