Download Warm-up: Put the following equations to slope

Tuesday September 23, 2014
Warm-up:
Put the following equations to slope-intercept
form then use 2 points to graph
1. 4x - 3y = 8
8 x – 6y = 16
2. 2x + y = 4
2x + y = 1
Tuesday September 23, 2014
Warm-up:
Put the following equations to slope-intercept
form then use 2 points to graph
1. 4x - 3y = 8
8 x – 6y = 16
2. 2x + y = 4
2x + y = 1
Tuesday September 23, 2014
3.1 Solving Linear Systems using Graph
Objective: To solve a system of Linear equations
EQ: How many ways can we solve a system of linear equations
Definition:
• A system of linear equations consists of 2 or more
equations
•
The solution of a system of linear equations can be 1 of
the following 3 cases:
1.
Exactly 1 solution
Example1: When slope (m) and y-intercept (b) are given
Lines intersect at 1 point
Write the equation for the line in(consistent
the graph and independent)
Solution: Use Slope-Intercept form:
According
to the graph,
= -1
2. Infinitely
manybsolutions
and m = (-3 - -1) / (3 – 0) = -2/3
Lines coincide
(consistent and dependent)
Therefore, the equation is y = (-2/3) x - 1
3. No solutions
Lines are parallel
(inconsistent)
Tuesday September 23, 2014
3.1 Solving Linear Systems using Graph
There are 4 ways to solve a system of linear equations
1. Graphing
2. Substitution
3. Elimination
4. Multiplication/Elimination
Steps to solve a linear system using graphing:
1.
2.
3.
4.
Put each equation in slope-intercept form (y = mx+b)
Use y-intercept to plot 1st point and use m to plot 2nd point
Look for solution(s)
Substitute the solution in each equation to check for error
Example: Graph the linear system and estimate the solution
then check the solution algebraically
4x + y
=8
(1)
2x – 3y
= 18
(2)
Solution:
1. Put equations in y = mx + b form
y = -4x + 8
(1)
y = (2/3)x - 6
(2)
2. In (1), m = -4, b = 8
In (2), m = 2/3, b = 6
3. Solution is (3, -4)
4. Check:
Wednesday September 24, 2014
Warm-up:
Wednesday September 24, 2014
3.2Solving Linear Systems by Substitution
Steps to solve a linear system using substitution method:
1.
2.
3.
Solve one of the equation for one of its variables
Substitute the expression from step 1 into the other
equation and solve for the other variable
Substitute the value from step 2 into the revised equation
from step 1 and solve for the other variable
Example: Solve the system using the substitution method
2x + 5y
= -5
(1)
x + 3y
=3
(2)
Solution:
Step 1. Solve (2) for x
x + 3y
=3
x
= -3y + 3
(2)
(3)
Step 2. Substitute (3) into (1)
2x
+ 5y = -5 (1)
2 (-3y +3) + 5 y = -5
-6 y + 6 + 5 y = -5
6y = -5
y = 11
Step 3. Substitute y =11 back to either (1) or (2)
x + 3y
=3
(2)
x + 3 (11) = 3
x + 33
=3
x
= -30
 The solution is (-30, 11)
 Check:
Wednesday September 24, 2014
3.2Solving Linear Systems by Elimination
Steps to solve a linear system using elimination method:
1.
2.
Add or subtract the equations to eliminate one of the
variables. Then solve for the other variable
Substitute the expression from step 1 into either the
original equation and solve for the other variable.
Example: Solve the system using the elimination method
6x - 14y = 20
(1)
6x - 8y
=8
(2)
Solution:
Step 1. Subtract (1) and (2) to eliminate x
6x - 14y = 20
(1)
6x - 8y
=8
(2)
------------------------6y = 12
y =-2
Step 2. Substitute y = -2 into either (1) or (2)
6x - 8y
= 8
(2)
6x - 8 (-2) = 8
6x + 16
= 8
6x = 8 - 16
6x = -8
x = -8/6 = -4/3
The solution is (-4/3, -2)
 Check:
(1) 6(-4/3) - 14 (-2) = 20
-8
+ 28
= 20
Yes
(2) 6(-4/3) – 8 (-2) = 8
-8
+ 16
=8
Yes
Wednesday September 24, 2014
3.2Solving Linear Systems by Multiplication and
Elimination
Steps to solve a linear system using multiplication and
elimination method:
1. Multiply on or both of the equation by a constant to make
the coefficient of one variable in both equations the same
2. Add or subtract the equations to eliminate one of the
variables. Then solve for the other variable
3. Substitute the expression from step 1 into either the
original equation and solve for the other variable.
