1. Algebra 1 Ch 7.5 – Special Linear Systems

2. Objective  I can identify linear systems with one solution, no solution or many solutions.  I can model/show real-life problems using a linear system.

3. Before we begin…  In the last couple of lessons when we solved the liner systems of equations the result was one solution…that is not always the case…  There are instances where the result can be no solution or many solutions…  The goal of this lesson is to be able to solve the system of equations and interpret the results…

4. Linear Systems – 1 Solution  At this point you should be familiar with what a graph of a linear system with 1 solution looks like.  Essentially, it is a graph where the lines intersect. The intersection point is the solution to the system of equations…  Graphically, it looks something like this… y x

5. Linear Systems – No Solution  You can recognize the graph of a linear system with no solution because the lines do not intersect.  In other words, the graph will be of parallel lines.  No point on either line will be the solution to the linear system  It looks something like this… y x

6. Linear Systems – Many Solutions  You can recognize the graph of a linear system with many solutions because the lines will be on top of each other.  It means that all points on the line will be a solution to the linear system  It looks something like this… y x

7. Graphing vs. Algebraic Solutions  When a linear system is graphed it is easy to interpret the results based upon what the graph looks like  As you have already learned, graphing is not the only way to solve linear systems…  It is equally easy to interpret the results when solving a linear system algebraically…  When the variables are eliminated and you are left with a false statement, that means that the system has no solution (regardless of the values of x and y)  When the variables are eliminated and you are left with a true statement, that means the system has many solutions (regardless of the values of x and y)  Let’s look at some examples…

8. Example #1  Using the substitution method we will solve the following linear system and interpret the results 2x + y = 5 Equation #1 2x + y = 1 Equation #2

9. Example #1 (Continued) 2x + y = 5 Equation #1 2x + y = 1 Equation #2 In this example I can solve either equation for y. I choose to solve equation #2 for y and then substitute the resulting expression into equation #1. 2x + y = 1 Equation # 2 -2x y = -2x + 1 Equation #1 2x + y = 5 2x + (-2x + 1) = 5 1 = 5 False In this example, the variable was eliminated and the resulting statement is false. Therefore, there is no solution to this system of linear equations. 1 ≠ 5

10. Something to think about…  Mathematical Reasoning – The previous example uses proof by contradiction. That is you assume something is true, you show that the assumption leads to a contradiction or false statement, and conclude that the opposite of what you assumed is true.

11. Example #2  Using linear combinations we will solve the linear system and then interpret the results. -2x + y = 3 Equation #1 -4x + 2y = 6 Equation #2

12. Example # 2 (Continued) -2x + y = 3 Equation #1 -4x + 2y = 6 Equation #2 After analyzing the equations, I choose to multiply equation #1 by -2. Then add the equations. -2x + y = 3 Equation #1 (-2x + y = 3 )(-2) 4x - 2y = 6 Combination -4x + 2y = 6 0 = 0 In this example, the variables were eliminated and the resulting statement is true. Therefore, there are many solutions to this system of linear equations. True

13. Comments  At this point it is expected that you can solve systems of linear equations using a variety of methods…  It is not enough to be able to mechanically solve the linear systems…you are also expected to be able to interpret the results…  Interpreting the results and/or applying the results to other situations are called higher order thinking skills…  Yes…we want and expect you to be able to think at a higher order!

14. Comments  On the next couple of slides are some practice problems…The answers are on the last slide…  Do the practice and then check your answers…If you do not get the same answer you must question what you did…go back and problem solve to find the error…  If you cannot find the error bring your work to me and I will help…

15. Your Turn  Solve the linear systems (use any method). State the number of solutions. 1. 2x + y = 5and-6x – 3y = -15 2. -6x + 2y = 4and-9x + 3y = 12 3. 2x + y = 7and3x – y = -2 4. -x + y = 7 and2x – 2y = -18 5. -4x + y = -8and-12x + 3y = -24

16. Your Turn 6. -4x + y = -8and2x – 2y = -14 7. -7x + 7y = 7and 2x – 2y = -18 8. 4x + 4y = -8and2x + 2y = -4 9. 2x + y = -4and4x – 2y = 8 10. 6x – 2y = 4and-4x + 2y = -8/3

17. Your Turn Solutions 1. Many solutions 2. No solutions 3. 1 solution 4. No solution 5. Many solutions 6. 1 solution (5, 12) 7. No solution 8. Many solutions 9. 1 solution (0, -4) 10. 1 solution (2/3, 0)