Finding The Solution Of System Of Equations Y=-7x+5 And Y=2x-6
In mathematics, systems of equations represent a fundamental concept with wide-ranging applications across various fields, from engineering and physics to economics and computer science. A system of equations comprises two or more equations that share common variables. The solution to a system of equations is the set of values for the variables that satisfy all equations simultaneously. This means that when these values are substituted into each equation, the equation holds true. Finding these solutions is a crucial skill in algebra and beyond, enabling us to model and solve real-world problems involving multiple interconnected variables.
This article delves into the process of determining whether an ordered pair is a solution to a given system of equations. We will specifically focus on the system:
y = -7x + 5
y = 2x - 6
and explore how to verify if different ordered pairs, such as A (20/9, -95/9), B (2/9, 31/9), and others, satisfy both equations. Understanding this process is essential for solving more complex systems of equations and for grasping the underlying principles of algebraic solutions. By the end of this guide, you will have a clear understanding of how to check potential solutions and a solid foundation for tackling more advanced algebraic problems.
Verifying Solutions to Systems of Equations
The core concept in determining whether an ordered pair is a solution to a system of equations lies in the simultaneous satisfaction of all equations within the system. An ordered pair, typically represented as (x, y), provides specific values for the variables x and y. To verify if this pair is a solution, we substitute these values into each equation in the system. If the ordered pair makes all equations true, then it is indeed a solution to the system. Conversely, if the ordered pair fails to satisfy even one equation, it is not a solution.
The process is straightforward yet crucial. First, we identify the x and y values from the ordered pair. Then, we substitute these values into each equation in the system. Next, we simplify each equation to check if the left-hand side (LHS) equals the right-hand side (RHS). If the LHS equals the RHS for all equations, the ordered pair is a solution. If not, the ordered pair is not a solution. This method provides a concrete way to confirm solutions and avoid errors in more complex problem-solving scenarios. Understanding this verification process is fundamental for solving systems of equations and is a cornerstone of algebraic proficiency.
Step-by-Step Verification Process
To effectively verify if an ordered pair is a solution to a system of equations, a structured approach is necessary. This step-by-step process ensures accuracy and clarity in the solution. Let's break down the method into manageable steps:
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Identify the Ordered Pair and Equations: Begin by clearly identifying the ordered pair (x, y) and the system of equations you are working with. This initial step sets the stage for the verification process.
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Substitute the x and y Values: Substitute the x-value and the y-value from the ordered pair into each equation in the system. This involves replacing the variables x and y with their corresponding numerical values.
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Simplify Each Equation: After substituting the values, simplify each equation independently. This involves performing any necessary arithmetic operations to reduce the equation to its simplest form.
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Check for Equality: For each equation, check if the left-hand side (LHS) is equal to the right-hand side (RHS). If the LHS equals the RHS, the equation is satisfied. If not, the equation is not satisfied.
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Determine if the Ordered Pair is a Solution: If the ordered pair satisfies all equations in the system, then it is a solution to the system. If the ordered pair fails to satisfy even one equation, it is not a solution.
By following these steps meticulously, you can confidently verify potential solutions to systems of equations. This process is essential for accuracy and is a fundamental skill in algebra and beyond.
Analyzing the Given System of Equations
The system of equations we are working with is:
y = -7x + 5
y = 2x - 6
This system consists of two linear equations, each representing a straight line on a coordinate plane. The solution to this system is the point (x, y) where these two lines intersect. Algebraically, this point is the ordered pair that satisfies both equations simultaneously. The first equation, y = -7x + 5, has a slope of -7 and a y-intercept of 5. This means that the line slopes downward steeply from left to right, crossing the y-axis at the point (0, 5). The second equation, y = 2x - 6, has a slope of 2 and a y-intercept of -6. This line slopes upward more gently and crosses the y-axis at (0, -6).
Understanding the characteristics of these equations helps in visualizing the solution. Graphically, the solution is the intersection point of these two lines. Analytically, we find the solution by solving the system of equations, which involves finding the values of x and y that satisfy both equations. We can do this through substitution, elimination, or other algebraic methods. The ordered pairs we will analyze are potential solutions to this system, and we will use the step-by-step verification process described earlier to determine if they are indeed solutions. This analysis sets the stage for evaluating the given ordered pairs and finding the correct solution to the system.
Evaluating Ordered Pair A (20/9, -95/9)
To determine if the ordered pair A (20/9, -95/9) is a solution to the system of equations,
y = -7x + 5
y = 2x - 6
we will substitute x = 20/9 and y = -95/9 into each equation and check if both equations hold true.
Substituting into the First Equation
For the first equation, y = -7x + 5, we substitute x = 20/9 and y = -95/9:
-95/9 = -7(20/9) + 5
Now, we simplify the right-hand side (RHS):
-7(20/9) = -140/9
So the equation becomes:
-95/9 = -140/9 + 5
To add 5 to -140/9, we convert 5 to a fraction with a denominator of 9:
5 = 45/9
Now we have:
-95/9 = -140/9 + 45/9
Combining the fractions on the RHS:
-140/9 + 45/9 = -95/9
Thus, the equation is:
-95/9 = -95/9
The first equation holds true for the ordered pair A (20/9, -95/9).
