Solving Systems Of Equations By Substitution Answer Key

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

Solving systems of equations is a fundamental concept in algebra with wide-ranging applications in various fields, from physics and engineering to economics and computer science. One of the most common and straightforward methods for solving these systems is the substitution method. This guide will provide a comprehensive understanding of the substitution method, covering its principles, steps, and practical applications, complete with a detailed answer key for various example problems.

Understanding Systems of Equations

Before diving into the substitution method, let's establish a clear understanding of what a system of equations is. A system of equations is a collection of two or more equations with the same set of variables. The goal is to find the values of these variables that satisfy all equations simultaneously. These values represent the point(s) of intersection between the graphs of the equations. Systems of equations can have:

One unique solution: The lines intersect at a single point.
Infinitely many solutions: The lines are coincident (overlap completely).
No solution: The lines are parallel and never intersect.

The Substitution Method: A Step-by-Step Approach

The substitution method involves solving for one variable in one equation and substituting that expression into the other equation. This reduces the system to a single equation with one variable, which can then be solved. Here's a step-by-step guide:

Step 1: Solve for one variable in one equation. Choose the equation that is easiest to solve for one variable. Look for equations where a variable has a coefficient of 1 or -1, simplifying the process.

Step 2: Substitute the expression from Step 1 into the other equation. Replace the chosen variable in the second equation with the expression you found in Step 1. This will create a new equation with only one variable.

Step 3: Solve the resulting equation. Solve the equation from Step 2 for the remaining variable. This will give you the value of one of the variables.

Step 4: Substitute the value from Step 3 back into either of the original equations. Substitute the value you found in Step 3 into either of the original equations to solve for the other variable.

Step 5: Check your solution. Substitute both values into both original equations to verify that they satisfy both equations simultaneously. This step is crucial to ensure accuracy.

Examples with Detailed Solutions (Answer Key)

Let's work through several examples to illustrate the substitution method effectively.

Example 1:

Solve the following system of equations:

x + y = 5 2x - y = 1

Solution:

Solve for one variable: The first equation is easiest to solve for x: x = 5 - y
Substitute: Substitute x = 5 - y into the second equation: 2(5 - y) - y = 1
Solve: Simplify and solve for y: 10 - 2y - y = 1 => -3y = -9 => y = 3
Substitute back: Substitute y = 3 into x = 5 - y: x = 5 - 3 => x = 2
Check: Substitute x = 2 and y = 3 into both original equations: 2 + 3 = 5 (True) 2(2) - 3 = 1 (True)

Therefore, the solution is x = 2, y = 3.

Example 2:

Solve the following system of equations:

3x + 2y = 7 x - y = 1

Solution:

Solve for one variable: The second equation is easily solved for x: x = y + 1
Substitute: Substitute x = y + 1 into the first equation: 3(y + 1) + 2y = 7
Solve: Simplify and solve for y: 3y + 3 + 2y = 7 => 5y = 4 => y = 4/5
Substitute back: Substitute y = 4/5 into x = y + 1: x = 4/5 + 1 => x = 9/5
Check: Substitute x = 9/5 and y = 4/5 into both original equations: 3(9/5) + 2(4/5) = 27/5 + 8/5 = 35/5 = 7 (True) 9/5 - 4/5 = 5/5 = 1 (True)

Therefore, the solution is x = 9/5, y = 4/5.

Example 3: A system with no solution.

x + y = 3 x + y = 5

Solution: Attempting to solve this system using substitution will quickly reveal an inconsistency. Solving the first equation for x gives x = 3 - y. Substituting this into the second equation gives:

3 - y + y = 5

This simplifies to 3 = 5, which is a false statement. This means the system has no solution. Graphically, these are parallel lines.

Example 4: A system with infinitely many solutions.

x + y = 3 2x + 2y = 6

Solution: Solving the first equation for x gives x = 3 - y. Substituting this into the second equation gives:

2(3 - y) + 2y = 6 6 - 2y + 2y = 6 6 = 6

This is a true statement, but it doesn't give us a specific value for x or y. This indicates that the two equations are equivalent (one is a multiple of the other), and there are infinitely many solutions. Graphically, these lines are coincident.

Advanced Applications and Considerations

The substitution method is particularly useful when one of the equations can be easily solved for one variable. However, for more complex systems or those involving non-linear equations, other methods like elimination or graphing might be more efficient.

Non-linear Systems: The substitution method can also be applied to systems of non-linear equations, such as those involving quadratic or exponential functions. The process remains similar, but the resulting equation to solve might be more complex.

Systems with Three or More Variables: The substitution method can be extended to systems with three or more variables, but the process becomes significantly more involved. It often requires multiple steps of substitution and can be prone to errors.

Conclusion

The substitution method is a powerful and versatile tool for solving systems of equations. By following the step-by-step process outlined in this guide and practicing with various examples, you can master this essential algebraic technique. Remember to always check your solution to ensure accuracy and to understand the different possibilities: one unique solution, infinitely many solutions, or no solution. Understanding these possibilities is crucial for interpreting the results within the context of the problem you are trying to solve. With practice and a solid understanding of the principles, you can confidently tackle even more complex systems of equations.