Solving System Of Equations: Approximate Solutions

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Hey guys! Let's dive into solving systems of equations and finding approximate solutions. It might sound intimidating, but trust me, we'll break it down so it's super easy to understand. We're going to tackle the system:

y=βˆ’74x+52y=-\frac{7}{4}x+\frac{5}{2}

y=34xβˆ’3y=\frac{3}{4}x-3

and figure out which points could be the approximate solutions. So grab your thinking caps, and let's get started!

Understanding Systems of Equations

First off, what even is a system of equations? Simply put, it’s a set of two or more equations that we’re trying to solve simultaneously. In our case, we have two linear equations. The solution to a system of equations is the point (or points) where the lines intersect on a graph. This point satisfies both equations, meaning if you plug the x and y values into either equation, it will hold true. When we talk about approximate solutions, we're often dealing with scenarios where the intersection point isn't a neat, whole number. Think decimals and fractions – that's where approximations come into play. Finding these approximations often involves a mix of algebraic manipulation and a bit of educated guessing, especially when we're given multiple-choice options.

Methods for Finding Solutions

There are several ways to solve a system of equations. The most common methods are graphing, substitution, and elimination. Graphing involves plotting both lines on a coordinate plane and visually identifying the point of intersection. This method is great for visualizing the solution, but it can be less precise when dealing with non-integer solutions. Substitution involves solving one equation for one variable and then substituting that expression into the other equation. This eliminates one variable and allows you to solve for the remaining one. Elimination, also known as the addition method, involves manipulating the equations so that when you add them together, one of the variables cancels out. This method is particularly useful when the coefficients of one variable are multiples of each other.

Why Approximations?

Now, why are we focusing on approximate solutions? In many real-world scenarios, the solutions to equations aren't always nice, round numbers. They might be messy decimals or fractions, which can make finding the exact solution algebraically a bit challenging. Additionally, in some contexts, an approximate solution is perfectly acceptable. For example, if we're modeling physical quantities, a slight deviation from the exact value might not significantly impact the outcome. Understanding how to find and evaluate approximate solutions is a valuable skill in mathematics and its applications. It's about striking a balance between precision and practicality, which is something we often encounter in problem-solving.

Analyzing the Given Equations

Okay, let's get back to our system of equations. We have:

y=βˆ’74x+52y=-\frac{7}{4}x+\frac{5}{2}

y=34xβˆ’3y=\frac{3}{4}x-3

Notice that both equations are in slope-intercept form (y = mx + b), which makes it easy to see the slope and y-intercept of each line. This form is incredibly useful because it gives us a quick snapshot of how the lines behave on a graph. The slope (m) tells us the steepness and direction of the line, while the y-intercept (b) tells us where the line crosses the y-axis. This visual understanding is a great starting point for figuring out where the lines might intersect.

Slope and Y-intercept

For the first equation, $y=-\frac{7}{4}x+\frac{5}{2}$, the slope is -7/4, and the y-intercept is 5/2 (or 2.5). This means the line is decreasing (since the slope is negative) and crosses the y-axis at 2.5. For the second equation, $y=\frac{3}{4}x-3$, the slope is 3/4, and the y-intercept is -3. This line is increasing (positive slope) and crosses the y-axis at -3. Just by looking at these slopes and y-intercepts, we can start to get a sense of where these lines might intersect. One line is heading downwards from a positive y-intercept, while the other is heading upwards from a negative y-intercept. This suggests they will intersect somewhere in the first or fourth quadrant.

Visualizing the Intersection

Imagine plotting these two lines on a graph. The first line starts high on the y-axis and slopes downward rather steeply. The second line starts low on the y-axis and slopes upward more gently. The point where they cross is the solution to the system. Because the slopes have opposite signs, we know they will intersect at exactly one point. Visualizing this intersection is a powerful tool, especially when we're dealing with approximations. It helps us to mentally check whether our calculated solutions make sense. If we calculate a solution that is far from where we visualize the intersection, it's a red flag that we might have made a mistake. So, always try to build a mental picture of what's happening graphically.

Solving the System Algebraically

Now, let's solve this system algebraically to find the exact solution, which we can then use to check our approximations. Since both equations are already solved for y, the substitution method seems like a straightforward approach. We can set the two expressions for y equal to each other:

- rac{7}{4}x+ rac{5}{2} = rac{3}{4}x-3

This equation now only has one variable, x, which we can solve. Let's get all the x terms on one side and the constants on the other.

