Download Precalc Notes Ch.7

Notes #3-___
Date:______
7.1 Solving Systems of Two Equations (568)
The solution to a system of linear equations is the ordered
pair (x, y) where the lines intersect! A solution can be
substituted into both equations to make true statements.
Consistent – solution(s)
Dependent (∞)
Inconsistent – Ø
Independent (1 or Ø)
These lines coincide
so they are a.k.a
coincident.
Substitution
1. Solve an equation for one of the variables (if necessary).
2. Substitute the expression for the variable and solve.
3. Substitute into either equation to get the other value.
y = 2x + 3
1. Not necessary.
3x + 2y = -1 2. 3x + 2(2x + 3) = -1
3x + 4x + 6 = -1
7x + 6 = -1,7x = -7, x = -1
3. y = 2(-1) + 3 = -2 + 3 = 1 Ans: (-1, 1)
Ex.1 Solve using substitution:
a) 3x + 2y = 14
b) -3x + y = 7
x – 2y = 10
-6x + 2y = -8
You do not have to
substitute for a single
variable (you can use
an expression).
Graphing
Linear Combination
or Elimination
Ex.2 Solving nonlinear systems:
a) y = x2 – 1
b) x + y = 6
 x y 
2
x2 + y2 = 13

x y
x
6
Ex.3 Graph and determine the point of intersection.:
a) 3x + y = 4
b) 3x + 2y = 6
x – 2y = 6
y = ln(x – 1)
1.
2.
3.
4.
Arrange like terms in columns (if necessary).
Multiply so that one set of variables are opposites.
Add and then solve.
Substitute into either equation to get the other value.
y = 2x + 3
3x + 2y = -1
1. -2x both sides
-2x + y = 3
3x + 2y = -1
2. Eliminate y if we multiply eq.#1 by -2
-2(-2x + y = 3) = 4x – 2y = -6
3x + 2y = -1
3. 7x = -7
x = -1
4. y = 2(-1) + 3 =
-2 + 3 = 1
Ans: (-1, 1)
Ex.4 Solve using elimination:
a) Find the product of the x & y coordinates of the
solution:
3x – 2y = 4
2x + 5y = -2
b) 2x – 4y = 8
x – 2y = 4
Ex.5 Three gallons of a mixture is 60% water by volume.
Determine the number of gallons of water that must be
added to bring the mixture to 75% water.
Ex.6 A car radiator contains 10 quarts of a 30% antifreeze
solution. How many quarts will have to be replaced
with pure antifreeze if the resulting solution is to be
50% antifreeze?
Ex.7 A machine takes 3 minutes to form a bowl and 2
minutes to form a plate. The material costs $0.25 for
a bowl and $0.20 for a plate. How many bowls and
plates were made if the machine ran for 8 hours and
$44 was spent on materials?
Ex.8 A company determines its supply and demand
equations for a product to be: S: p = x2 + 4x + 40
D: p = 50 – 3x
Find the equilibrium point in terms of x (thousands of
the item) and p (the price). What does it mean?
What relationship does the company want to have
between the supply and the demand?
Summary:
Notes #3-___
Date:______
7.2 Matrix Algebra Day 1 (579-583)
Order (dimensions) 3 × 2
(rows) r × c (columns)
 1  1


A = 0 5 
 3 2 
If r = c then it is a square matrix
entry/element
a32 = -2 (row 3, column 2)
Ex.1 Determine the order:
Is it a square matrix?
Identify:
b33 =
b24 =
2 5 1 4


B= 4 8 0 6
 3 1 2 1
In addition & subtraction, matrices must have the same order.
Add/subtract the corresponding elements. You multiply all
elements of a matrix by a scalar multiple (think distributive).
Ex.2
 2 0   4 8
7 1  2 0 2  =

 

Scalar Multiple
Identities:
Addition:
3+0=3
Multiplication:
3·1=3
Matrix Addition:
 2 14 
 1 0   

 
  2 14
   1 0 
 

Matrix Multiplication (not commutative AB ≠ BA):
Ex.3 Find the product AB:
B
A
2 5 1 4
 2 4 1 

5 8 0   4 8 0 6 

  3 1 2 1


2×3
3×4
Must be = to ×
Order of answer 2 × 4.
2(2)+4(4)+(-1)(-3)
10 + 32 + (-1) =
2+0+2=
4+16+3=23
41
4
Move me!
Move me!
Move me!
11
12
13
21
22
23
14
24
Ex.4 Find the product:
1 0 
2 1 3 

 1 1 0   2 1 

 3 2


Ex.5 Redo Ex.3 & Ex.4 on the graphing calculator. List the
steps in your notes.
Identity:
3+0=3
Inverse:
3 + (-3) = 0
3·1=3
3·
1 2 0 0 1 2
3 4  0 0  3 4

 
 

1 2  1 2 0 0
3 4   3 4  0 0

 
 

 3 1 
 2 1 


 3 1 
 2 1 


  3 1
   2 1
 

(opposite)
1
=1
3
(reciprocal)
 1 0 
  0 1
 

Inverse – not a negative exponent
1 1
 3 1
-1  
Ex.6 Verify: A  
and
A

 2 3 
 2 1


a b 
Determinant: If A = 
, then det(A) = A = ad – bc

c d 
Ex.7 Find the value of:
7 9
a)
3 6
 2 4
b) det 

 3 7
Ex.8 Find the inverse of a 2 × 2:
 3 1
A= 

 2 1
1. Find det(A), no A-1 if det(A) = 0.
2. Exchange a & d.
3. Opposite signs on b & c.
4. Multiply by the reciprocal of det(A).
 4 1 
Ex.9 Find the inverse of: 

 6 2
Summary:
Notes #3-___
Date:______
7.2 Matrix Algebra Day 2 (584-589)
A matrix that has a nonzero determinant is called a
nonsingular matrix (which means it has an inverse).
The minor (short for minor determinant) of an element of an
nth order determinant can be found by deleting the row
and column containing the element.
2
4
A = 0 3
6 5
1
2
The minor of a11 is
4
3 2
5
4
.
The cofactor of an element is (-1)i+j Mij. So the cofactor of
a11 is (-1)1+1(-12 – 10) = (-1)2(-22) = 1(-22) = -22.
  
Another way to deal with the signs
is to follow the pattern that a11 is +
and adjacent elements (horizontal
& vertical) are the opposite sign.
  
  
To evaluate a determinant of the nth order, multiple the
elements of a row or column by their corresponding
cofactors:
-2(+)
3 2
5
4
+ 0(-)
4
2
1 4
+ (-6)(+)
4
3
1
5
=
Ex.1 Find the value of:
4
2 3
a) 5 1 0
2 1 11
Lattice method:
 0 1 0 2 
 5 6 0 3

b) det 
 4 5 2 6 


 0 3 0 4 
4
2 3
4
2
5
1
5
1
0
2 1 11 2 1
-(6 + 0 + 110) + (44 + 0 + -15)
-116 + 29 = -87
Ex.2 Find the inverse of a 2 × 2:
 3 1
A= 

 2 1
Adjoin the identity and perform row
ops to get the identity on the left:
 3 1 1 0
2 1 0 1


If you get the 0s first you will have less fractions to
deal with.
Switching rows is
also a row operation.
Row operations are basically linear combinations.
Same as 2 by 2.
 0 2 1
Ex.3 Find the inverse:  1 1 1 
 2 3 0 
Cramer’s Rule:
ax + by = c
dx + ey = f
If you get
c
x
a
b
Dx f e

D a b
d e
y
Dy
D

d f
a b
d
e
0
#
you have coincident lines (planes) & is Ø.
0
0
Ex.4 Use Cramer’s Rule to solve: 5x – 3y = 4
2x + 4y = 1
Ex.5 Use Cramer’s Rule to solve for z:
a) x – 2y – z = 4
b)
x– y+ z= 8
4x + y + z = 7
2x + 3y – z = -2
x + 3y – 4z = -1
3x – 2y – 9z = 9
Summary:
c
S R1 & R2
-2 R1 + R3
R3 – 2R2
-5R2 + R3
-.5 R3
-R3 + R1
R2 + R1
0 2 1 1 0 0
 1 1 1 0 1 0 


 2 3 0 0 0 1 
 1 1 1 0 1 0 
 0 2 1 1 0 0 


 2 3 0 0 0 1 
 2 2  2 0 2 0
2
3
0
0
0
1
0
5 2 0 2 1
1 1 1 0 1 0 
 0 2 1 1 0 0 


0 5 2 0 2 1 
0
5
2
0
2 1
0 4
2
2
0
0
1
0
2 2 1
0 1 0
 1 1 1
0 1 0 2 2 1 


0 5 2 0 2 1 
0 5
0
10 10
5
0
5
2
0
2
1
0
0
2 10
8
4
1 1 1 0 1 0 
0 1 0 2 2 1 


0 0 1 5 4 2 
0
0
1 5 4 2
1 1
1
0 1
1 1
0
5 5 2
1 1 0 5 5 2 
0 1 0 2 2 1 


0 0 1 5 4 2 
1 0 0 3 3 1
0 1 0 2 2 1 


0 0 1 5 4 2 
0
 3 3 1
A-1 =  2 2 1 
 5 4 2 
0
Notes #3-___
Date:______
7.3 Multivariate Linear Systems & Row Operations (594)
Triangular Form (leading coefficients = 1):
Ex.1 Use back substitution:
It is easier to get
the 0s first, if
getting a 1 leads to
fractions.
Verify with sub or
elimination.
Gaussian Elimination:
Interchange rows or use
elimination to change rows.
1x – 2y – 2z = 7
1y – 2z = -7
1z = 3
1. Exactly One Solution
2. Infinite Solutions
3. No Solutions
Ex.2 2x + y – z = 0
x – y + 2z = -3
-x + 2y – 3z = 7
x–y+z=8
2x + 3y – z = -2
3x – 2y – 9z = 9
Augmented
Matrix
Row Echelon Form (REF):
Ex.3 3x – 6y + 3z = 7
2x + y + 4z = 2
x – 2y + z = 4
1 1 1 8 
 2 3 1 2 


