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Section 8.1
Solving Systems of
Linear Equations by
Graphing
OBJECTIVES
A
Solve a system of two
equations in two
variables by graphing.
OBJECTIVES
B
Determine whether a
system of equations is
consistent, inconsistent, or
dependent.
OBJECTIVES
C
Solve an application.
SOLVING SYSTEMS OF
SIMULTANEOUS
EQUATIONS
There are three possible
solutions.
POSSIBLE SOLUTION 1
Consistent and independent
equations:
The graphs of the equations
intersect at one point, whose
coordinates give the solution
of the system.
POSSIBLE SOLUTION 2
Inconsistent equations:
The graphs of the equations
are parallel lines; there is no
solution for the system
POSSIBLE SOLUTION 3
Dependent equations:
The graphs of the equations
coincide (are the same).
There are infinitely many
solutions for the system.
COMPARISONS
Intersecting Lines
Have different slopes
COMPARISONS
Intersecting Lines
Have one solution
COMPARISONS
Intersecting Lines
Form a consistent system
COMPARISONS
Parallel Lines
Have the same slopes
COMPARISONS
Parallel Lines
Have different y-intercepts
COMPARISONS
Parallel Lines
Have no solution
COMPARISONS
Parallel Lines
Form inconsistent systems
COMPARISONS
Coinciding Lines
Have the same slope
COMPARISONS
Coinciding Lines
Have the same y-intercept
COMPARISONS
Coinciding Lines
Have infinite solutions
COMPARISONS
Coinciding Lines
Form a dependent system
A HELPFUL HINT
For y = m1x + b1
y = m2 x + b2
When m1  m2, the lines intersect.
When m1 = m2 and b1 = b2,
there is only one line.
A HELPFUL HINT
For y = m1x + b1
y = m2 x + b2
When m1 = m2 and b1  b2,
the lines are parallel.
Chapter 8
Solving Systems of Linear
Equations and Inequalities
Section 8.1
Chapter 8
Solving Systems of Linear
Equations and Inequalities
Section 8.1
Exercise #1
Use the graphical method to solve the system.
x + 2y = 4
2y – x = 0
y
5
–4
6
–5
x
Use the graphical method to solve the system.
x
y
x + 2y = 4
0
2
2y – x = 0
4
0
y
5
–4
6
–5
x
Use the graphical method to solve the system.
x
y
x + 2y = 4
0
0
2y – x = 0
4
2
y
5
(2, 1)
–4
6
–5
x
Chapter 8
Solving Systems of Linear
Equations and Inequalities
Section 8.1
Exercise #2
Use the graphical method to solve the system.
y – 2x = 2
2y – 4x = 8
y
6
–5
5
–4
x
Use the graphical method to solve the system.
x
y
y – 2x = 2
0
2
2y – 4x = 8
–1 0
y
6
–5
5
–4
x
Use the graphical method to solve the system.
x
y
y – 2x = 2
0
4
2y – 4x = 8
–2 0
y
6
Inconsistent:
No Solution.
–5
5
–4
x
Section 8.2
Solving Systems of Equations
by Substitution
OBJECTIVES
A
Solve a system of
equations in two
variables.
OBJECTIVES
B
Determine whether a
system of equations is
consistent, inconsistent, or
dependent.
OBJECTIVES
C
Solve an application.
PROCEDURE:
Solving a system of equations
by the Substitution Method.
1. Solve one of the
equations for x or y.
PROCEDURE:
Solving a system of equations
by the Substitution Method.
2. Substitute the resulting
expression into the other
equation.
PROCEDURE:
Solving a system of equations
by the Substitution Method.
3. Solve the new equation for
the variable.
PROCEDURE:
Solving a system of equations
by the Substitution Method.
4. Substitute the value of the
variable and solve to get
the value for the second
variable.
PROCEDURE:
Solving a system of equations
by the Substitution Method.
5. Check the solution by