Example: Solve the system using the elimination method
3x - 7y
= 10
(1)
6x - 8y
=8
(2)
Solution:
Step 1. Multiply (1) with 2
2 (3x - 7y = 10) --- 6 x - 14y = 20
Step 2. Subtract (1) and (2) to eliminate x
6x - 14y = 20
(1)
6x - 8y
=8
(2)
------------------------6y = 12
y =-2
Step 2. Substitute y = -2 into either (1) or (2)
6x - 8y
= 8
(2)
6x - 8 (-2) = 8
6x + 16
= 8
6x = 8 - 16
6x = -8
x = -8/6 = -4/3
The solution is (-4/3, -2)
 Check:
(1) 6(-4/3) - 14 (-2) = 20
(2) 6(-4/3) – 8 (-2) = 8
-8
+ 28
= 20
-8
+ 16
=8
Yes
Yes
Monday September 29, 2014
Warm-up:
Graph and shade the solution area
-x + y > 4
x +y<3
Monday September 29, 2014
Warm-up:
Graph and shade the solution area
-x + y > 4
x +y<3
Monday September 29, 2014
3.3 Graphing Systems of Linear Inequalities
Objective: To graph systems of linear inequalities in slopeintercept form and standard form
EQ: Is graphing a system of linear inequalities any different
than just a linear inequality?
Steps to graph a system of inequalities
1. Write each inequalities in slope-intercept form (y=mx+b) by
solving for y
2. Use b to plot y-intercept then use m to plot the 2nd point
for the 1st inequality
3. Use dashed line for > or <. Use solid line for ≥ or ≤
4. Shade the appropriate region
5. On the same graph, repeat steps 2,3, and 4 to plot and
shade for the 2nd inequality
6. Solution is where the graph is double shaded.
7. Check the solution algebraically
Example1: Graph
y > -2x -5
y≤ x+3
Solution:
Step 1:
Step 2: for line 1, m = -2 and b = -5
for line 2, m = 1 and b = 3
Step 3: line 1 is dashed, line 2 is solid
Step 4: Shade the regions using check points
Monday September 29, 2014
Example1: Graph the system of inequalities
2x + 3y < 6
y ≥ -(2/3)x + 4
Solution:
1. Write in y=mx form
(1)
3y = -2x +6
y = -(2/3) + 2
(2)
y = - (2/3) x + 4
2. In (1), m = -(2/3), b =2
In (2), m =-(2/3), b =4
3. No solution
No intersection
 No solution
Tuesday September 30, 2014
Warm-up:
Solve by elimination:
x + 2y = -1
3x – y = 18
(1)
(2)
Tuesday September 30, 2014
Warm-up:
Solve by elimination:
x + 2y = -1
3x – y = 18
(1)
(2)
Solution:
1. Multiply (1) with 3:
3x + 6y = -3
- 3x - y = 18
--------------------7 y = -21
y = -3
2. Substitute
3x – (-3)
3x + 3
3x
x
y = -3 to (2)
= 18
= 18
= 15
=5
Tuesday September 30, 2014
3.4Solving Linear Systems in Three Variables:
Objective: To solve linear systems with 3 variables
EQ: How many different ways can 3 linear equations in 3
variables intersect?
Steps to solve a linear system using elimination method:
1.
2.
3.
4.
5.
Pick any 2 equations and eliminate 1 variable equation (4)
Pick 2 different equations and eliminate the same variable
equation (5)
Use equations (4) and (5) to solve for 1 of its variables.
Substitute the known variable to either (4) or (5) to find
the other variable.
