Solving 4x + 3y = 9 And 2x + 5y = 15 A Detailed Guide
Solving systems of linear equations is a fundamental concept in mathematics with wide-ranging applications in various fields, including engineering, economics, and computer science. A system of linear equations consists of two or more linear equations involving the same variables. The solution to such a system is the set of values for the variables that satisfy all equations simultaneously. In this article, we will delve into a comprehensive discussion on solving the given system of linear equations:
4x + 3y = 9
2x + 5y = 15
We will explore several methods for finding the solution, including substitution, elimination, and matrix methods, providing a step-by-step explanation of each approach. By understanding these techniques, you will gain a solid foundation for tackling more complex mathematical problems and real-world applications.
1. Introduction to Systems of Linear Equations
Before we dive into the solution methods, let's first understand what systems of linear equations are and why they are important. A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. The variables are raised to the power of one, and there are no products or other functions of the variables. A system of linear equations is a collection of two or more linear equations with the same set of variables.
The general form of a linear equation in two variables, x and y, is:
Ax + By = C
where A, B, and C are constants, and A and B are not both zero. A solution to a linear equation in two variables is an ordered pair (x, y) that satisfies the equation. Geometrically, a linear equation in two variables represents a straight line in the Cartesian plane. The solution to a system of two linear equations in two variables is the point (x, y) where the two lines intersect.
Systems of linear equations arise in numerous applications. For example, in economics, they can be used to model supply and demand curves. In engineering, they can be used to analyze electrical circuits or structural systems. In computer graphics, they are used to perform transformations and projections.
2. Methods for Solving Systems of Linear Equations
There are several methods for solving systems of linear equations, each with its advantages and disadvantages. We will explore three common methods:
- Substitution
- Elimination
- Matrix Methods
2.1 Solving by Substitution
The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This eliminates one variable, resulting in a single equation with one variable that can 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.
Let's apply the substitution method to our system of equations:
4x + 3y = 9 (Equation 1)
2x + 5y = 15 (Equation 2)
Step 1: Solve one equation for one variable.
We can choose either equation and solve for either variable. Let's solve Equation 2 for x:
2x = 15 - 5y
x = (15 - 5y) / 2
Step 2: Substitute the expression into the other equation.
Now, substitute this expression for x into Equation 1:
4((15 - 5y) / 2) + 3y = 9
Step 3: Solve for the remaining variable.
Simplify and solve for y:
2(15 - 5y) + 3y = 9
30 - 10y + 3y = 9
-7y = -21
y = 3
Step 4: Substitute back to find the other variable.
Substitute y = 3 back into the expression for x:
x = (15 - 5(3)) / 2
x = (15 - 15) / 2
x = 0
Therefore, the solution to the system of equations using the substitution method is x = 0 and y = 3.
2.2 Solving by Elimination
The elimination method involves manipulating the equations so that the coefficients of one variable are opposites. When the equations are added together, that variable is eliminated, resulting in a single equation with one variable. This equation can be solved for the remaining variable, and then that value can be substituted back into either of the original equations to find the value of the eliminated variable.
Let's apply the elimination method to our system of equations:
4x + 3y = 9 (Equation 1)
2x + 5y = 15 (Equation 2)
Step 1: Multiply the equations to make the coefficients of one variable opposites.
To eliminate x, we can multiply Equation 2 by -2:
-2(2x + 5y) = -2(15)
-4x - 10y = -30
Now we have:
4x + 3y = 9 (Equation 1)
-4x - 10y = -30 (Modified Equation 2)
Step 2: Add the equations to eliminate one variable.
Add Equation 1 and the modified Equation 2:
(4x + 3y) + (-4x - 10y) = 9 + (-30)
-7y = -21
Step 3: Solve for the remaining variable.
Solve for y:
y = 3
Step 4: Substitute back to find the other variable.
Substitute y = 3 back into either Equation 1 or Equation 2. Let's use Equation 1:
4x + 3(3) = 9
4x + 9 = 9
4x = 0
x = 0
Therefore, the solution to the system of equations using the elimination method is x = 0 and y = 3.
