Solving Linear Systems By Substitution Answer Key

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

Solving systems of linear equations is a fundamental concept in algebra with broad applications in various fields, from physics and engineering to economics and computer science. One of the most common methods for solving these systems is the substitution method. This comprehensive guide will walk you through the substitution method, providing detailed explanations, examples, and a comprehensive answer key to solidify your understanding.

What is a System of Linear Equations?

A system of linear equations is a set of two or more linear equations with the same variables. A linear equation is an equation that can be written in the form ax + by = c, where a, b, and c are constants, and x and y are variables. The goal when solving a system of linear equations is to find the values of the variables that satisfy all equations simultaneously. This point represents the intersection of the lines represented by each equation.

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 process 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 for one of the variables in terms of the other. Select the equation and variable that makes this step easiest. Often, this will involve an equation with a coefficient of 1 or -1 for one of the variables.

Step 2: Substitute the Expression into the Other Equation

Substitute the expression you found in Step 1 into the other equation. This will create a new equation with only one variable.

Step 3: Solve for the Remaining Variable

Solve the new equation from Step 2 for the remaining variable. This will give you the value of one of the variables in the system.

Step 4: Substitute Back to Find the Other Variable

Substitute the value you found in Step 3 back into either of the original equations (or the expression from Step 1) to solve for the other variable.

Step 5: Check Your Solution

Substitute both values back into both original equations to verify that they satisfy both equations simultaneously. This step is crucial to ensure the accuracy of your solution.

Examples with Detailed Solutions and Answer Key

Let's work through several examples to illustrate the substitution method. Each example will include a detailed solution and the final answer will be included in the answer key at the end.

Example 1:

Solve the system:

• x + y = 5
• x - y = 1

Solution:

1. Solve for one variable: From the first equation, we can easily solve for x: x = 5 - y

2. Substitute: Substitute this expression for x into the second equation: (5 - y) - y = 1

3. Solve for the remaining variable: Simplify and solve for y: 5 - 2y = 1 => -2y = -4 => y = 2

4. Substitute back: Substitute y = 2 back into x = 5 - y: x = 5 - 2 = 3

5. Check: Substitute x = 3 and y = 2 into both original equations:

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

Example 2:

Solve the system:

• 2x + y = 7
• x - 3y = 4

Solution:

1. Solve for one variable: Let's solve for x in the second equation: x = 3y + 4

2. Substitute: Substitute this into the first equation: 2(3y + 4) + y = 7

3. Solve for the remaining variable: Simplify and solve for y: 6y + 8 + y = 7 => 7y = -1 => y = -1/7

4. Substitute back: Substitute y = -1/7 into x = 3y + 4: x = 3(-1/7) + 4 = 25/7

5. Check: Substitute x = 25/7 and y = -1/7 into both original equations (this step involves fraction arithmetic and is left as an exercise for the reader to verify).

Example 3:

Solve the system:

• 3x + 2y = 11
• x - y = 2

Solution:

1. Solve for one variable: From the second equation, we can solve for x: x = y + 2

2. Substitute: Substitute this into the first equation: 3(y + 2) + 2y = 11

3. Solve for the remaining variable: Simplify and solve for y: 3y + 6 + 2y = 11 => 5y = 5 => y = 1

4. Substitute back: Substitute y = 1 into x = y + 2: x = 1 + 2 = 3

5. Check: Substitute x = 3 and y = 1 into both original equations:

   • 3(3) + 2(1) = 11 (True)
   • 3 - 1 = 2 (True)

Example 4 (Involving Fractions):

Solve the system:

• (1/2)x + y = 3
• x - 2y = 1

Solution:

1. Solve for one variable: From the second equation, we can solve for x: x = 2y + 1

2. Substitute: Substitute this into the first equation: (1/2)(2y + 1) + y = 3

3. Solve for the remaining variable: Simplify and solve for y: y + 1/2 + y = 3 => 2y = 5/2 => y = 5/4

4. Substitute back: Substitute y = 5/4 into x = 2y + 1: x = 2(5/4) + 1 = 7/2

5. Check: Substitute x = 7/2 and y = 5/4 into both original equations (again, verifying this step is left as an exercise for the reader).

Dealing with Special Cases

Sometimes, systems of linear equations will have special solutions:

• No Solution: If, during the substitution process, you arrive at a contradiction (e.g., 0 = 5), then the system has no solution. This means the lines represented by the equations are parallel and never intersect.

• Infinite Solutions: If, during the substitution process, you arrive at an identity (e.g., 0 = 0), then the system has infinitely many solutions. This means the two equations represent the same line.

Answer Key

Here's a summary of the solutions to the examples above:

• Example 1: x = 3, y = 2
• Example 2: x = 25/7, y = -1/7
• Example 3: x = 3, y = 1
• Example 4: x = 7/2, y = 5/4

Advanced Applications and Extensions

The substitution method forms the basis for understanding more advanced techniques for solving systems of linear equations, including:

• Solving systems with three or more variables: The substitution method can be extended to solve larger systems, although it can become more complex.

• Matrix methods: Matrices provide a more efficient way to solve larger systems of linear equations.

• Gaussian elimination: This is a systematic method for transforming a system of linear equations into an equivalent system that is easier to solve.

Conclusion

The substitution method is a powerful and versatile tool for solving systems of linear equations. While it might become cumbersome for very large systems, mastering this method is crucial for developing a strong foundation in algebra and its applications. Remember to always check your solutions to ensure accuracy. Practice regularly with a variety of examples, including those with fractions and special cases, to build confidence and proficiency in solving linear systems by substitution. This comprehensive guide, along with the detailed examples and answer key, should equip you to confidently tackle these types of problems.