Substitute the 2 known variables into any of the original
equations to solve for the 3rd variable.
Example: Solve the system:
x+ y+z =3
4x + 4y + 4z = 7
3x - y + 2z = 5
Solution:
(1)
(2)
(3)
Step 1. Pick any 2 equations and eliminate 1 variable:
Multilply (1) with 4 and subtract (2)
4x + 4y + 4z = 12
(4)
4x + 4y + 4z = 7
(2)
----------------------------0 =5
 The system has no solution
Tuesday September 30, 2014
3.4Solving Linear Systems in Three Variables:
Example: Solve the system using the elimination method
4x + 2y + 3z = 1
(1)
2x - 3y + 5z = -14
(2)
6x - y + 4z = -1
(3)
Solution:
Step 1. Multiply (3) with 2 and add to (1) to eliminate y
4x + 2y + 3z = 1
(1)
12x - 2y + 8z = -2 (2)
-----------------------16x
+ 11z = -1 (4)
Step 2. Multiply (3) with 3 and subtract to (2) to also eliminate y
2x - 3y + 5z = -14
(2)
- { 18x – 3y + 12z = -3 } (3)
--------------------------------16x
- 7z = -11 (5)
Step 3. Add (4) and (5) to eliminate x
16x + 11z = -1
(4)
-16x - 7z = -11 (5)
-----------------------------4z = -12
z
= -3
Use (5) to solve for x:
- 16x + 7(-3) = -11
- 16x - 21
= -11
- 16x
= -32
x = 2
Step4: Substitute z = -3 and x = 2 to (3)
6(2) - y + 4(-3) = -1
-y
= -1 – 12 +12
y
= -1
The solution is (2, 1, -3)
Tuesday September 30, 2014
3.4Solving Linear Systems in Three Variables:
Example: Solve the system:
x+ y+z =4
x+ y-z =4
3x + 3y + z = 12
Solution:
(1)
(2)
(3)
Step 1. Pick any 2 equations and eliminate 1 variable:
Add (1) and (2)
x+y+z =4
(1)
+{ x + y - z = 4 } (2)
----------------------------2x +2y
=8
(4)
Step2: Pick 2 different equations and also eliminate z:
x+ y-z =4
(2)
3x + 3y + z = 12
(3)
-------------------------------4x + 4y
= 16
(5)
Step3: Use the new equations to eliminate the next variable
Multiply (4) with 2 and subtract with (5)
4x + 4y
= 16
4x + 4y
= 16
(5)
---------------------------0
=0
The system has infinitely many solutions.
Tuesday September 30, 2014
3.4Solving Linear Systems in Three Variables:
Example: Solve the system using the substitution method
2a + b + c = 8
(1)
- a + 3b – 2c = 3
(2)
-a +b -c =0
(3)
Solution:
Step 1. Rewrite 1 of the equations for 1 variable.
-a +b -c =0
(3)
b = a + c (4)
Step 2. Substitute (4) into (1)
2a + b + c = 8
(1)
2a + (a+c) +c = 8
2a + a + c + c = 8
3a + 2c
=8
(5)
Step 3. Substitute (4) into (2)
- a + 3b – 2c = 3
(2)
- a + 3(a+c) – 2c = 3
- a + 3a + 3c – 2c = 3
2a
+c
= 3 (6)
Step4: Use (5) and (6) to solve for 1 variable
3a + 2c
=8
(5)
2a + c
=3
(6)
Multiply (6) with 2
3a + 2c
=8
(5)
4a + 2c
=6
(6)
--------------------------------a
=2
a
=-2
Step 5: Substitute a = -2 into (5) or (6)
2(-2) + c = 3
-4
+c=3
c=7
Step 6: Substitute a = -2 and c = 7 into (3)
b=a+c
= -2 + 7 = 5
The solution is (-2, 5, 7)
Tuesday September 30, 2014
Warm-up:
Solve by elimination:
x + 2y = -1
3x – y = 18
(1)
(2)
Friday October 9, 2015
Warm-up:
The school that Lisa goes to is selling tickets to the
annual talent show.
