Solving Systems Of Equations X-y=5, 2x+y=7, -x+3y=4 A Comprehensive Guide

Aug 6, 2025 by ADMIN 74 views

Introduction to Solving Systems of Equations

Hey guys! Let's dive into the fascinating world of solving systems of equations. In this article, we're going to tackle a specific system: x - y = 5, 2x + y = 7, and -x + 3y = 4. Solving systems of equations is a fundamental skill in mathematics, and it pops up in various real-world applications, from engineering to economics. A system of equations is simply a set of two or more equations that share the same variables. The goal? Find the values for those variables that satisfy all equations simultaneously. Think of it as a puzzle where all the pieces (equations) need to fit together perfectly. There are several methods we can use, including substitution, elimination, and graphing. Each method has its strengths, and the best one to use often depends on the specific equations you're dealing with. We'll explore these methods as we solve our system. Now, why is this important? Well, imagine you're trying to figure out how much to charge for a product. You might have one equation representing your costs and another representing your revenue. Solving the system helps you find the break-even point, where your costs equal your revenue. Or, consider a scenario in physics where you're analyzing the motion of two objects. You might have equations describing their positions over time. Solving the system can tell you when and where the objects will meet. So, whether you're a student learning the basics or a professional applying these concepts in your field, mastering systems of equations is a valuable asset. Let's jump in and break down this particular system step by step!

Understanding the Equations: x - y = 5, 2x + y = 7, -x + 3y = 4

Okay, let's take a closer look at the system of equations we're working with: x - y = 5, 2x + y = 7, and -x + 3y = 4. Each of these equations represents a straight line when graphed on a coordinate plane. The solution to the system is the point (or points) where these lines intersect. In this case, we have three equations, which means we're looking for a single point that lies on all three lines simultaneously. Let's break down each equation individually to get a better handle on what they mean.

First, we have x - y = 5. This equation tells us that the difference between x and y is always 5. If we rearrange it to solve for y, we get y = x - 5. This form is called slope-intercept form, and it makes it easy to visualize the line. The slope is 1 (for every increase of 1 in x, y also increases by 1), and the y-intercept is -5 (the line crosses the y-axis at -5). This line slopes upward as you move from left to right.

Next, we have 2x + y = 7. If we solve for y here, we get y = -2x + 7. This line has a slope of -2 (for every increase of 1 in x, y decreases by 2), and the y-intercept is 7. This line slopes downward, and it's steeper than the first line because the slope is larger in magnitude.

Finally, we have -x + 3y = 4. Solving for y, we get 3y = x + 4, and then y = (1/3)x + 4/3. This line has a slope of 1/3 (for every increase of 3 in x, y increases by 1), and the y-intercept is 4/3 (approximately 1.33). This line is less steep than the other two, and it slopes upward.

Now, what do these equations tell us collectively? They describe three different lines in the same coordinate plane. To solve the system, we need to find the (x, y) coordinates that satisfy all three equations. This means the point must lie on all three lines. Graphically, this is the point where all three lines intersect. If there is such a point, the system has a unique solution. If the lines don't all intersect at a single point, the system might have no solution (if the lines are parallel or don't all meet) or infinitely many solutions (if the equations represent the same line).

Understanding these individual equations is the first step in solving the system. Next, we'll explore different methods to find the actual solution, the (x, y) values that make all three equations true.

Methods for Solving Systems of Equations

Alright, let's explore some methods for solving systems of equations. There are several approaches we can take, but we'll focus on two common ones: the substitution method and the elimination method. Each has its own advantages, and the best choice often depends on the structure of the equations.

Substitution Method

The substitution method involves solving one equation for one variable and then substituting that expression into another equation. This reduces the system to a single equation with one variable, which is much easier to solve. Once you find the value of that variable, you can substitute it back into one of the original equations to find the value of the other variable.

Here's how it works in general:

Solve one equation for one variable: Choose one of the equations and solve it for either x or y. Pick the equation and variable that look easiest to isolate. For example, if one equation has a lone x or y term, that's a good candidate.
Substitute into another equation: Take the expression you found in step 1 and substitute it into one of the other equations. It's crucial to substitute into a different equation than the one you solved in step 1. Otherwise, you'll just end up with a trivial identity.
Solve the resulting equation: After the substitution, you'll have an equation with only one variable. Solve this equation to find the value of that variable.
Substitute back to find the other variable: Take the value you found in step 3 and substitute it back into any of the original equations (or the expression you found in step 1) to find the value of the other variable.
Check your solution: Finally, plug both values you found into all the original equations to make sure they satisfy all of them. This step is important to catch any errors you might have made along the way.