Combining Like Terms

First, let’s add (7/4)x to both sides:

52=34x+74xβˆ’3\frac{5}{2} = \frac{3}{4}x + \frac{7}{4}x - 3

Combining the x terms, we get:

52=104xβˆ’3\frac{5}{2} = \frac{10}{4}x - 3

Since 10/4 simplifies to 5/2, we have:

52=52xβˆ’3\frac{5}{2} = \frac{5}{2}x - 3

Next, let’s add 3 to both sides. To do this, we'll rewrite 3 as 6/2 so that we have a common denominator:

52+62=52x\frac{5}{2} + \frac{6}{2} = \frac{5}{2}x

112=52x\frac{11}{2} = \frac{5}{2}x

Isolating x

Now, to isolate x, we'll multiply both sides by the reciprocal of 5/2, which is 2/5:

25β‹…112=x\frac{2}{5} \cdot \frac{11}{2} = x

x=115x = \frac{11}{5}

So, we've found that x = 11/5, which is equal to 2.2. This is our x-coordinate for the solution to the system. Now we need to find the corresponding y-coordinate. We can plug this value of x into either of our original equations. Let's use the second equation because it looks a bit simpler:

y=34xβˆ’3y = \frac{3}{4}x - 3

Substitute x = 11/5:

y=34β‹…115βˆ’3y = \frac{3}{4} \cdot \frac{11}{5} - 3

Solving for y

Multiply the fractions:

y=3320βˆ’3y = \frac{33}{20} - 3

To subtract 3, we need a common denominator, so we'll rewrite 3 as 60/20:

y=3320βˆ’6020y = \frac{33}{20} - \frac{60}{20}

y=βˆ’2720y = -\frac{27}{20}

So, y = -27/20, which is equal to -1.35. Therefore, the exact solution to the system of equations is (11/5, -27/20), or (2.2, -1.35).

Evaluating Possible Approximations

Now that we have the exact solution, (2.2, -1.35), we can evaluate which of the given points are possible approximations. This is where our understanding of approximations comes into play. We're looking for points that are reasonably close to the exact solution. Remember, an approximation doesn't have to be perfect; it just needs to be in the vicinity of the true answer.

We have two options to consider:

  1. (1.9, 2.5)
  2. (2.2, -1.4)

Let's take each option one at a time and compare it to our exact solution.

Analyzing Option 1: (1.9, 2.5)

The first point, (1.9, 2.5), has an x-coordinate of 1.9 and a y-coordinate of 2.5. Let's compare these to our exact solution, (2.2, -1.35).

  • X-coordinate: 1. 9 is somewhat close to 2.2, but the difference is noticeable (0.3).
  • Y-coordinate: 2. 5 is significantly different from -1.35. This is a large discrepancy.

Given the substantial difference in the y-coordinates, (1.9, 2.5) is not a good approximation for the solution to the system. The y-coordinate is way off, suggesting this point is not near the intersection of the two lines. So, we can rule out this option.

Analyzing Option 2: (2.2, -1.4)

Now let's consider the second point, (2.2, -1.4). The x-coordinate is 2.2, and the y-coordinate is -1.4. Comparing these to our exact solution (2.2, -1.35):

  • X-coordinate: 2. 2 is exactly the same as the x-coordinate of our exact solution.
  • Y-coordinate: -1. 4 is very close to -1.35. The difference is only 0.05, which is quite small.

This point looks like a much better approximation. The x-coordinate matches perfectly, and the y-coordinate is only slightly off. In many practical situations, a difference of 0.05 would be considered a very good approximation.

Conclusion

After carefully analyzing both options and comparing them to our exact solution (2.2, -1.35), we can confidently say that the point (2.2, -1.4) is a possible approximation for the solution to the system of equations. The other option, (1.9, 2.5), is not a good approximation due to the significant difference in the y-coordinate.

So there you have it, guys! We've successfully navigated through a system of equations, found the exact solution, and identified a good approximation. Remember, understanding approximations is key in many mathematical and real-world scenarios. Keep practicing, and you'll become a pro at solving these types of problems!