Algebra 1

Ch 7.5 – Special Linear Systems

Objective Students will identify linear systems with

one solution, no solution or many solutions.

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…

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

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

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

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…

Example #1 Using the substitution method we will

solve the following linear system and interpret the results

2x + y = 5 Equation #12x + y = 1 Equation #2

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 -2x

y = -2x + 1

Equation #12x + 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

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.

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

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

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

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!

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…

Your Turn Solve the linear systems (use any

method). State the number of solutions.

1. 2x + y = 5 and -6x – 3y = -152. -6x + 2y = 4 and -9x + 3y = 123. 2x + y = 7 and 3x – y = -2 4. -x + y = 7 and 2x – 2y = -185. -4x + y = -8 and -12x + 3y = -24

Your Turn6. -4x + y = -8 and 2x – 2y = -14

7. -7x + 7y = 7 and 2x – 2y = -18

8. 4x + 4y = -8 and 2x + 2y = -4

9. 2x + y = -4 and 4x – 2y = 8

10. 6x – 2y = 4 and -4x + 2y = -8/3

Your Turn Solutions1. Many solutions2. No solutions3. 1 solution4. No solution5. Many solutions

6. 1 solution (5, 12)7. No solution8. Many solutions9. 1 solution (0, -4)10. 1 solution (2/3, 0)

Summary

A key tool in making learning effective is being able to summarize what you learned in a lesson in your own words…

In this lesson we talked about special types of linear systems. Therefore, in your own words summarize this lesson…be sure to include key concepts that the lesson covered as well as any points that are still not clear to you…

I will give you credit for doing this lesson…please see the next slide…

Credit I will add 25 points as an assignment grade for you working

on this lesson… To receive the full 25 points you must do the following:

Have your name, date and period as well a lesson number as a heading.

Do each of the your turn problems showing all work Have a 1 paragraph summary of the lesson in your own

words Please be advised – I will not give any credit for work

submitted: Without a complete heading Without showing work for the your turn problems Without a summary in your own words…