Solving Systems Of Linear Equations By Substitution Answer Key

Solving Systems of Linear Equations by Substitution: A Comprehensive Guide with Answer Key

Solving systems of linear equations is a fundamental concept in algebra with wide-ranging applications in various fields, from physics and engineering to economics and computer science. While several methods exist for solving these systems, substitution is a particularly powerful and intuitive approach, especially for beginners. This comprehensive guide will walk you through the process of solving systems of linear equations by substitution, providing a detailed explanation, examples, and a comprehensive answer key to practice problems.

Understanding Systems of Linear Equations

A system of linear equations is simply a set of two or more linear equations with the same variables. A linear equation is an equation where the highest power of the variable is 1. For example:

• 2x + y = 5
• x - y = 1

This is a system of two linear equations with two variables, x and y. Solving this system means finding the values of x and y that satisfy both equations simultaneously. Graphically, this represents finding the point of intersection between the two lines represented by the equations.

The Substitution Method: A Step-by-Step Guide

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This eliminates one variable, allowing you to solve for the remaining variable. Here's a step-by-step guide:

Step 1: Solve for one variable in one equation.

Choose one of the equations and solve it for one of the variables. It's best to choose the equation and variable that will require the least amount of algebraic manipulation. For example, in the system:

• x + 2y = 5
• 3x - y = 1

It's easier to solve the first equation for x: x = 5 - 2y

Step 2: Substitute the expression into the other equation.

Substitute the expression you found in Step 1 into the other equation. In our example, substitute x = 5 - 2y into the second equation:

• 3(5 - 2y) - y = 1

Step 3: Solve for the remaining variable.

Simplify and solve the resulting equation for the remaining variable.

• 15 - 6y - y = 1
• 15 - 7y = 1
• -7y = -14
• y = 2

Step 4: Substitute the value back into either original equation.

Substitute the value you found in Step 3 back into either of the original equations to solve for the other variable. Let's use the first equation:

• x + 2(2) = 5
• x + 4 = 5
• x = 1

Step 5: Check your solution.

Substitute both values (x and y) into both original equations to verify that they satisfy both equations.

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

Therefore, the solution to the system of equations is x = 1, y = 2.

Examples with Detailed Solutions and Answer Key

Let's work through several examples to solidify your understanding. Each example will include a step-by-step solution. Remember to always check your answers!

Example 1:

• 2x + y = 7
• x - y = 2

Solution:

1. Solve the second equation for x: x = y + 2
2. Substitute this into the first equation: 2(y + 2) + y = 7
3. Simplify and solve for y: 2y + 4 + y = 7 => 3y = 3 => y = 1
4. Substitute y = 1 back into x = y + 2: x = 1 + 2 = 3
5. Check: 2(3) + 1 = 7 (True); 3 - 1 = 2 (True)

Answer: x = 3, y = 1

Example 2:

• 3x - 2y = 4
• x + y = 6

Solution:

1. Solve the second equation for x: x = 6 - y
2. Substitute this into the first equation: 3(6 - y) - 2y = 4
3. Simplify and solve for y: 18 - 3y - 2y = 4 => -5y = -14 => y = 14/5
4. Substitute y = 14/5 back into x = 6 - y: x = 6 - 14/5 = 16/5
5. Check: 3(16/5) - 2(14/5) = 4 (True); 16/5 + 14/5 = 6 (True)

Answer: x = 16/5, y = 14/5

Example 3:

• 4x + 2y = 10
• x - y = 1

Solution:

1. Solve the second equation for x: x = y + 1
2. Substitute this into the first equation: 4(y + 1) + 2y = 10
3. Simplify and solve for y: 4y + 4 + 2y = 10 => 6y = 6 => y = 1
4. Substitute y = 1 back into x = y + 1: x = 1 + 1 = 2
5. Check: 4(2) + 2(1) = 10 (True); 2 - 1 = 1 (True)

Answer: x = 2, y = 1

Example 4 (Involving Fractions):

• (1/2)x + y = 3
• x - 2y = 2

Solution:

1. Solve the second equation for x: x = 2y + 2
2. Substitute this into the first equation: (1/2)(2y + 2) + y = 3
3. Simplify and solve for y: y + 1 + y = 3 => 2y = 2 => y = 1
4. Substitute y = 1 back into x = 2y + 2: x = 2(1) + 2 = 4
5. Check: (1/2)(4) + 1 = 3 (True); 4 - 2(1) = 2 (True)

Answer: x = 4, y = 1

Practice Problems with Answer Key

Here are some practice problems to test your understanding. Try to solve them on your own before checking the answer key below.

Practice Problem 1:

• x + y = 5
• x - y = 1

Practice Problem 2:

• 2x + 3y = 12
• x - y = 1

Practice Problem 3:

• 3x + y = 7
• x + 2y = 4

Practice Problem 4:

• (2/3)x + y = 4
• x - 3y = 3

Answer Key for Practice Problems:

Practice Problem 1: x = 3, y = 2

Practice Problem 2: x = 3, y = 2

Practice Problem 3: x = 2, y = 1

Practice Problem 4: x = 6, y = 1

Advanced Applications and Considerations

While the examples above focus on systems with two variables, the substitution method can be extended to systems with three or more variables. However, the process becomes significantly more complex and time-consuming. For larger systems, other methods like elimination or matrix methods are generally more efficient.

Additionally, it's important to note that not all systems of linear equations have a unique solution. Some systems have infinitely many solutions (dependent systems), while others have no solutions (inconsistent systems). When solving a system by substitution, you'll encounter these situations if you arrive at an equation that is always true (e.g., 0 = 0) or always false (e.g., 1 = 0).

Conclusion

Solving systems of linear equations by substitution is a valuable algebraic skill. By mastering this technique, you'll be equipped to tackle a wide range of mathematical problems and real-world applications. Remember the step-by-step process, practice regularly, and always check your solutions to ensure accuracy. This comprehensive guide, along with the provided examples and practice problems, should provide a strong foundation for your understanding of this crucial concept. Continue practicing, and you’ll become proficient in solving even more complex systems of linear equations.