Solving Linear Equations: Finding Points & Coordinates

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Hey guys! Let's dive into some cool math problems. We're going to explore a linear equation, 3xβˆ’5y=83x - 5y = 8, and figure out how to find specific points on the line it represents. This is super helpful for understanding how these equations work and how to graph them. We will find values for x and y and discover the coordinates to graph this equation.

Finding x When y = 2

Alright, first things first. We're asked to find the value of x when y equals 2. This means we'll substitute y with 2 in our equation, and then solve for x. It's like a puzzle, but with numbers and equations! Are you ready?

So, our equation is 3xβˆ’5y=83x - 5y = 8. We know y = 2, so let's plug that in:

3xβˆ’5(2)=83x - 5(2) = 8

Now, let's simplify that a bit. Multiply -5 by 2, and we get:

3xβˆ’10=83x - 10 = 8

Next, to isolate the x term, we'll add 10 to both sides of the equation. Remember, whatever we do on one side, we have to do on the other to keep things balanced!

3xβˆ’10+10=8+103x - 10 + 10 = 8 + 10

This simplifies to:

3x=183x = 18

Almost there! To find the value of x, we need to divide both sides by 3:

3x/3=18/33x / 3 = 18 / 3

Which gives us:

x=6x = 6

So, when y = 2, the value of x is 6. This means the point (6, 2) lies on the line. Pretty cool, huh? This is a fundamental concept in coordinate geometry, allowing us to pinpoint specific locations on a line. Grasping this helps us visualize the relationship between x and y in a linear equation, offering a clear understanding of its behavior across a graph. It's like knowing the secret code to unlock the position of a point! This process emphasizes the importance of substitution and the application of inverse operations to isolate variables. This foundation is essential for more complex mathematical concepts and problem-solving scenarios, especially in areas like calculus and physics.

This basic understanding is expanded upon in more advanced topics, like understanding how the slope of the line affects its direction and how to graph multiple linear equations to find intersections. In essence, mastering these skills is key to unlocking a deeper comprehension of how mathematical principles apply in the real world.

Remember, understanding that each point on a line satisfies the original equation is crucial. This understanding is the cornerstone of many graphical and analytical methods, enabling you to derive insights and solve problems in mathematics, science, and engineering. Thus, the ability to isolate variables and apply operations is indispensable for handling more complex equations.

Finding y When x = 1

Now, let's switch gears and find the value of y when x = 1. We'll follow a similar process to the one we used before, but this time we'll substitute x with 1 and solve for y. This exercise shows how changing one variable affects the other.

Our equation is still 3xβˆ’5y=83x - 5y = 8. This time, we know x = 1, so let's substitute that in:

3(1)βˆ’5y=83(1) - 5y = 8

Simplify:

3βˆ’5y=83 - 5y = 8

To isolate the y term, we'll first subtract 3 from both sides:

3βˆ’3βˆ’5y=8βˆ’33 - 3 - 5y = 8 - 3

This simplifies to:

βˆ’5y=5-5y = 5

Finally, to find y, we'll divide both sides by -5:

βˆ’5y/βˆ’5=5/βˆ’5-5y / -5 = 5 / -5

Which gives us:

y=βˆ’1y = -1

So, when x = 1, the value of y is -1. Therefore, the point (1, -1) lies on the line. This shows that for every value of x you input into a linear equation, you get a unique y value, creating an ordered pair that represents a specific point on the line. The power of substitution and simplification is highlighted here.

It is also very important to note that the process of finding these points isn't just an isolated mathematical exercise; rather, it’s a precursor to understanding more advanced concepts. This approach is similar to more sophisticated methods used in scientific fields and technology, like finding the intersections of multiple lines, which is crucial in areas like computer graphics, data analysis, and robotics. So, understanding how these individual points interact and relate to each other is fundamental for comprehending larger mathematical concepts and applications.

Showing Two Coordinate Points

Okay, let's recap! We've already found two coordinate points that lie on the line 3xβˆ’5y=83x - 5y = 8: (6, 2) and (1, -1). Finding two points is the bare minimum to draw a line on a graph; if you have two points, you can connect them and extend the line indefinitely. Let's recap how we found them: For (6, 2) we set y to 2 and solved for x. For (1, -1) we set x to 1 and solved for y. Using the correct x and y values to solve for the other is key to solving the problem. Let's make sure we've got the correct values!

Point 1: (6, 2)

Let’s plug these values back into our original equation to make sure it works:

3(6)βˆ’5(2)=18βˆ’10=83(6) - 5(2) = 18 - 10 = 8. Great, it works!

Point 2: (1, -1)

Let’s check this one too:

3(1)βˆ’5(βˆ’1)=3+5=83(1) - 5(-1) = 3 + 5 = 8. Excellent! This confirms that both points satisfy the equation.

These two points are all we need to visualize and draw our line accurately on a graph. To graph it, we'd simply plot these points on a coordinate plane and draw a straight line through them, extending infinitely in both directions. Understanding how to find multiple points allows you to accurately represent any linear equation on a graph. The beauty of these equations is that they are infinitely long, allowing us to find multiple points.

This provides a basic understanding of how to plot a linear equation and is fundamental for understanding how different equations and lines interact. This is one of the foundations for comprehending more complex mathematical scenarios like systems of equations and can also be applied to real-world applications such as analyzing trends in data, designing systems, and making predictions based on observed relationships. Moreover, this knowledge forms the base of understanding the concept of linear algebra and vector space.

We could find many other points that are part of the equation, as we can find points that do not follow this equation. But these points are all we need to get started and understand this math problem!