Determining The Number Of Solutions In A Linear System
Understanding Linear Systems and Their Solutions
When dealing with linear systems, a crucial aspect is determining the number of solutions they possess. A linear system is a set of two or more linear equations containing the same variables. The solutions to such a system are the values that, when substituted for the variables, satisfy all equations simultaneously. There are three possible outcomes when solving a linear system: one unique solution, no solution, or an infinite number of solutions. In this article, we will delve into the given system of equations to ascertain the nature of its solutions. Determining the number of solutions involves analyzing the equations to see if they intersect at a single point, are parallel and never intersect, or are the same line, leading to infinite solutions. The given linear system is:
y = (2/3)x + 2
6x - 4y = -10
To determine the number of solutions, we need to analyze these equations. We can start by manipulating the second equation to see if it resembles the first equation or contradicts it. The first equation is already in slope-intercept form, which is y = mx + b, where m represents the slope and b represents the y-intercept. This form makes it easy to visualize and compare the lines. In our case, the first equation has a slope of 2/3 and a y-intercept of 2. Now, let's work on the second equation to bring it into a more comparable form. The goal is to isolate y on one side of the equation. We'll subtract 6x from both sides, and then divide by -4 to solve for y. This process will help us clearly see the slope and y-intercept of the second line and compare it to the first. By bringing both equations into the same format, we can quickly determine if the lines are the same, parallel, or intersecting, which will tell us how many solutions the system has. This analytical approach is a standard technique for solving linear systems and understanding their graphical representations.
Analyzing the Given System of Equations
To identify the number of solutions for the given linear system, we need to manipulate the equations to a comparable form, typically the slope-intercept form (y = mx + b). The first equation, y = (2/3)x + 2, is already in slope-intercept form, making it easy to discern its slope (2/3) and y-intercept (2). This form allows us to visualize the line's orientation and position on the coordinate plane. Now, let's transform the second equation, 6x - 4y = -10, into slope-intercept form as well. This involves isolating y on one side of the equation. First, subtract 6x from both sides of the equation to get -4y = -6x - 10. Next, divide both sides by -4 to solve for y: y = (3/2)x + 5/2. Now we have both equations in the same form:
y = (2/3)x + 2
y = (3/2)x + 5/2
Comparing the two equations in slope-intercept form reveals crucial information about their relationship. The slopes are different (2/3 and 3/2), indicating that the lines are not parallel and will intersect at some point. The y-intercepts are also different (2 and 5/2), further confirming that the lines are distinct. Since the slopes are different, the lines will intersect at exactly one point, which means the linear system has a single, unique solution. This intersection point represents the values of x and y that satisfy both equations simultaneously. To find this solution, one would typically use methods such as substitution or elimination. However, for the purpose of this question, we've already determined the nature of the solution—that it is a single, unique solution—by analyzing the slopes and y-intercepts of the lines. This analysis is a fundamental technique in linear algebra for understanding the behavior of linear systems.
Determining the Number of Solutions
After transforming the equations into slope-intercept form, we have:
y = (2/3)x + 2
y = (3/2)x + 5/2
As observed, the slopes of the two lines, 2/3 and 3/2, are different. This single fact definitively tells us that the lines are not parallel and are not the same line. Parallel lines have the same slope, and identical lines have both the same slope and the same y-intercept. Since the slopes here are distinct, the two lines will intersect at exactly one point. This intersection point represents the unique solution to the linear system. The y-intercepts, which are 2 and 5/2, respectively, also differ, reinforcing that the lines are not the same. If the lines had the same slope but different y-intercepts, they would be parallel and there would be no solution. If they had the same slope and the same y-intercept, they would be the same line, and there would be an infinite number of solutions, as every point on the line would satisfy both equations. However, in our case, the differing slopes guarantee a single point of intersection and, therefore, one unique solution.
The implications of having one solution are significant in various applications of linear systems, such as in engineering, economics, and computer science. It means there is a specific set of values for the variables that satisfies all conditions represented by the equations. This unique solution can be found through several methods, including graphing, substitution, and elimination. The graphical method involves plotting the lines and identifying the point where they intersect. The substitution method involves solving one equation for one variable and substituting that expression into the other equation. The elimination method involves adding or subtracting multiples of the equations to eliminate one variable. All these methods will lead to the same unique solution, confirming our initial determination based on the slopes and y-intercepts. Therefore, understanding how the slopes and y-intercepts relate to the number of solutions is a fundamental concept in solving linear systems.
Verifying the Solution Options
Based on our analysis, we know that the linear system has one solution. This eliminates option C, which states "no solution," and option D, which states "infinite number of solutions." We are left with two options that claim a single solution, but with different coordinates:
- A. one solution: (-0.6,-1.6)
- B. one solution: (-0.6,1.6)
To determine which of these options is correct, we need to verify which point satisfies both equations. We can do this by substituting the x and y values from each option into both equations and checking if the equations hold true. Let's start with option A, the point (-0.6, -1.6). Substitute these values into the first equation:
y = (2/3)x + 2
-1.6 = (2/3)(-0.6) + 2
-1.6 = -0.4 + 2
-1.6 = 1.6
This equation is not true, so option A is incorrect. Now, let's try option B, the point (-0.6, 1.6). Substitute these values into the first equation:
y = (2/3)x + 2
1. 6 = (2/3)(-0.6) + 2
2. 6 = -0.4 + 2
3. 6 = 1.6
This equation holds true. Now, we need to verify this point in the second equation to ensure it's a solution for the entire linear system:
6x - 4y = -10
6(-0.6) - 4(1.6) = -10
-3.6 - 6.4 = -10
-10 = -10
This equation also holds true. Therefore, the point (-0.6, 1.6) satisfies both equations, making option B the correct solution. This verification process is crucial to ensure accuracy, especially when multiple solution options are provided. By substituting the potential solutions back into the original equations, we can confirm whether they are indeed valid solutions for the linear system. This method is a fundamental technique in algebra for checking the correctness of solutions.
Conclusion
In conclusion, by analyzing the slopes and y-intercepts of the given linear system, we determined that there is one unique solution. We then verified the provided options by substituting the coordinates into the original equations and found that the point (-0.6, 1.6) satisfies both equations. Therefore, the correct answer is B. one solution: (-0.6, 1.6). This exercise demonstrates the importance of understanding the relationship between the graphical representation of linear equations and the algebraic solutions to linear systems. It also highlights the practical methods for verifying potential solutions to ensure accuracy. The ability to analyze and solve linear systems is a fundamental skill in mathematics with wide-ranging applications in various fields.
