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 process of solving linear systems using substitution, providing detailed explanations, examples, and, importantly, an answer key to help you solidify your understanding.

Understanding Linear Systems

Before diving into the substitution method, let's clarify what a system of linear equations is. A linear system consists of two or more linear equations, each involving the same variables. A linear equation is an equation where the highest power of the variables is 1. For example:

• 2x + y = 7
• x - 3y = 4

This is a system of two linear equations with two variables, x and y. The goal is to find the values of x and y that satisfy both equations simultaneously. Graphically, this represents the point of intersection of 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 reduces the system to a single equation with one variable, which can be easily solved. Here's a step-by-step guide:

Step 1: Solve for One Variable

Choose one of the equations and solve it for one of the variables. It's best to choose an equation and a variable that will result in the least amount of fractions. 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

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

• 3(5 - 2y) - y = 1

Step 3: Solve for the Remaining Variable

Solve the resulting equation for the remaining variable. This will usually involve expanding, combining like terms, and applying basic algebraic operations.

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

Step 4: Substitute Back

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. Let's substitute y = 2 into the equation x = 5 - 2y:

• x = 5 - 2(2)
• x = 5 - 4
• x = 1

Step 5: Check Your Solution

Always check your solution by substituting the values of x and y back into both original equations to ensure they satisfy both.

• Equation 1: x + 2y = 5 => 1 + 2(2) = 5 (True)
• Equation 2: 3x - y = 1 => 3(1) - 2 = 1 (True)

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

Handling Special Cases

Not all linear systems have a unique solution. There are two special cases to be aware of:

1. Inconsistent Systems (No Solution): These systems have parallel lines that never intersect. When solving by substitution, you'll arrive at a contradiction, such as 0 = 5. This indicates there is no solution.

2. Dependent Systems (Infinitely Many Solutions): These systems have lines that coincide (they are the same line). When solving by substitution, you'll arrive at an identity, such as 0 = 0. This means there are infinitely many solutions.

Examples with Answer Key

Let's work through several examples to further solidify your understanding. Remember to follow the five steps outlined above.

Example 1:

• 2x + y = 8
• x - 2y = -1

Solution: Solve the second equation for x: x = 2y - 1. Substitute into the first equation: 2(2y - 1) + y = 8. Solve for y: y = 2. Substitute y = 2 back into x = 2y -1: x = 3. Solution: x = 3, y = 2

Example 2:

• 3x - 2y = 7
• x + y = 2

Solution: Solve the second equation for x: x = 2 - y. Substitute into the first equation: 3(2 - y) - 2y = 7. Solve for y: y = -1/5. Substitute y = -1/5 back into x = 2 - y: x = 11/5. Solution: x = 11/5, y = -1/5

Example 3:

• x + y = 3
• 2x + 2y = 4

Solution: Solve the first equation for x: x = 3 - y. Substitute into the second equation: 2(3 - y) + 2y = 4. This simplifies to 6 = 4, which is a contradiction. Solution: No Solution

Example 4:

• x + 2y = 4
• 2x + 4y = 8

Solution: Solve the first equation for x: x = 4 - 2y. Substitute into the second equation: 2(4 - 2y) + 4y = 8. This simplifies to 8 = 8, which is an identity. Solution: Infinitely Many Solutions

Advanced Applications and Extensions

The substitution method, while seemingly straightforward, forms the basis for understanding more complex techniques in linear algebra. It lays the groundwork for solving larger systems of equations using methods like Gaussian elimination and matrix operations. Understanding this fundamental method is crucial for mastering these advanced techniques.

Conclusion

The substitution method provides a powerful and efficient way to solve systems of linear equations. By carefully following the steps outlined above and practicing with numerous examples, you'll develop a strong understanding of this essential algebraic concept. Remember to always check your solutions and be aware of the special cases of inconsistent and dependent systems. Mastering the substitution method will not only enhance your algebraic skills but also prepare you for more advanced mathematical concepts. Through consistent practice and a thorough understanding of the underlying principles, you will be well-equipped to tackle any linear system presented to you. Remember to always check your answers to ensure accuracy. Consistent practice is key to mastering this valuable skill.