Identifying Points On A Linear Equation Graph 8x + 2y = 24

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In mathematics, understanding how to identify points that lie on the graph of a linear equation is a fundamental skill. This article delves into the process of determining whether a given point satisfies a linear equation, providing a step-by-step approach and illustrative examples. We will focus on the equation 8x+2y=248x + 2y = 24 and explore how to verify if specific points, such as (βˆ’1,8)(-1, 8), (2,8)(2, 8), (6,βˆ’12)(6, -12), and (8,2)(8, 2), lie on its graph. By the end of this guide, you will have a solid understanding of how to solve such problems efficiently.

Understanding Linear Equations and Their Graphs

Before we dive into the specifics, let's establish a clear understanding of what linear equations represent and how their graphs are formed. 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. These equations are called "linear" because they describe a straight line when plotted on a coordinate plane. The general form of a linear equation in two variables, xx and yy, is Ax+By=CAx + By = C, where AA, BB, and CC are constants.

When we graph a linear equation, we are essentially plotting all the points (x,y)(x, y) that satisfy the equation. Each point on the line represents a solution to the equation, and conversely, any solution to the equation corresponds to a point on the line. The graph of a linear equation is a visual representation of the relationship between the variables xx and yy. Understanding this relationship is crucial for solving problems that involve identifying points on the graph.

To determine whether a point (x,y)(x, y) lies on the graph of a linear equation, we substitute the xx and yy coordinates into the equation. If the equation holds true after the substitution, then the point lies on the graph; otherwise, it does not. This simple yet powerful method forms the basis for solving a wide range of problems in algebra and coordinate geometry. In the following sections, we will apply this method to the equation 8x+2y=248x + 2y = 24 and test the given points.

Step-by-Step Method to Check if a Point Lies on the Graph

The core concept behind determining whether a point lies on the graph of a linear equation is substitution. We substitute the xx and yy coordinates of the point into the equation and check if the equation holds true. If it does, the point lies on the graph; if not, it does not. Here’s a detailed step-by-step method:

  1. Write down the linear equation: Begin by clearly stating the linear equation you are working with. In our case, the equation is 8x+2y=248x + 2y = 24.
  2. Identify the point: Note the coordinates of the point you want to test. A point is represented as (x,y)(x, y), where xx is the x-coordinate and yy is the y-coordinate.
  3. Substitute the coordinates: Replace xx and yy in the linear equation with the corresponding coordinates of the point. Be careful to substitute the correct values; the xx-coordinate replaces xx, and the yy-coordinate replaces yy.
  4. Simplify the equation: Perform the arithmetic operations (multiplication, addition, etc.) to simplify both sides of the equation.
  5. Check for equality: After simplification, check if the left-hand side (LHS) of the equation is equal to the right-hand side (RHS). If LHS = RHS, the equation holds true, and the point lies on the graph. If LHS β‰  RHS, the equation does not hold true, and the point does not lie on the graph.

This method is straightforward and can be applied to any linear equation and any given point. It provides a systematic way to verify whether a point is a solution to the equation and thus lies on its graph. In the subsequent sections, we will apply this method to the specific points provided in the problem and determine which of them lie on the graph of 8x+2y=248x + 2y = 24.

Testing the Point (-1, 8)

Let’s begin by testing the point (βˆ’1,8)(-1, 8) to see if it lies on the graph of the equation 8x+2y=248x + 2y = 24. We will follow the step-by-step method outlined earlier.

  1. Write down the linear equation: The linear equation is 8x+2y=248x + 2y = 24.
  2. Identify the point: The point we are testing is (βˆ’1,8)(-1, 8), where x=βˆ’1x = -1 and y=8y = 8.
  3. Substitute the coordinates: Substitute x=βˆ’1x = -1 and y=8y = 8 into the equation: 8(βˆ’1)+2(8)=248(-1) + 2(8) = 24
  4. Simplify the equation: Perform the multiplication: βˆ’8+16=24-8 + 16 = 24 Now, add the numbers on the left-hand side: 8=248 = 24
  5. Check for equality: The left-hand side (LHS) is 88, and the right-hand side (RHS) is 2424. Since 8β‰ 248 β‰  24, the equation does not hold true.

Therefore, the point (βˆ’1,8)(-1, 8) does not lie on the graph of the equation 8x+2y=248x + 2y = 24. The substitution and simplification clearly show that the coordinates of this point do not satisfy the given equation. This process highlights the importance of accurate substitution and arithmetic operations in determining whether a point lies on a graph. In the next section, we will test another point using the same method.

Testing the Point (2, 8)

Next, we will test the point (2,8)(2, 8) to determine if it lies on the graph of the equation 8x+2y=248x + 2y = 24. We will again use the step-by-step method to ensure accuracy.