Substituting into the Second Equation
Next, we substitute x = 20/9 and y = -95/9 into the second equation, y = 2x - 6:
-95/9 = 2(20/9) - 6
Simplifying the RHS:
2(20/9) = 40/9
The equation becomes:
-95/9 = 40/9 - 6
To subtract 6 from 40/9, we convert 6 to a fraction with a denominator of 9:
6 = 54/9
Now we have:
-95/9 = 40/9 - 54/9
Combining the fractions on the RHS:
40/9 - 54/9 = -14/9
Thus, the equation is:
-95/9 = -14/9
This equation is not true. Therefore, the ordered pair A (20/9, -95/9) does not satisfy the second equation.
Conclusion for Ordered Pair A
Since the ordered pair A (20/9, -95/9) satisfies the first equation but not the second equation, it is not a solution to the system of equations. For an ordered pair to be a solution, it must satisfy all equations in the system, and in this case, it fails to do so.
Evaluating Ordered Pair B (2/9, 31/9)
Next, we will evaluate ordered pair B (2/9, 31/9) to determine if it is a solution to the system of equations:
y = -7x + 5
y = 2x - 6
We will substitute x = 2/9 and y = 31/9 into each equation and check if both equations hold true.
Substituting into the First Equation
For the first equation, y = -7x + 5, we substitute x = 2/9 and y = 31/9:
31/9 = -7(2/9) + 5
Now, we simplify the right-hand side (RHS):
-7(2/9) = -14/9
So the equation becomes:
31/9 = -14/9 + 5
To add 5 to -14/9, we convert 5 to a fraction with a denominator of 9:
5 = 45/9
Now we have:
31/9 = -14/9 + 45/9
Combining the fractions on the RHS:
-14/9 + 45/9 = 31/9
Thus, the equation is:
31/9 = 31/9
The first equation holds true for the ordered pair B (2/9, 31/9).
Substituting into the Second Equation
Next, we substitute x = 2/9 and y = 31/9 into the second equation, y = 2x - 6:
31/9 = 2(2/9) - 6
Simplifying the RHS:
2(2/9) = 4/9
The equation becomes:
31/9 = 4/9 - 6
To subtract 6 from 4/9, we convert 6 to a fraction with a denominator of 9:
6 = 54/9
Now we have:
31/9 = 4/9 - 54/9
Combining the fractions on the RHS:
4/9 - 54/9 = -50/9
Thus, the equation is:
31/9 = -50/9
This equation is not true. Therefore, the ordered pair B (2/9, 31/9) does not satisfy the second equation.
Conclusion for Ordered Pair B
Since the ordered pair B (2/9, 31/9) satisfies the first equation but not the second equation, it is not a solution to the system of equations. For an ordered pair to be a solution, it must satisfy all equations in the system, and in this case, it fails to do so.
Solving the System of Equations Algebraically
To find the actual solution to the system of equations,
y = -7x + 5
y = 2x - 6
we can use the substitution method. Since both equations are already solved for y, we can set them equal to each other:
-7x + 5 = 2x - 6
Now, we solve for x. First, add 7x to both sides:
5 = 9x - 6
Next, add 6 to both sides:
11 = 9x
Finally, divide by 9:
x = 11/9
Now that we have the value of x, we can substitute it into either equation to find the value of y. Let's use the second equation, y = 2x - 6:
y = 2(11/9) - 6
Simplify:
y = 22/9 - 6
Convert 6 to a fraction with a denominator of 9:
6 = 54/9
Now we have:
y = 22/9 - 54/9
Combine the fractions:
y = -32/9
Thus, the solution to the system of equations is the ordered pair (11/9, -32/9). This algebraic solution confirms the actual point of intersection between the two lines represented by the equations.
In this comprehensive guide, we explored the process of verifying whether an ordered pair is a solution to a system of equations. We emphasized the importance of substituting the x and y values from the ordered pair into each equation in the system and checking if all equations hold true. Through detailed examples, we demonstrated this step-by-step verification process, highlighting its critical role in confirming solutions.
We specifically analyzed the system of equations:
y = -7x + 5
y = 2x - 6
and evaluated ordered pairs A (20/9, -95/9) and B (2/9, 31/9). We found that neither of these pairs satisfied both equations simultaneously, and therefore, neither was a solution to the system. Additionally, we solved the system algebraically using the substitution method, determining the actual solution to be (11/9, -32/9).
This exploration underscores the significance of a systematic approach to verifying solutions and the value of algebraic methods in finding accurate answers. Understanding these concepts is crucial for success in algebra and various related fields. By mastering the techniques discussed in this article, you will be well-equipped to tackle more complex problems involving systems of equations and enhance your mathematical proficiency.