Transcript Document 7190472

3.7 - Graphing Linear Inequalities
Graphing Inequalities in Two Variables
Are the ordered pairs a solution to the problem?
1
y  x 1
3
 3, 2
 1,3
1
2   3  1
3
yes
2  2
1
3   1  1
3
4
3 
no
3


3.7 - Graphing Linear Inequalities
Graphing Inequalities in Two Variables
Are the ordered pairs a solution to the problem?
1
y  x 1
3
3,0
 4, 3
1
0   3  1
3
0  0 yes
1
3   4   1
3
1
3 
no
3




3.7 - Graphing Linear Inequalities
Graphing Inequalities in Two Variables
Are the ordered pairs a solution to the problem?
1
y  x 1
3
 3, 2
 1,3
3,0
 4, 3
2  2 yes
4
3
no
3
yes
00
1
3 
no
3

no
yes
no


.
3.7 - Graphing LinearInequalities
Graphing Inequalities in Two Variables
Graph the solution.
1
y  x 1
4

1, 4
3
4  
4
no
 2, 2 
1
2
2
yes

3.7 - Graphing Linear Inequalities
Graphing Inequalities in Two Variables
Graph the solution.
3 x  2 y  10
3
y   x5
2
 4,3
3  1
no
 0, 0 
05
yes


3.7 - Graphing Linear Inequalities
Graphing Inequalities in Two Variables
Graph the solution.
y  3x

 2, 2
2  6
no
3,3
39
yes

3.7 - Graphing Linear Inequalities
Graphing Inequalities in Two Variables
Graph the solution.
 3 x  5 y  15  0
3
y  x3
5

 0, 0 
0  3
yes
 2, 3
4
3 4
5
no

4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Graphing
A system of linear equations allows the relationship between two or
more linear equations to be compared and analyzed.
3 x  3 y  0

 x  2y
 y  7 x 1

 y4
 x y 0

2 x  y  10
7

 y  9 x2

2

y   x  4
3

 y  2x


4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Graphing
Determine whether (3, 9) is a solution of the following system.
5 x  2 y  3

 y  3x
5  3  2  9  3
9  3  3
15 18  3
3  3
99
Both statements are true, therefore (3, 9) is a solution
to the given system of linear equations.
4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Graphing
Determine whether (3, -2) is a solution of the following system.
2 x  y  8

x  3y  4
2  3   2  8
3  3 2  4
62 8
88
36  4
3  4
Both statements are not true, therefore (3, -2) is not a
solution to the given system of linear equations.
4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Graphing
 y  2x  4

1

 y   3 x  3
Solution :  3, 2
2  2  3  4
22
1
2    3  3
3
22

4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Graphing
 y3

 y  4 x  1

Solution :  1,3
33
3  4  1  1
33
4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Graphing
 x y 3

2 x  y  0
y  x 3

 y  2 x
Solution : 1, 2

1   2   3
2 1   2   0
33
00
4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Graphing
Special Systems of Linear Equations
Consistent system has at least one solution.
Inconsistent system has no solution.
Independent equations have different graphs.
Dependent equations have identical graphs.
Consistent system
Independent equations

4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Graphing
Special Systems of Linear Equations
Consistent system has at least one solution.
Inconsistent system has no solution.
Independent equations have different graphs.
Dependent equations have identical graphs.
Inconsistent system
Independent equations
4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Graphing
Special Systems of Linear Equations
Consistent system has at least one solution.
Inconsistent system has no solution.
Independent equations have different graphs.
Dependent equations have identical graphs.
Consistent system
Dependent equations
4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Substitution
5 x  2 y  3

 y  3x
5 x  2 y  3
5x  23x   3
5x  6 x  3
 x  3
x3
y  3x
y  33
y9
Solution
3,9
4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Substitution
2x  y  8
2x  y  8
2 x  y  8

2 3 y  4  y  8
2
x

0

8
x

3
y

4

 6y 8  y  8
2x  8
x  3y  4
7y 8  8
x4
7y  0
x  3 y  4
Solution
y0
4,0
4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Elimination
7 5  39
8 2  33
10  6  36  20
27  6  50  29
15  3  6  12
17 12  14  9
4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Elimination
 3x  y  1

4 x  y  6
3x  y  1
4x  y  6
7
7x
7x  7
x 1
3x  y  1
31  y  1
3 y 1
 y  2
y2
Solution
1,2
4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Elimination
5 x  4 y  1

10 x  2 y  3
5 x  4 y  1
210 x  2 y  3
5 x  4 y  1
20 x  4 y  6
25x  5
1
x
5
5 x  4 y  1
1
5   4 y  1
5
1  4 y  1
 4 y  2
1
y
2
Solution
1 1
 , 
5 2
4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Elimination
2 x  5 y  6

3 x  4 y  9
32 x  5 y  6
 23x  4 y  9
2x  5 y  6
2 x  50  6
2x  6
x3
6 x  15 y  18
Solution
 6 x  8 y  18
3,0
7y  0
y0
4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Elimination
Solution
 4 x  7 y  10

x, y  4x  7 y  10
 8 x  14 y  20
or
24 x  7 y  10
x, y   8x  14 y  20
 8 x  14 y  20
8 x  14 y  20
 8 x  14 y  20
00
(lines are the same)
4.1 - Systems of Linear Equations (2 variables)
Solving Systems of Linear Equations by Elimination
3 x  9 y  5

 x  3y  2
3x  9 y  5
 3x  3 y  2
3x  9 y  5
 3 x  9 y  6
0  1
No
Solution
(lines are parallel)
4.3 - Systems of Linear Equations and Problem
Solving
The consumption of red meat and poultry are defined by the given
equations, where x represents the number of years since 2003 and y
represents the pounds per year consumed. In what year will the
consumption be the same?
red meat :
poultry :
y  0.05 x  111.8
y  0.75 x  71
Substitution Method
0.75x  71  0.05x  111.8
 0.05x
 0.05x
0.70x  71  111.8
 71  71
0.7 x  40.8
4.3 - Systems of Linear Equations and Problem
Solving
The consumption of red meat and poultry are defined by the given
equations, where x represents the number of years since 2003 and y
represents the pounds per year consumed. In what year will the
consumption be the same?
red meat :
poultry :
y  0.05 x  111.8
y  0.75 x  71
Substitution Method
0.7 x  40.8
0.7
0.7
x  58.3 years
2003  58.3  2061
4.3 - Systems of Linear Equations and Problem
Solving
A first number is seven greater than a second number. Twice the
first number is four more than three times the second number.
What are the numbers? 1st number is x, 2nd number is y
x  y7
2x  3y  4
Substitution Method
2 y  7  3 y  4
2 y  14  3 y  4
 2y  4  2y 4
10  y
x  y7
x  10  7
x  17
Solution
17,10
4.3 - Systems of Linear Equations and Problem
Solving
Two trains leave Tulsa, one traveling north and the other south.
After four hours, they are 376 miles apart. If one train is traveling
ten miles per hour faster than the other, what is the speed of each
train?
Train
Rate
Time
Distance
North
x
y
4
4
4x
4y
South
4 x  4 y  376
x  y  10
Substitution Method
4 y  10  4 y  376
4 y  40  4 y  376
8 y  40  376
8 y  336
y  42 mph
x  y  10
x  42  10
x  52 mph
4.3 - Systems of Linear Equations and Problem
Solving
One solution contains 20% acid and a second solution contains 50%
acid. How many ounces of each solution should be mixed in order to
have sixty ounces of a 30% solution?
Solution
Ounces
Decimal
Pure Acid
20%
x
y
60
0.2
0.5
0.3
0.2x
0.5y
(60)(0.3)
50%
30%
0.2 x  0.5 y  600.3
0.2 x  0.5 y  18
x  y  60
4.3 - Systems of Linear Equations and Problem
Solving
One solution contains 20% acid and a second solutions contains 50%
acid. How many ounces of each solution should be mixed in order to
have sixty ounces of a 30% solution?
2 x  5 y  180
x  y  60
Elimination Method
2 x  5 y  180
 2x  y  60
2 x  5 y  180
 2 x  2 y  120
3 y  60
3 y  60
y  20
x  y  60
x  20  60
x  40
40 ounces of the 20% solution
20 ounces of the 50% solution
4.5 – Systems of Linear Inequalities
Graphing Inequalities in Two Variables
Graph the Union.
2

 y  x 1

3

 y  x  2
4.5 – Systems of Linear Inequalities
Graphing Inequalities in Two Variables
Graph the solution (Graph the intersection).
2

 y  x 1

3

 y  x  2
4.5 – Systems of Linear Inequalities
Graphing Inequalities in Two Variables
Graph the union.
1

y   x  2

2

 x  2
4.5 – Systems of Linear Inequalities
Graphing Inequalities in Two Variables
Graph the solution. (Graph the intersection)
1

y   x  2

2

 x  2
4.5 – Systems of Linear Inequalities
Graphing Inequalities in Two Variables
Graph the solution. (Graph the intersection)
 y  2x  3  0

  x  3y  9
 y  2x  3

y  1 x3

3