  1. Write down the linear equation: The equation remains 8x+2y=248x + 2y = 24.
  2. Identify the point: The point we are testing is (2,8)(2, 8), where x=2x = 2 and y=8y = 8.
  3. Substitute the coordinates: Substitute x=2x = 2 and y=8y = 8 into the equation: 8(2)+2(8)=248(2) + 2(8) = 24
  4. Simplify the equation: Perform the multiplication: 16+16=2416 + 16 = 24 Now, add the numbers on the left-hand side: 32=2432 = 24
  5. Check for equality: The left-hand side (LHS) is 3232, and the right-hand side (RHS) is 2424. Since 32β‰ 2432 β‰  24, the equation does not hold true.

Thus, the point (2,8)(2, 8) does not lie on the graph of the equation 8x+2y=248x + 2y = 24. The result of the substitution and simplification shows that this point is not a solution to the equation. This reinforces the method's reliability in identifying points that do not satisfy a given linear equation. In the following section, we will test another point, continuing our systematic approach.

Testing the Point (6, -12)

Now, let's test the point (6,βˆ’12)(6, -12) to see if it lies on the graph of the equation 8x+2y=248x + 2y = 24. We will follow our established method to verify this point.

  1. Write down the linear equation: The equation is 8x+2y=248x + 2y = 24.
  2. Identify the point: The point we are testing is (6,βˆ’12)(6, -12), where x=6x = 6 and y=βˆ’12y = -12.
  3. Substitute the coordinates: Substitute x=6x = 6 and y=βˆ’12y = -12 into the equation: 8(6)+2(βˆ’12)=248(6) + 2(-12) = 24
  4. Simplify the equation: Perform the multiplication: 48+(βˆ’24)=2448 + (-24) = 24 Now, add the numbers on the left-hand side: 24=2424 = 24
  5. Check for equality: The left-hand side (LHS) is 2424, and the right-hand side (RHS) is 2424. Since 24=2424 = 24, the equation holds true.

Therefore, the point (6,βˆ’12)(6, -12) lies on the graph of the equation 8x+2y=248x + 2y = 24. This result confirms that this point is a solution to the equation. It highlights the importance of careful arithmetic and substitution in accurately determining whether a point satisfies a linear equation. In the next section, we will test the final point in our list.

Testing the Point (8, 2)

Finally, we will test the point (8,2)(8, 2) to determine whether it lies on the graph of the equation 8x+2y=248x + 2y = 24. As before, we will use our step-by-step method to ensure accuracy.

  1. Write down the linear equation: The equation is 8x+2y=248x + 2y = 24.
  2. Identify the point: The point we are testing is (8,2)(8, 2), where x=8x = 8 and y=2y = 2.
  3. Substitute the coordinates: Substitute x=8x = 8 and y=2y = 2 into the equation: 8(8)+2(2)=248(8) + 2(2) = 24
  4. Simplify the equation: Perform the multiplication: 64+4=2464 + 4 = 24 Now, add the numbers on the left-hand side: 68=2468 = 24
  5. Check for equality: The left-hand side (LHS) is 6868, and the right-hand side (RHS) is 2424. Since 68β‰ 2468 β‰  24, the equation does not hold true.

Therefore, the point (8,2)(8, 2) does not lie on the graph of the equation 8x+2y=248x + 2y = 24. The substitution and simplification show that this point is not a solution to the equation. This completes our testing of the provided points. In the concluding section, we will summarize our findings and reinforce the method we have used.

Conclusion and Summary of Findings

In this article, we addressed the problem of determining which of the points (βˆ’1,8)(-1, 8), (2,8)(2, 8), (6,βˆ’12)(6, -12), and (8,2)(8, 2) lies on the graph of the linear equation 8x+2y=248x + 2y = 24. We employed a systematic step-by-step method, which involved substituting the coordinates of each point into the equation and checking for equality. Our findings can be summarized as follows:

  • The point (-1, 8) does not lie on the graph because the substitution resulted in 8β‰ 248 β‰  24.
  • The point (2, 8) does not lie on the graph because the substitution resulted in 32β‰ 2432 β‰  24.
  • The point (6, -12) lies on the graph because the substitution resulted in 24=2424 = 24.
  • The point (8, 2) does not lie on the graph because the substitution resulted in 68β‰ 2468 β‰  24.

Therefore, the only point among the given options that lies on the graph of the equation 8x+2y=248x + 2y = 24 is (6,βˆ’12)(6, -12). This exercise demonstrates the importance of accurate substitution and simplification in verifying whether a point is a solution to a linear equation. The method we have used is applicable to any linear equation and any given point, providing a reliable way to solve such problems.

Understanding how to identify points on the graph of a linear equation is a fundamental concept in algebra and coordinate geometry. By mastering this skill, you can solve a wide range of problems and gain a deeper understanding of the relationship between equations and their graphical representations. This article has provided a comprehensive guide to this process, equipping you with the knowledge and tools necessary to tackle similar problems with confidence.