substituting the numerical
values of the variables in
both equations.
Chapter 8
Solving Systems of Linear
Equations and Inequalities
Section 8.2
Exercise #3
Use the method of substitution to solve the
system (if possible).
x = 6 – 3y
x + 3y = 6
2x + 6y = 8
2  6 – 3 y  + 6y = 8
12 – 6y + 6y = 8
12 = 8
12  8
No Solution (inconsistent)
Chapter 8
Solving Systems of Linear
Equations and Inequalities
Section 8.2
Exercise #4
Use the method of substitution to solve the
system (if possible).
x = 6 – 3y
x + 3y = 6
3x + 9y = 18
3  6 – 3 y  + 9y = 18
18 – 9y + 9y = 18
18 = 18 identity
Dependent (infinitely many solutions).
Section 8.3
Solving Systems of
Equations by Elimination
OBJECTIVES
A
Solve a system of
equations in two
variables.
OBJECTIVES
B
Determine whether a
system of equations is
consistent, inconsistent, or
dependent.
OBJECTIVES
C
Solve an application.
ELIMINATION METHOD
One or both equations in a
system of simultaneous
equations can be multiplied
(or divided) by any nonzero
number to obtain an
equivalent system.
ELIMINATION METHOD
In the equivalent system, the
coefficients of x (or y) are
opposites, thus eliminating x
or y when the equations are
added.
Chapter 8
Solving Systems of Linear
Equations and Inequalities
Section 8.3
Exercise #5
Solve the system (if possible).
2x + 3y = – 8 Multiply by –1
3x + y = – 5 Multiply by 3
9x + 3y = – 15
+ –2x – 3y = 8
7x = – 7
x = –1
Solve the system (if possible).
2x + 3y = – 8
3x + y = – 5
3  –1 + y = – 5
–3 + y = – 5
y = –2
(–1, –2)
x = –1
Chapter 8
Solving Systems of Linear
Equations and Inequalities
Section 8.3
Exercise #6
Solve the system (if possible).
3x – 2y = 6
Multiply by 2
– 6x + 4y = – 2
+ 6x – 4y = 12
0 = 10
0 ° 10
No solution (inconsistent)
Section 8.4
Coin, General, Motion, and
Investment Problems
OBJECTIVES
A
Solve word problems
involving coins.
B
Solve word problems of a
general nature.
OBJECTIVES
C
Solve word problems
using the distance
formula: D = RT.
OBJECTIVES
D
Solve word problems
involving the interest
formula: I = Pr
Chapter 8
Solving Systems of Linear
Equations and Inequalities
Section 8.4
Exercise #8
Eva has $2 in nickels and dimes. She has twice as many
dimes as nickels. How many nickels and how many
dimes does she have?
Let n = number of nickels.
d = number of dimes.
0.05n + 0.10d = 2.00
d = 2n
Substitute 2n for d
0.05 n + 0.10  2n  = 2.00
0.05n + 0.20n = 2.00 Multiply by 100
Eva has $2 in nickels and dimes. She has twice as many
dimes as nickels. How many nickels and how many
dimes does she have?
5n + 20n = 200
25n = 200 Divide by 25
n=8
d = 2n Substitute 8 for n
d = 2 8
d = 16
She has 8 nickels and 16 dimes.
Chapter 8
Solving Systems of Linear
Equations and Inequalities
Section 8.4
Exercise #9
The sum of two numbers is 140. Their difference is 90.
What are the numbers?
Let x = the larger number.
y = the smaller number.
x + y = 140
+ x – y = 90
2x = 230 Divide by 2
x = 115
115 + y = 140
y = 25
The numbers are
115 and 25.
Chapter 8
Solving Systems of Linear
Equations and Inequalities
Section 8.4
Exercise #10
A plane flies 600 miles with a tailwind in 2 hours. It takes
the same plane 3 hours to fly the 600 miles
when flying against the wind. What is the
plane’s speed in still air?
Let r = plane's speed in still air.
w = wind's speed.
r + w = speed with tail wind.
r – w = speed against wind.