- On the first day of ticket sales
the school sold 4 senior citizen tickets and 5 student
tickets for a total of $102.
- The school took in $126 on the second day by selling
7 senior citizen tickets and 5 student tickets.
What is the price each of one senior citizen ticket and
one student ticket?
Friday October 9, 2015
Warm-up:
The school that Lisa goes to is selling tickets to the
annual talent show.
- On the first day of ticket sales
the school sold 4 senior citizen tickets and 5 student
tickets for a total of $102.
- The school took in $126 on the second day by selling
7 senior citizen tickets and 5 student tickets.
What is the price each of one senior citizen ticket and
one student ticket?
senior citizen ticket: $8, student ticket: $14
Friday October 9, 2015
3.5 Matrices-Basic Operations:
Objective: To perform basic operations with matrices
EQ: How can we organize sports data?
Matrix is a rectangular arrangement of numbers in row and
columns.
 Dimension: m x n (row x column)
 Each number is an element
Two matrices are equal iff:
1. Their dimensions are the same
2. Corresponding elements are equal
Adding and Subtracting Matrices –
(only when dimensions are the same)
Simply add or subtract
corresponding elements in
the corresponding positions
Example:
1.
2.
Friday October 9, 2015
3.5 Matrices-Basic Operations:
Scalar Multiplication
Simply multiply each element with the scalar
Example:
Tuesday October 13, 2015
Warm-up: Find x and y
Tuesday October 13, 2015
Warm-up: Find x and y
Solution:
Therefore:
 The solution is x = -2 and y = 4
Tuesday October 13, 2015
3.6 Multiply Matrices:
Objective: To perform basic operations with matrices
EQ: How can we organize sports data?
The product of 2 matrices A and B is defined iff:
the number of columns in A is equal to the number of rows in B
Example: State whether the product AB is defined:
1. A: 5x2 and B:2x2
defined, AB: 5x2
2. A: 3x2 and B:3x2
undefined
3. A: 4x3 and B:3x2
defined, AB: 4x2
4. A: 3x4 and B:3x2
undefined
3.6 Multiply Matrices:
Friday October 9, 2015
Multiplying Matrices –
Multiply each element in the ith row of A to jth column of B
Note: The product AB ≠ BA
Example: Find product AB then find product BA
if
Solution:
Find AB: since A: 2x2 and B: 2x2, AB: 2x2
1. AB =
2. AB =
3. AB =
4. AB =
5. AB =
Find BA:
3.6 Multiply Matrices:
Monday October 19, 2015
Example: Use matrices to calculate the total cost of 2 hockey
teams.
Sticks
($60)
Pucks
($2)
Uniforms
($35)
Women’sTeam
14
30
18
Men’s Team
16
25
20
Solution: Let’s write the equipment list and the cost list in
matrix form:
So, the total cost of equipment for each team can be found by
multiply the equipment matrix (E) by the cost matrix (C ).
Since E: 2x3 and C: 3x1, then EC: 2x1
 The total cost for women’s team is $1530
 The total cost for men’s team is $1710
Tuesday October 20, 2015
Warm-up: Find: A(B+C) where
Tuesday October 20, 2015
Warm-up: Use calculator to find: A(B+C) where
Solution:
Tuesday October 20, 2015
Warm-up: Multiply. Give reason why if not
possible.
Tuesday October 20, 2015
Warm-up: Multiply. Give reason why if it’s not.
Wednesday October 21, 2015
3.8 Inverse Matrices:
Objective: To use inverse matrices to solve linear systems
EQ: How can we solve linear systems using matrices?
Definition:
Identity matrix: a matrix with 1’s on the main diagonal and 0’s
Inverse Matrices: Matrices A and B are inverses of each other
if their product, AB and BA, is an identity matrix.
To find the inverse of matrix A
Example:
Solution:
The inverse of a 3x3 matrix is difficult to compute by hand.
Instead, for matrices that greater then 2x2, use calculator.
Wednesday October 21, 2015
3.8 Inverse Matrices:
Using an Inverse Matrix to Solve a Linear System:
Example:
Solution: