6.1 Solving Systems by Graphing:
System of Linear Equations: Two or more
linear equations
Solution of a System of Linear Equations:
Any ordered pair that makes all the
equations in a system true.
Trend line: Line on a scatter plot, drawn
near the points, that shows a correlation
Consistent: System of equations that has
at least one solution.
1) Could have the same
or different slope but
they intersect.
2) The point where
they meet is a solution
Consistent Independent: System of
equations that has EXACTLY one solution.
1) Have different slopes
2) Only intersect once
3) The point of
intersection is the
solution.
Consistent Dependent: System of
equations that has infinitely many
solutions.
1) Have same slopes
2) Same y-intercepts
3) Each point is a
solution.
Inconsistent: System of equations that
has no solutions.
1) Have same slopes
2) different y-intercepts
3) No solutions
Remember:
Remember:
GOAL:
SOLVING A SYSTEM BY GRAPHING:
To solve a system by graphing we must:
1) Write the equations in slope-intercept
form (y=mx+b)
2) Graph the equations
3) Find the point of intersection
4) Check
Ex:
What is the solution of the system?
Use a graph to check your answer.
 2 x  y  4

y  x2
SOLUTION:
1) Write the equations in slope-intercept
form (y=mx+b)
2 x 
y
y  4
 x2
 y  2x  4
 y  x2
SOLUTION: 2) Graph the equations
y  x2
 y  2x  4
SOLUTION: 3) Find the solution
 y  2x  4  y  x  2
Looking at the
graph, we see
that these two
equations
intersect at the
point : (-2, 0)
SOLUTION: 4) Check
We know that (-2,0) is the solution
from our graph.
 y  2x  4
 0  2(  2 )  4
y
 x2
0 
0 
4  4
0 
2  2
0 
0
0
TRUE
TRUE
YOU TRY IT:
What is the solution of the system?
Use a graph to check your answer.
 x  y  2

 y  2  3x
SOLUTION:
1) Write the equations in slope-intercept
form (y=mx+b)
 x 
y  2
y2
3x
 y  x2
 y  3x  2
SOLUTION: 2) Graph the equations
 y  3x  2
y  x2
SOLUTION: 3) Find the solution
y  x  2  y  3x  2
Looking at the
graph, we see
that these two
equations
intersect at the
point : (2,4)
SOLUTION: 4) Check
We know that (2,4) is the solution
from our graph.
y  x2
4  2  2
y
 3x  2
4 
4 
4
TRUE
3( 2 )  2
4 
62
4  4
TRUE
SYSTEM WITH INFINITELY MANY
SOLUTIONS: Using the same procedure
we can see that sometimes the system
will give us infinitely many solutions
(any point will make the equations
true).
Ex: What is the solution to the system?
Use a graph.
2 y  x  2


1
y  x1

2
SOLUTION:
1) Write the equations in slope-intercept
form (y=mx+b)
2 y 
x  2 2y  x 2 y 
1

 y  2  3x  y  x  1
2

1
2
x1
SOLUTION: 2) Graph the equations
1

y  x1
2

1

y  x1
2

Notice: Every
point of one line
is on the other.
SYSTEM WITH NO SOLUTIONS:
Using the same procedure we can
see that sometimes the system will give
us infinitely many solutions (any point
will make the equations true).
Ex: What is the solution to the system?
Use a graph.
−𝟐𝒙 + 𝒚 = 𝟐
𝟐𝒙 − 𝒚 = 𝟏
SOLUTION:
1) Write the equations in slope-intercept
form (y=mx+b)
−𝟐𝒙 + 𝒚 = 𝟐 → 𝐲 = 𝟐𝐱 + 𝟐
𝟐𝒙 − 𝒚 = 𝟏 → −𝒚 = −𝟐𝒙 + 𝟏
𝒚 = 𝟐𝒙 − 𝟏
SOLUTION: 2) Graph the equations
𝒚 = 𝟐𝒙 + 𝟐
𝒚 = 𝟐𝒙 − 𝟏
Notice: These
lines will never
intersect. NO
SOLUTIONS.
WRITING A SYSTEM OF EQUATIONS:
Putting ourselves in the real world,
we must be able to solve problems
using systems of equations.
Ex:
One satellite radio service charges
$10.00 per month plus an activation fee
of $20.00. A second service charges $11
per month plus an activation fee of $15.
For what number of months is the cost
of either service the same?
SOLUTION: Looking at the data we
must be able to do 5 things:
1) Relate- Put the problem in simple
terms.
Cost = service charge + monthly dues
2) Define- Use variables to represent
change:
Let C = total Cost
Let x = time in months
SOLUTION: (continue)
3) Write- Create two equations to
represent the events.
Satellite 1: C = $10 x + $20
Satellite 2: C = $11 x + $15
4) Graph the equations: Remember to
put them in slope/intercept form (y =
mx + b)
The two equations are already in
y=mx+b form.
SOLUTION: 4) Continue
𝒚 = 𝟏𝟎𝒙 + 𝟐𝟎
𝒚 = 𝟏𝟏𝒙 + 𝟏𝟓
100
Cost
90
80
70
60
50
40
30
20
10
1
2
3
4
5
6
Months
SOLUTION:
5) Interpret the solution.
Cost
Notice:
100
90
These lines 80
70
intersect at 60
𝒚 = 𝟏𝟎𝒙 + 𝟐𝟎
50
at (5, 70).
40
𝒚 = 𝟏𝟏𝒙 + 𝟏𝟓
30
This means 20
10
that the two
1 2 3 4 5 6
satellite services
Months
will cost the same in 5 months and $70.
YOU TRY IT:
Scientists studied the weights of
two alligators over a period of 12
months. The initial weight and growth
rate of each alligator are shown below.
After how many months did the two
alligators weight the same?
SOLUTION: Looking at the data, Here
are the 5 things we must do:
1) Relate- Put the problem in simple
terms.
Total Weight = initial weight + growth
per month.
2) Define- Use variables to represent
change:
Let W = Total weight
Let x = time in months
SOLUTION: (continue)
3) Write- Create two equations to
represent the events.
Alligator 1: W = 1.5x + 4
Alligator 2: W= 1.0 x + 6
4) Graph the equations: Remember to
put them in slope/intercept form (y =
mx + b)
The two equations are already in
y=mx+b form.
SOLUTION: 4) Continue
W = 𝟏. 𝟓𝒙 + 𝟒
𝒚 = 𝟏𝒙 + 𝟔
Weight
10
9
8
7
6
5
4
3
2
1
1
2
3
4
5
6
Months
SOLUTION:
5) Interpret the solution.
Weight
10
Notice:
9
8
These lines
7
intersect at 6
5
at (4, 10)
4
3
This means 2
1
that the two
1 2 3 4 5 6
Alligators will
Months
Weight 10 lbs after 4 months.
VIDEOS:
Solve by Graphing
https://www.khanacademy.org/math/algebra/syst
ems-of-eq-and-ineq/fast-systems-ofequations/v/solving-linear-systems-by-graphing
CLASSWORK:
Page 363-365
Problems: As many as needed
to master the
concept.