2 r + w = 600 Divide by 2


3 r – w = 600
A plane flies 600 miles with a tailwind in 2 hours. It takes
the same plane 3 hours to fly the 600 miles
when flying against the wind. What is the
plane’s speed in still air?
Let r = plane's speed in still air.
w = wind's speed.
r + w = speed with tail wind.
r – w = speed against wind.
r + w = 300


3 r – w = 600 Divide by 3
A plane flies 600 miles with a tailwind in 2 hours. It takes
the same plane 3 hours to fly the 600 miles
when flying against the wind. What is the
plane’s speed in still air?
Let r = plane's speed in still air.
w = wind's speed.
r + w = speed with tail wind.
r – w = speed against wind.
r + w = 300 
 add
+ r – w = 200 
2r = 500
r = 250
The plane’s speed in still air is 250 mi/hr.
Let r = plane's speed in still air.
w = wind's speed.
r + w = speed with tail wind.
r – w = speed against wind.
r + w = 300 
 add
+ r – w = 200 
2r = 500
r = 250
Chapter 8
Solving Systems of Linear
Equations and Inequalities
Section 8.4
Exercise #11
Herbert invests $10,000, part at 5% and part at 6%.
How much money is invested at each rate if his
annual interest is $568?
Let x = amount invested at 5%; 0.05x = interest
y = amount invested at 6%; 0.06y = interest
x + y = 10,000 Multiply by –5
0.05x + 0.06y = 568
Herbert invests $10,000, part at 5% and part at 6%.
How much money is invested at each rate if his
annual interest is $568?
Let x = amount invested at 5%; 0.05x = interest
y = amount invested at 6%; 0.06y = interest
– 5x – 5y = – 50,000
0.05x + 0.06y = 568 Multiply by 100
Herbert invests $10,000, part at 5% and part at 6%.
How much money is invested at each rate if his
annual interest is $568?
Let x = amount invested at 5%; 0.05x = interest
y = amount invested at 6%; 0.06y = interest
– 5x – 5y = – 50,000
+ 5x + 6y =
56,800
y=
6,800

add
x + 6,800 = 10,000
x = 3,200
$3,200 is invested at 5% and
$6,800 is invested at 6%.
Let x = amount invested at 5%; 0.05x = interest
y = amount invested at 6%; 0.06y = interest
– 5x – 5y = – 50,000
+ 5x + 6y =
56,800
y=
6,800

add
x + 6,800 = 10,000
x = 3,200
Section 8.5
Systems of Linear
Inequalities
OBJECTIVES
A
Solve a system of linear
inequalities by graphing.
PROCEDURE:
Solving a System of
Inequalities.
Graph each inequality on the
same set of axes using the
following steps.
PROCEDURE:
Solving a System of
Inequalities.
1. Graph the line that is the
boundary of the region.

If the inequality involves or
 draw a solid line; if it involves
< or > draw a dashed line.
PROCEDURE:
Solving a System of
Inequalities.
2. Use any point (a,b) not on
the line as a test point.
Substitute the values of a
and b for x and y.
PROCEDURE:
Solving a System of
Inequalities.
If a true statement results,
shade the side of the line
containing the test point.
PROCEDURE:
Solving a System of
Inequalities.
If a false statement results,
shade the other side.
PROCEDURE:
Solving a System of
Inequalities.
The solution set is the set of
points that satisfies all the
inequalities in the system.
Chapter 8
Solving Systems of Linear
Equations and Inequalities
Section 8.5
Exercise #12
Graph the solution set of the system.
x – 2y > 4 and 2x – y  6
y
3
–3
7
–7
x
Graph the solution set of the system.
x – 2y > 4
x – 2y = 4
Let x = 4, y = 0
y
Let x = 0, y = – 2
3
– 3 (0, –2)
(4, 0)
–7
7
x
Graph the solution set of the system.
2x – y  6
2x – y = 6
Let x = 3, y = 0
y
Let x = 0, y = – 6
3
(3, 0)
– 3 (0, –2)
(4, 0)
(0, –6)
–7
7
x
Graph the solution set of the system.
x – 2y > 4 and 2x – y  6
y
3
(3, 0)
– 3 (0, –2)
(4, 0)
(0, –6)
–7
7
x