Solving Linear Systems By Substitution Worksheet Answers

Solving linear systems by substitution is a fundamental concept in algebra, offering a powerful method for finding the values of variables that satisfy multiple linear equations simultaneously. The substitution method shines when one equation is easily solved for one variable in terms of the other, simplifying the process of finding solutions. In this comprehensive guide, we will delve into the intricacies of solving linear systems by substitution, providing a detailed explanation, step-by-step instructions, practical examples, and answers to common questions.

Understanding Linear Systems

Before diving into the substitution method, it’s crucial to understand what a linear system is. A linear system, also known as a system of linear equations, is a set of two or more linear equations that share the same variables. A solution to a linear system is a set of values for the variables that makes all equations in the system true simultaneously. Graphically, the solution represents the point(s) where the lines intersect.

Types of Solutions

Linear systems can have three types of solutions:

• Unique Solution: The system has exactly one solution, representing a single point of intersection between the lines.
• No Solution: The system has no solution, meaning the lines are parallel and never intersect.
• Infinite Solutions: The system has infinitely many solutions, indicating that the equations represent the same line.

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 detailed breakdown of the steps:

Step 1: Solve one equation for one variable.

• Choose one of the equations and isolate one of the variables. Look for an equation where one variable has a coefficient of 1 or -1, as this will simplify the process.
• For example, given the system:
  • x + y = 5
  • 2x - y = 1
• We can easily solve the first equation for x:
  • x = 5 - y

Step 2: Substitute the expression into the other equation.

• Take the expression you found in step 1 and substitute it into the other equation (the one you didn’t use in step 1).
• In our example, substitute x = 5 - y into the second equation:
  • 2(5 - y) - y = 1

Step 3: Solve for the remaining variable.

• Now you have an equation with only one variable. Solve for that variable using algebraic techniques.
• Continuing our example:
  • 10 - 2y - y = 1
  • 10 - 3y = 1
  • -3y = -9
  • y = 3

Step 4: Substitute the value back into the expression from step 1.

• Now that you know the value of one variable, substitute it back into the expression you found in step 1 to find the value of the other variable.
• In our example, substitute y = 3 into x = 5 - y:
  • x = 5 - 3
  • x = 2

Step 5: Check your solution.

• To ensure your solution is correct, substitute the values you found for both variables into both original equations. If both equations are true, then your solution is correct.
• In our example, check x = 2 and y = 3 in the original equations:
  • x + y = 5 => 2 + 3 = 5 (True)
  • 2x - y = 1 => 2(2) - 3 = 1 => 4 - 3 = 1 (True)
• Therefore, the solution to the system is x = 2 and y = 3, or the ordered pair (2, 3).

Example Problems with Solutions

Let's work through several example problems to solidify your understanding of the substitution method.

Example 1:

Solve the following system of equations:

• y = 2x + 1
• 3x + y = 11

Solution:

1. Solve for one variable: The first equation is already solved for y: y = 2x + 1
2. Substitute: Substitute y = 2x + 1 into the second equation: 3x + (2x + 1) = 11
3. Solve for x:
   • 3x + 2x + 1 = 11
   • 5x + 1 = 11
   • 5x = 10
   • x = 2
4. Substitute back: Substitute x = 2 into y = 2x + 1:
   • y = 2(2) + 1
   • y = 4 + 1
   • y = 5
5. Check:
   • y = 2x + 1 => 5 = 2(2) + 1 => 5 = 5 (True)
   • 3x + y = 11 => 3(2) + 5 = 11 => 6 + 5 = 11 (True)

The solution is x = 2 and y = 5, or (2, 5).

Example 2:

Solve the following system of equations:

• 2x + y = 7
• x - 2y = -4

Solution:

1. Solve for one variable: Solve the second equation for x: x = 2y - 4
2. Substitute: Substitute x = 2y - 4 into the first equation: 2(2y - 4) + y = 7
3. Solve for y:
   • 4y - 8 + y = 7
   • 5y - 8 = 7
   • 5y = 15
   • y = 3
4. Substitute back: Substitute y = 3 into x = 2y - 4:
   • x = 2(3) - 4
   • x = 6 - 4
   • x = 2
5. Check:
   • 2x + y = 7 => 2(2) + 3 = 7 => 4 + 3 = 7 (True)
   • x - 2y = -4 => 2 - 2(3) = -4 => 2 - 6 = -4 (True)

The solution is x = 2 and y = 3, or (2, 3).

