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 applications spanning various fields like physics, economics, and computer science. One of the most common and straightforward methods for solving these systems is the substitution method. This comprehensive guide will walk you through the process, provide examples with detailed explanations, and offer an answer key for practice problems. We'll cover everything from basic linear systems to more complex scenarios, ensuring you master this crucial algebraic technique.

Understanding Systems of Equations

A system of equations is a set of two or more equations with the same variables. The goal is to find the values of the variables that satisfy all equations simultaneously. These values represent the point(s) of intersection between the graphs of the equations. For example:

• Linear System: A system where all equations are linear (highest power of the variable is 1). Example:

  x + y = 5
  x - y = 1
  

• Non-linear System: A system where at least one equation is not linear. Example:

  x² + y = 4
  x + y = 2
  

The Substitution Method: Step-by-Step

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:

Step 1: Solve for One Variable

Choose one equation and solve it for one variable in terms of the other. Select the equation and variable that makes this step easiest. Often, this involves looking for an equation where a variable already has a coefficient of 1 or -1.

Step 2: Substitute

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.

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.

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 accuracy and catch any potential errors.

Examples with Detailed Explanations

Let's work through several examples to solidify your understanding.

Example 1: A Simple Linear System

Solve the system:

x + y = 7
x - y = 1

Solution:

1. Solve for one variable: Let's solve the first equation for x: x = 7 - y

2. Substitute: Substitute this expression for x into the second equation: (7 - y) - y = 1

3. Solve for the remaining variable: Simplify and solve for y: 7 - 2y = 1 => -2y = -6 => y = 3

4. Substitute back: Substitute y = 3 into the equation x = 7 - y: x = 7 - 3 = 4

5. Check:

   • x + y = 4 + 3 = 7 (Correct)
   • x - y = 4 - 3 = 1 (Correct)

Therefore, the solution is x = 4, y = 3.

Example 2: A System with Fractions

Solve the system:

(1/2)x + y = 3
x - y = 1

Solution:

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

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

3. Solve for the remaining variable: (1/2)y + (1/2) + y = 3 => (3/2)y = (5/2) => y = 5/3

4. Substitute back: Substitute y = 5/3 into x = y + 1: x = (5/3) + 1 = 8/3

5. Check: (This step involves fraction arithmetic, ensure you verify the solution carefully)

Therefore, the solution is x = 8/3, y = 5/3.

Example 3: A Non-linear System

Solve the system:

x² + y = 5
x + y = 1

Solution:

1. Solve for one variable: Solve the second equation for y: y = 1 - x

2. Substitute: Substitute this into the first equation: x² + (1 - x) = 5

3. Solve for the remaining variable: x² - x - 4 = 0. This is a quadratic equation. We can use the quadratic formula to solve for x:

   x = [1 ± √(1 - 4(1)(-4))] / 2 = [1 ± √17] / 2

   This gives us two possible values for x: x₁ = (1 + √17)/2 and x₂ = (1 - √17)/2

4. Substitute back: Substitute each value of x back into y = 1 - x to find the corresponding y values.

5. Check: Verify both solutions by substituting them back into both original equations.

Therefore, the solutions are x₁ = (1 + √17)/2, y₁ = (1 - √17)/2 and x₂ = (1 - √17)/2, y₂ = (1 + √17)/2.

Practice Problems with Answer Key

Here are some practice problems to test your understanding. Try solving them yourself before checking the answer key.

Problem 1:

2x + y = 5
x - y = 1

Problem 2:

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

Problem 3:

(1/3)x + y = 2
x - 2y = 1

Problem 4: (Non-linear)

x² - y = 2
x + y = 2

Answer Key

Problem 1: x = 2, y = 1

Problem 2: x = 1, y = 2

Problem 3: x = 3, y = 1

Problem 4: x = 2, y = 0 and x = -1, y = 3

Advanced Topics and Considerations

While the examples above focus on relatively straightforward systems, the substitution method can be applied to more complex scenarios. These include:

• Systems with three or more variables: The process extends logically, though the calculations become more involved. You would solve one equation for one variable, substitute into the other equations, and continue the process until you have solved for all variables.

• Systems with non-linear equations of higher degree: These systems may require more sophisticated algebraic techniques or numerical methods to solve.

• Systems with special cases: Some systems have no solution (inconsistent systems), while others have infinitely many solutions (dependent systems). Recognizing these cases is crucial. An inconsistent system will lead to a contradiction during the solving process (e.g., 0 = 5), while a dependent system will result in an identity (e.g., 0 = 0).

Conclusion

Mastering the substitution method is essential for success in algebra and related fields. By following the steps outlined above and practicing with various examples, you can develop proficiency in solving systems of equations, a skill applicable to a wide range of mathematical problems. Remember to always check your solutions to ensure accuracy. Consistent practice is key to building your confidence and understanding of this fundamental algebraic technique. Continue practicing with diverse problems to solidify your comprehension and tackle even the most challenging systems with ease.