System Of Equations Examples With Answers

Systems of Equations: Examples with Answers and Solutions

Solving systems of equations is a fundamental concept in algebra with wide-ranging applications in various fields, including science, engineering, economics, and computer science. This comprehensive guide will delve into different methods for solving systems of equations, providing numerous examples with detailed answers and explanations. We'll cover linear systems, both with two and three variables, and touch upon non-linear systems. Mastering these techniques is crucial for success in higher-level mathematics and related disciplines.

Understanding Systems of Equations

A system of equations is a collection 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 solution or solutions to the system. The number of solutions can vary depending on the nature of the equations.

Types of Systems

• Linear Systems: These involve equations where the variables are raised to the power of one (no exponents). They represent straight lines when graphed. Linear systems can have one solution (intersecting lines), infinitely many solutions (coincident lines), or no solution (parallel lines).

• Non-Linear Systems: These involve equations where at least one variable has an exponent other than one (e.g., quadratic equations, cubic equations). They represent curves when graphed. Non-linear systems can have multiple solutions.

Solving Linear Systems of Equations

We'll explore three primary methods for solving linear systems:

1. Graphing Method

This method involves graphing each equation on a coordinate plane. The point(s) where the lines intersect represent the solution(s) to the system. This method is visually intuitive but can be less accurate for complex systems or solutions involving fractions or decimals.

Example: Solve the system:

x + y = 5 x - y = 1

Solution: Graphing these two equations reveals an intersection point at (3, 2). Therefore, the solution is x = 3 and y = 2. You can verify this by substituting these values into both equations.

2. Substitution Method

This algebraic 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. Then, substitute the found value back into either original equation to solve for the other variable.

Example: Solve the system:

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

Solution:

1. Solve for x in the second equation: x = 3y + 4
2. Substitute this expression for x into the first equation: 2(3y + 4) + y = 7
3. Solve for y: 6y + 8 + y = 7 => 7y = -1 => y = -1/7
4. Substitute y = -1/7 back into either original equation to solve for x: x - 3(-1/7) = 4 => x = 4 - 3/7 = 25/7

Therefore, the solution is x = 25/7 and y = -1/7.

3. Elimination Method (Addition Method)

This method involves manipulating the equations (multiplying by constants) so that when you add the equations together, one variable is eliminated. This allows you to solve for the remaining variable. Then, substitute the found value back into either original equation to solve for the other variable.

Example: Solve the system:

3x + 2y = 11 x - 2y = 1

Solution: Notice that the 'y' terms have opposite coefficients. Adding the two equations directly eliminates 'y':

(3x + 2y) + (x - 2y) = 11 + 1 4x = 12 x = 3

Substitute x = 3 into either original equation to solve for y:

3(3) + 2y = 11 9 + 2y = 11 2y = 2 y = 1

Therefore, the solution is x = 3 and y = 1.

Solving Linear Systems with Three Variables

Linear systems with three variables require a more systematic approach. We can extend the elimination method to solve these systems. The goal is to eliminate one variable at a time, reducing the system to a simpler two-variable system that can be solved using the methods described above.

Example: Solve the system:

x + y + z = 6 2x - y + z = 3 x + 2y - z = 3

Solution:

1. Eliminate z: Add the first and third equations: 2x + 3y = 9
2. Eliminate z again: Add the first and second equations: 3x + 2z = 9
3. Solve the two-variable system: Now we have two equations with two variables (x and y): 2x + 3y = 9 3x + 2z = 9 (Note: This should have been 3x + z = 9. There was an error in the original elimination step. Correcting this will lead to the correct solution)

Let's correct the step and proceed:

1. Eliminate z: Add the first and third equations: 2x + 3y = 9
2. Eliminate z again: Subtract the third equation from the second equation: x - 3y = 0 => x = 3y
3. Substitute x = 3y into 2x + 3y = 9: 2(3y) + 3y = 9 => 9y = 9 => y = 1
4. Substitute y = 1 into x = 3y: x = 3(1) = 3
5. Substitute x = 3 and y = 1 into any original equation to solve for z: 3 + 1 + z = 6 => z = 2

Therefore, the solution is x = 3, y = 1, and z = 2.

Solving Non-Linear Systems of Equations

Solving non-linear systems often involves a combination of substitution and other algebraic techniques. The number of solutions can vary greatly. Graphical methods can be helpful in visualizing the solutions.

Example: Solve the system:

x² + y = 5 x + y = 3

Solution:

1. Solve the linear equation for y: y = 3 - x
2. Substitute this expression for y into the quadratic equation: x² + (3 - x) = 5
3. Solve the quadratic equation: x² - x - 2 = 0 This factors to (x - 2)(x + 1) = 0. Therefore, x = 2 or x = -1.
4. Substitute each value of x back into y = 3 - x to find the corresponding y values: If x = 2, y = 3 - 2 = 1 If x = -1, y = 3 - (-1) = 4

Therefore, the solutions are (2, 1) and (-1, 4).

Conclusion

Mastering the art of solving systems of equations is a critical skill in mathematics and its applications. This guide has provided a detailed overview of various methods, from graphing to substitution and elimination, for both linear and non-linear systems. Remember to always verify your solutions by substituting the values back into the original equations. Practice is key to building proficiency in this area. By understanding these techniques and working through numerous examples, you will gain the confidence to tackle more complex problems and apply these skills in diverse contexts.