Example 3:

Solve the following system of equations:

• 4x + 3y = 10
• x - y = 1

Solution:

1. Solve for one variable: Solve the second equation for x: x = y + 1
2. Substitute: Substitute x = y + 1 into the first equation: 4(y + 1) + 3y = 10
3. Solve for y:
   • 4y + 4 + 3y = 10
   • 7y + 4 = 10
   • 7y = 6
   • y = 6/7
4. Substitute back: Substitute y = 6/7 into x = y + 1:
   • x = (6/7) + 1
   • x = (6/7) + (7/7)
   • x = 13/7
5. Check:
   • 4x + 3y = 10 => 4(13/7) + 3(6/7) = 10 => (52/7) + (18/7) = 10 => 70/7 = 10 (True)
   • x - y = 1 => (13/7) - (6/7) = 1 => 7/7 = 1 (True)

The solution is x = 13/7 and y = 6/7, or (13/7, 6/7).

When Does Substitution Work Best?

The substitution method is most effective when:

• One of the equations is already solved for one variable.
• One of the variables has a coefficient of 1 or -1 in one of the equations, making it easy to isolate.

If neither of these conditions is met, the elimination method might be a more efficient approach.

Common Mistakes to Avoid

• Forgetting to substitute back: After solving for one variable, remember to substitute the value back into one of the equations to find the value of the other variable.
• Substituting into the same equation: Make sure to substitute the expression into the other equation, not the one you used to solve for the variable.
• Incorrectly distributing: When substituting an expression into an equation, be careful to distribute any coefficients correctly.
• Arithmetic errors: Double-check your calculations to avoid making mistakes in the algebraic steps.

Practice Problems

Here are a few practice problems for you to try on your own. The answers are provided below, but try to solve them yourself first!

1. y = 3x - 2 x + 2y = 8
2. x - y = 4 2x + y = 5
3. 4x - y = 1 y = x + 2

Answers:

1. (2, 4)
2. (3, -1)
3. (-1/3, 5/3)

Advanced Applications

The principles of solving linear systems by substitution extend to more complex scenarios, including:

• Systems with three or more variables: While substitution can become cumbersome, it's still applicable. You would solve for one variable in one equation and substitute that expression into the other equations, reducing the system until you can solve for the remaining variables.
• Non-linear systems: In some cases, substitution can be used to solve non-linear systems, where the equations are not linear. However, this often leads to more complex algebraic manipulations.
• Real-world applications: Linear systems are used to model various real-world scenarios, such as:
  • Mixture problems: Determining the amounts of different ingredients needed to create a mixture with specific properties.
  • Investment problems: Calculating the returns on different investments.
  • Distance-rate-time problems: Solving for unknown distances, rates, or times.

Frequently Asked Questions (FAQ)

• Q: Can I use the substitution method for any system of linear equations?

  • A: Yes, the substitution method can be used for any system of linear equations. However, it is most efficient when one of the equations is easily solved for one variable.

• Q: What if I get a false statement, like 0 = 1, when solving a system of equations?

  • A: This indicates that the system has no solution. The lines are parallel and do not intersect.

• Q: What if I get a true statement, like 0 = 0, when solving a system of equations?

  • A: This indicates that the system has infinitely many solutions. The equations represent the same line.

• Q: Is there a specific variable I should always solve for first?

  • A: No, you can solve for any variable you choose. However, it's generally easier to solve for a variable with a coefficient of 1 or -1.

• Q: Can I use a calculator to help me solve systems of equations?

  • A: Yes, many calculators have built-in functions for solving systems of equations. However, it's important to understand the underlying concepts and steps involved in the substitution method.

Conclusion

Solving linear systems by substitution is a valuable skill in algebra and has wide-ranging applications in various fields. By following the step-by-step guide, practicing with example problems, and avoiding common mistakes, you can master this technique and confidently solve a wide range of linear systems. Remember to always check your solutions to ensure accuracy. The ability to solve linear systems effectively opens doors to understanding and solving more complex mathematical problems.