2.3 Solving by Matrix Methods
Matrix methods provide a more systematic approach for solving systems of linear equations, especially when dealing with larger systems. These methods involve representing the system of equations in matrix form and then using techniques from linear algebra to solve for the variables.
The system of equations:
4x + 3y = 9
2x + 5y = 15
can be represented in matrix form as:
| 4 3 | | x | = | 9 |
| 2 5 | | y | | 15 |
This can be written more compactly as:
AX = B
where:
- A is the coefficient matrix:
| 4 3 |\ | 2 5 | - X is the variable matrix:
| x |\ | y | - B is the constant matrix:
| 9 |\ | 15 |
To solve for X, we can use the following formula:
X = A^(-1)B
where A^(-1) is the inverse of matrix A.
Step 1: Find the inverse of the coefficient matrix A.
For a 2x2 matrix | a b |\ | c d |, the inverse is given by:
A^(-1) = (1 / (ad - bc)) | d -b |\ | -c a |
For our matrix A = | 4 3 |\ | 2 5 |:
Determinant of A (ad - bc) = (4 * 5) - (3 * 2) = 20 - 6 = 14
A^(-1) = (1 / 14) | 5 -3 |\ | -2 4 |
Step 2: Multiply the inverse matrix A^(-1) by the constant matrix B.
X = A^(-1)B
X = (1 / 14) | 5 -3 | | 9 |
| -2 4 | | 15 |
X = (1 / 14) | (5 * 9) + (-3 * 15) |\ | (-2 * 9) + (4 * 15) |
X = (1 / 14) | 45 - 45 |\ | -18 + 60 |
X = (1 / 14) | 0 |\ | 42 |
X = | 0 |\ | 3 |
Therefore, the solution to the system of equations using matrix methods is x = 0 and y = 3.
3. Verification of the Solution
To ensure the correctness of our solution, we can substitute the values x = 0 and y = 3 back into the original equations:
Equation 1:
4(0) + 3(3) = 9
0 + 9 = 9
9 = 9 (True)
Equation 2:
2(0) + 5(3) = 15
0 + 15 = 15
15 = 15 (True)
Since the solution satisfies both equations, we can confidently conclude that x = 0 and y = 3 is the correct solution to the system of equations.
4. Graphical Interpretation
As mentioned earlier, each linear equation in two variables represents a straight line in the Cartesian plane. The solution to a system of two linear equations is the point where the two lines intersect. We can graph the two equations to visualize the solution.
Equation 1: 4x + 3y = 9 can be rewritten as y = (-4/3)x + 3, which has a slope of -4/3 and a y-intercept of 3.
Equation 2: 2x + 5y = 15 can be rewritten as y = (-2/5)x + 3, which has a slope of -2/5 and a y-intercept of 3.
When we graph these two lines, we see that they intersect at the point (0, 3), which confirms our algebraic solution.
5. Conclusion
In this article, we have thoroughly discussed the process of solving the system of linear equations:
4x + 3y = 9
2x + 5y = 15
We explored three different methods:
- Substitution
- Elimination
- Matrix Methods
Each method led us to the same solution: x = 0 and y = 3. We also verified the solution by substituting the values back into the original equations and provided a graphical interpretation to further solidify our understanding.
Mastering the techniques for solving systems of linear equations is crucial for success in various mathematical disciplines and real-world applications. By understanding the strengths and weaknesses of each method, you can choose the most efficient approach for a given problem. Whether you prefer the algebraic manipulation of substitution and elimination or the systematic approach of matrix methods, the ability to solve systems of linear equations is an invaluable skill.
This comprehensive exploration should equip you with the necessary knowledge and skills to confidently tackle similar problems and appreciate the broader applications of linear algebra in mathematics and beyond. Remember to practice regularly and explore different types of systems to further enhance your proficiency in this important area of mathematics.
