Algebra 1
Ch 7.5 – Special Linear Systems
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.
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

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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


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

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 #1
2x + 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.
Equation # 2
2x + y = 1
-2x
-2x
y = -2x + 1
Equation #1
2x + y = 5
2x + (-2x + 1) = 5
False
1=5
1≠5
In this example, the variable was eliminated and the
resulting statement is false. Therefore, there is no solution
to this system of linear equations.
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
-4x + 2y = 6
Equation #1
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.
Equation #1
-2x + y = 3
(-2) (-2x + y = 3 )
4x - 2y = 6
Combination
4x - 2y = 6
-4x + 2y = 6
0=0
True
In this example, the variables were eliminated and the
resulting statement is true. Therefore, there are many
solutions to this system of linear equations.
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
-6x + 2y = 4
2x + y = 7
-x + y = 7
-4x + y = -8
2.
3.
4.
5.
and
and
and
and
and
-6x – 3y = -15
-9x + 3y = 12
3x – y = -2
2x – 2y = -18
-12x + 3y = -24
Your Turn
6.
-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 Solutions
1.
2.
3.
4.
5.
Many solutions
No solutions
1 solution
No solution
Many solutions
1 solution (5, 12)
7. No solution
8. Many solutions
9. 1 solution (0, -4)
10. 1 solution (2/3, 0)
6.