Solve The System Of Equations Using The Substitution Method: 3x + 4y = 0 x - 5y = 0

Jun 18, 2025 by ADMIN 85 views

Solving Systems of Equations by Substitution: A Comprehensive Guide

In the realm of algebra, solving systems of equations is a fundamental skill. These systems, consisting of two or more equations with multiple variables, often appear in various mathematical and real-world applications. Among the several methods to tackle these systems, the substitution method stands out as a powerful and versatile technique. This article delves into the intricacies of the substitution method, providing a step-by-step guide along with illustrative examples to solidify your understanding.

Understanding the Substitution Method

The substitution method is an algebraic technique used to solve systems of equations by expressing one variable in terms of another. This method is particularly effective when one of the equations can be easily solved for one variable. The core idea is to substitute the expression for one variable into the other equation, effectively reducing the system to a single equation with a single variable. Solving this equation gives the value of one variable, which can then be substituted back into either of the original equations to find the value of the other variable.

Step-by-Step Guide to the Substitution Method

To effectively employ the substitution method, follow these steps:

1. Choose an Equation and Solve for One Variable

Begin by carefully examining the system of equations. Identify an equation where one of the variables can be easily isolated. This typically involves selecting an equation where the variable has a coefficient of 1 or -1. Solve this equation for the chosen variable. This will give you an expression for that variable in terms of the other variable. For instance, in the system:

2x + y = 5
x - 3y = -1

The first equation can be easily solved for y:

y = 5 - 2x

2. Substitute the Expression into the Other Equation

Now, take the expression you obtained in step 1 and substitute it into the other equation in the system. This step is crucial as it eliminates one variable, leaving you with an equation in a single variable. In our example, substitute y = 5 - 2x into the second equation:

x - 3(5 - 2x) = -1

3. Solve the Resulting Equation

After the substitution, you'll have an equation with only one variable. Solve this equation using standard algebraic techniques. This may involve simplifying, combining like terms, and isolating the variable. Continuing our example:

x - 15 + 6x = -1
7x - 15 = -1
7x = 14
x = 2

4. Substitute Back to Find the Other Variable

Once you've found the value of one variable, substitute it back into either of the original equations or the expression you derived in step 1. This will allow you to solve for the other variable. Using our example, substitute x = 2 into y = 5 - 2x:

y = 5 - 2(2)
y = 5 - 4
y = 1

5. Check Your Solution

To ensure accuracy, it's always a good practice to check your solution by substituting the values of both variables into both original equations. If the solution satisfies both equations, you've found the correct solution. In our example, substituting x = 2 and y = 1 into the original equations:

2(2) + 1 = 5 (True)
2 - 3(1) = -1 (True)

Thus, the solution is x = 2 and y = 1.

Illustrative Examples

Let's solidify our understanding with a couple more examples.

Example 1

Solve the following system using the substitution method:

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

Solve the first equation for x:
```
x = 7 - 2y
```
Substitute this expression for x into the second equation:
```
3(7 - 2y) - y = -3
```

Solve for y:

21 - 6y - y = -3
21 - 7y = -3
-7y = -24
y = 24/7

Substitute the value of y back into the expression for x:

x = 7 - 2(24/7)
x = 7 - 48/7
x = (49 - 48)/7
x = 1/7

The solution is x = 1/7 and y = 24/7.

Example 2

Consider the system:

4x - 3y = 8
x + 2y = -5

Solve the second equation for x:
```
x = -5 - 2y
```
Substitute this expression for x into the first equation:
```
4(-5 - 2y) - 3y = 8
```

Solve for y:

-20 - 8y - 3y = 8
-20 - 11y = 8
-11y = 28
y = -28/11

Substitute the value of y back into the expression for x:

x = -5 - 2(-28/11)
x = -5 + 56/11
x = (-55 + 56)/11
x = 1/11

The solution is x = 1/11 and y = -28/11.

Advantages of the Substitution Method

The substitution method offers several advantages:

Versatility: It can be applied to a wide range of systems of equations, including those with linear and nonlinear equations.
Simplicity: The steps involved are relatively straightforward, making it easier to grasp and apply.
Efficiency: When one variable can be easily isolated, the substitution method can be a quick and efficient way to find the solution.

Disadvantages of the Substitution Method

Despite its advantages, the substitution method also has some limitations:

Complexity: If neither variable can be easily isolated, the method can become cumbersome and may involve complex fractions or radicals.
Error-prone: The substitution process can be prone to errors if not performed carefully.
Alternative Methods: For certain systems, other methods like elimination or matrix methods might be more efficient.

A Detailed Solution Using the Substitution Method

Now, let's tackle the system provided in the original prompt using the substitution method. The system is:

3x + 4y = 0
x - 5y = 0

Step 1: Choose an Equation and Solve for One Variable

In this system, the second equation, x - 5y = 0, is the easiest to solve for x. Adding 5y to both sides, we get:

x = 5y

Step 2: Substitute the Expression into the Other Equation

Substitute the expression x = 5y into the first equation, 3x + 4y = 0:

3(5y) + 4y = 0

Step 3: Solve the Resulting Equation

Simplify and solve for y:

15y + 4y = 0
19y = 0
y = 0

Step 4: Substitute Back to Find the Other Variable

Substitute y = 0 back into the expression x = 5y:

x = 5(0)
x = 0

Step 5: Check Your Solution

Substitute x = 0 and y = 0 into both original equations:

3(0) + 4(0) = 0 (True)
0 - 5(0) = 0 (True)

Both equations are satisfied, so the solution is correct.

Conclusion

The substitution method is a valuable tool in the arsenal of algebraic techniques for solving systems of equations. By carefully following the steps outlined in this guide and practicing with examples, you can master this method and confidently tackle a wide range of problems. While it has its limitations, the substitution method remains a fundamental concept in algebra and a stepping stone to more advanced mathematical concepts. Understanding and mastering this method is crucial for success in mathematics and related fields. Remember to always check your solutions to ensure accuracy. Practice makes perfect, so work through various examples to solidify your understanding. The substitution method is not just a technique; it's a way of thinking about how equations relate to each other. By isolating variables and substituting expressions, we can simplify complex problems and find elegant solutions.