Solving Systems Of Linear Equations By Substitution Worksheet Answers

Solving Systems of Linear Equations by Substitution: A Comprehensive Guide with Worksheet Answers

Solving systems of linear equations is a fundamental concept in algebra, with applications spanning various fields like engineering, economics, and computer science. This comprehensive guide focuses on the substitution method, a powerful technique for finding the solution (the point of intersection) of two or more linear equations. We'll explore the method step-by-step, provide illustrative examples, and offer detailed solutions to a practice worksheet. Mastering this technique is crucial for progressing to more advanced mathematical concepts.

Understanding Systems 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 of the form ax + by = c, where a, b, and c are constants, and x and y are variables. The solution to a system of linear equations is the set of values for the variables that satisfy all equations simultaneously. Graphically, this represents the point(s) of intersection between the lines represented by each equation.

Systems of linear equations can have:

• One unique solution: The lines intersect at a single point.
• Infinitely many solutions: The lines are coincident (they are the same line).
• No solution: The lines are parallel (they never intersect).

The Substitution Method: A Step-by-Step Guide

The substitution method involves solving one equation for one variable and substituting that expression into the other equation. This eliminates one variable, allowing you to solve for the remaining variable. Here's a breakdown of the steps:

1. Solve one equation for one variable: Choose one equation and solve it for one of the variables. This is often easiest if one of the coefficients is 1 or -1. Isolate the chosen variable on one side of the equation.

2. Substitute the expression into the other equation: Substitute the expression you found in Step 1 into the other equation. This replaces the chosen variable with its equivalent expression, resulting in an equation with only one variable.

3. Solve the resulting equation: Solve the equation from Step 2 for the remaining variable.

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 equation from Step 1) to solve for the other variable.

5. Check your solution: Substitute both values into both original equations to verify that they satisfy both equations simultaneously.

Illustrative Examples

Let's work through a few examples to solidify our understanding.

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: Simplify and solve for y: 5 - 2y = 1 => -2y = -4 => y = 2

4. Substitute back: Substitute y = 2 into either original equation. Using the first equation: x + 2 = 5 => x = 3

5. Check: Substitute x = 3 and y = 2 into both equations: 3 + 2 = 5 (True) 3 - 2 = 1 (True)

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

Example 2:

Solve the system:

2x + y = 7 x - 3y = 4

Solution:

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

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

3. Solve: 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 = -3/7 + 28/7 = 25/7

5. Check: Substitute x = 25/7 and y = -1/7 into both equations (this step is crucial to ensure accuracy). The solution is x = 25/7, y = -1/7.

Example 3: A System with No Solution

Consider the system:

x + y = 3 x + y = 5

Notice that the left-hand sides of both equations are identical, but the right-hand sides are different. This indicates parallel lines, resulting in no solution. Trying the substitution method will lead to a contradiction.

Example 4: A System with Infinitely Many Solutions

Consider the system:

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

The second equation is simply a multiple of the first equation. This indicates that the lines are coincident. The substitution method will lead to an identity (e.g., 0 = 0), indicating infinitely many solutions. Any point on the line x + y = 3 satisfies the system.

Solving Systems of Linear Equations by Substitution Worksheet: Problems and Answers

Here's a worksheet with practice problems, followed by detailed solutions.

Worksheet Problems:

1. x + y = 8; x - y = 2
2. 2x + y = 5; x - 2y = 10
3. 3x - y = 7; x + 2y = 4
4. x + 2y = 1; 3x + 6y = 3
5. 2x - y = 4; 4x - 2y = 8
6. x + 3y = 5; 2x - y = 3
7. 4x - 2y = 6; 2x - y = 3
8. x + y = 0; 2x + 3y = 5
9. 3x + y = 11; x - 2y = -2
10. 5x - 2y = 19; x + y = 1

Worksheet Answers:

1. x = 5, y = 3 (Solve the first equation for x: x = 8 - y; substitute into the second equation: (8 - y) - y = 2; solve for y: y = 3; substitute y = 3 back into x = 8 - y to get x = 5.)

2. x = 4, y = -3 (Solve the first equation for y: y = 5 - 2x; substitute into the second equation: x - 2(5 - 2x) = 10; solve for x: x = 4; substitute x = 4 back into y = 5 - 2x to get y = -3.)

3. x = 2, y = -1 (Solve the second equation for x: x = 4 - 2y; substitute into the first equation: 3(4 - 2y) - y = 7; solve for y: y = -1; substitute y = -1 back into x = 4 - 2y to get x = 2.)

4. Infinitely many solutions. (The second equation is a multiple of the first equation. Any point on the line x + 2y = 1 is a solution.)

5. Infinitely many solutions. (The second equation is a multiple of the first equation.)

6. x = 2, y = 1 (Solve the first equation for x: x = 5 - 3y; substitute into the second equation: 2(5 - 3y) - y = 3; solve for y: y = 1; substitute y = 1 back into x = 5 - 3y to get x = 2.)

7. Infinitely many solutions. (The second equation is a multiple of the first equation.)

8. x = -5, y = 5 (Solve the first equation for x: x = -y; substitute into the second equation: 2(-y) + 3y = 5; solve for y: y = 5; substitute y = 5 back into x = -y to get x = -5.)

9. x = 3, y = 2 (Solve the second equation for x: x = 2y - 2; substitute into the first equation: 3(2y - 2) + y = 11; solve for y: y = 2; substitute y = 2 back into x = 2y - 2 to get x = 2.)

10. x = 3, y = -2 (Solve the second equation for y: y = 1 - x; substitute into the first equation: 5x - 2(1 - x) = 19; solve for x: x = 3; substitute x = 3 back into y = 1 - x to get y = -2.)

Frequently Asked Questions (FAQ)

Q: What if I get a contradiction when solving?

A: A contradiction (e.g., 2 = 5) means the system has no solution. The lines representing the equations are parallel.

Q: What if I get an identity when solving?

A: An identity (e.g., 0 = 0) means the system has infinitely many solutions. The lines are coincident.

Q: Which variable should I solve for first?

A: Choose the variable that is easiest to isolate. Look for equations where a coefficient is 1 or -1 to simplify the process.

Q: Can I use substitution with more than two equations?

A: Yes, but it becomes more complex. You'll need to solve for one variable in one equation and substitute it into the others, systematically eliminating variables until you find a solution.

Conclusion

The substitution method is a powerful and versatile technique for solving systems of linear equations. By following the steps outlined above and practicing with various examples, you can confidently solve a wide range of problems. Remember to always check your solutions by substituting them back into the original equations. This not only verifies your answer but also strengthens your understanding of the underlying concepts. Mastering this method is a significant step toward a deeper understanding of algebra and its applications. Consistent practice is key to building fluency and confidence in this essential algebraic skill.