Elimination Method

The elimination method (also known as the addition method) involves manipulating the equations so that when you add them together, one of the variables cancels out. This again reduces the system to a single equation with one variable.

Here's the general process:

Multiply equations to match coefficients: Look at the coefficients (the numbers in front of the variables) of either x or y in the equations. If necessary, multiply one or both equations by constants so that the coefficients of one of the variables are opposites (e.g., 2 and -2).
Add the equations: Add the two equations together. The variable with the matching coefficients should cancel out, leaving you with an equation in just one variable.
Solve the resulting equation: Solve the equation from step 2 to find the value of the remaining variable.
Substitute back to find the other variable: Substitute the value you found in step 3 into any of the original equations to find the value of the other variable.
Check your solution: As with the substitution method, check your solution by plugging the values into all the original equations.

Which method is better? It depends on the equations. If one equation is easily solved for one variable, substitution might be the way to go. If the coefficients of one variable are already opposites or can be easily made opposites, elimination might be more efficient. In our case, let's see how these methods work for the system x - y = 5, 2x + y = 7, and -x + 3y = 4.

Solving the System: x - y = 5, 2x + y = 7, -x + 3y = 4

Now, let's get our hands dirty and solve the system of equations x - y = 5, 2x + y = 7, and -x + 3y = 4 using both the substitution and elimination methods. This will give you a solid understanding of how each method works in practice.

Using the Elimination Method

The elimination method looks promising here because the second equation, 2x + y = 7, and the first equation, x - y = 5, have y terms with opposite signs. This means we can eliminate y by adding these two equations together.

Add the first two equations:

x - y = 5
2x + y = 7
---------
3x     = 12

Solve for x: Divide both sides by 3:

x = 4


Now we know that `x = 4`. Let's substitute this value back into one of the original equations to find `y`. We'll use the first equation, **x - y = 5**:

3.  **Substitute x = 4 into x - y = 5:**
    ```
    4 - y = 5
    ```

4.  **Solve for y:**
    Subtract 4 from both sides:
    ```
    -y = 1
    ```
    Multiply by -1:
    ```
y = -1

So, we've found a potential solution: x = 4 and y = -1. But remember, we have three equations, so we need to check if this solution satisfies all three.

Check the solution in all three equations:

Equation 1: x - y = 5
```
4 - (-1) = 4 + 1 = 5  (Correct)
```
Equation 2: 2x + y = 7
```
2(4) + (-1) = 8 - 1 = 7  (Correct)
```

Equation 3: -x + 3y = 4

-(4) + 3(-1) = -4 - 3 = -7  (Incorrect)

Uh oh! The solution (4, -1) satisfies the first two equations, but it doesn't satisfy the third equation. This means there's no single point that lies on all three lines. Therefore, the system has no solution.

Even though we found values for x and y that worked for the first two equations, the third equation acted as a constraint that our solution couldn't meet. This highlights an important aspect of systems of equations: all equations must be satisfied simultaneously for a solution to be valid.

Using the Substitution Method (for Demonstration)

Just for demonstration, let's see how the substitution method might look, even though we already know the system has no solution. We'll start with the first two equations and see where it leads us.

Solve the first equation for x:
```
x - y = 5
x = y + 5
```
Substitute into the second equation:
```
2x + y = 7
2(y + 5) + y = 7
```

Solve for y:

2y + 10 + y = 7
3y + 10 = 7
3y = -3
y = -1

Substitute back to find x:
```
x = y + 5
x = -1 + 5
x = 4
```

We arrive at the same potential solution (4, -1). But as we saw before, this solution doesn't satisfy the third equation. This further illustrates that even if a solution works for some equations in the system, it must work for all of them.

Conclusion: No Solution for x - y = 5, 2x + y = 7, -x + 3y = 4

So, guys, we've thoroughly investigated the system of equations x - y = 5, 2x + y = 7, and -x + 3y = 4. We used both the elimination and substitution methods to try and find a solution. While we found a potential solution that satisfied the first two equations, it failed to satisfy the third equation. This means that there is no single pair of (x, y) values that makes all three equations true simultaneously.

Therefore, the final answer is that the system has no solution. This can happen when the lines represented by the equations do not all intersect at a single point. In our case, the lines might intersect in pairs, but there's no common intersection point for all three.

This exercise highlights an important concept in solving systems of equations. It's not enough to find a solution that works for some of the equations; it must work for all of them. When dealing with real-world problems, this means that all the constraints or conditions represented by the equations must be met simultaneously.

I hope this detailed walkthrough has helped you understand how to approach and solve systems of equations. Remember, practice makes perfect! The more you work through different systems, the better you'll become at recognizing the best method to use and spotting potential pitfalls. Keep up the great work, and happy problem-solving!