 3 2 9 9 
3×4
Gauss-Jordan Elimination (Reduced REF):
Ex.4 3x + 2y + z = 5
x–y+z=4
x + 3y – 2z = -3
Ex.6
x + y – 3z = -1
y–z=0
-x + 2y
=1
Ex.5 2x + y – z = 1
3x + 3y + 2z = 4
x–y–z=0
Ex.7 Nonsquare Systems:
x – 2y – z = -5
2x + y + z = 5
Inverse matrices can be used to solve systems:
2x = 4 is solved by using inverses:
A is called the
coefficient matrix.
Ex.7
4 x  y  4
6x  2 y  3
1
1
2 x  4 , x = 2
2
2
 4 1 
 6 2


 x   4
 y    3
   
-1
-1
Ex.8 Solve A
using
· Ainverse
· X = Amatrices:
·B
a) x + 2y + z = -4
b) x + y + z + w = 4
2x – y + z = -4
2x – y + z
=0
x + 3y – z = -7
3x + 2y + z – w = 6
x – 2y – 2z + 2w = -1
(-3, -1, 1)
Ex.9 You borrow $10,000; some at 18%, 15% and 9%.
You borrow 3 times as much at 15% as you borrow at
18% and the interest is $1244.25. How much each?
Ex.10 Determine the quadratic function that contains the
points: (1, 3), (2, 10) & (-2, -6).
y = x2 + 4x – 2
Summary:
Notes #3-___
Date:______
7.4 Partial Fraction Decomposition (608)
I.
Not unique:
2 1

3 6
Intro:
Ex.1
1 1
2 1 1 3
2 3
5

    

=
3 2
2 3 2 3
6 6
6
How can we get back?
Ex.2
Unique?
3
2
=

x  4 x 3
=
5x 1
 x 4 x 3
PFD: writing rational expressions as the sum of simpler ones.
II. Distinct (no repeated) linear factors in the denominator.
Step #1: Factor the denominator.
Step #2: Write the factors as a sum.
Ex.3
5x 10
=
2 x 2 3x  2
=
A
B

2x 1 x  2
Step #3: How do we solve rational equations?
Multiply by the LCD.
Step #4: Solve the resulting system of equations.
Ex.4
16
t  4t
2
III. Repeated Linear Factors
x+1
 x  7 n
Ex.5

A1
A


2
x 7  x 7 

An
 x  7 n
x 2
x 2x 2  x
3
Strategic values of x
may help you solve
the system. See Ex.3.
IV. Irreducible Distinct Quadratic Factors
x 1
 x  2  x 2  x 1

8x 2 5x 18
Ex.6
x3 8
A
Bx C
 2
x  2 x  x 1
V. Irreducible Repeated Quadratic Factors
x 1


 x  2  x2  x 1
Ex.7
Summary:
2

x3  x 2 5x  2

x 2 3

2
A
Bx C
Dx  E
 2

x  2 x  x 1 x 2  x 1


2
Notes #3-___
Date:______
7.4 Partial Fractions Day 2 (613)
I.
Intro:
Ex.1 Write the terms for the PFD (don’t solve):

x 2  3x

x x  2 x 5  x  2 
2
3
=
3x 3  6 x 2  7 x  2
Ex.2 Decompose:
( x 2  2 x  2) 2
Ex.3 Use division to rewrite in the form q( x) 
x 4 3x 3  2 x 5
x 2  x 1
r ( x)
:
d ( x)
II. Improper Fractions (Degree of N > D):
r ( x)
and
d ( x)
4 x 3 9 x 2  2 x  6
r ( x)
then find the PFD of
: f(x) =
x ( x  2)
d ( x)
Ex.4 Use division to rewrite in the form q( x) 
Where is the low
battery warning?
Compare the rational f(x) graph to the terms of the PFD.
r ( x)
and
d ( x)
r ( x) 3 x 3  4 x 2  x  6
then find the PFD of
:
d ( x)
x 2 1
Ex.5 Use division to rewrite in the form q( x) 
Summary:
Notes #3-___
Date:______
7.5 Systems of Inequalities in Two Variables (617)
The graph of a linear inequality is a half-plane.
More lead?
Number Line
Open dot
<&>
Closed (solid) dot
<&≥
Coordinate Plane
Dotted line
Solid Line
Use a test point for each inequality.
Ex.1 Graph (are they bounded or unbounded):
a) x < 4, y ≥ -3
b) x ≥ 1, x < 4, y ≥ -1
and x – y ≥ 4
& y ≤ 1 x2 + 1.
2
Ex.2 Write a system of
inequalities:
Ex.3 Solve the system of inequalities.
1
a) y  x
b) y   x 1  3
2
y < -x2 – 2x + 2
x2 + y2 < 8
Linear Programming:
Constraints: a system of inequalities
Feasible Region: solution set
Objective Function: an equation applied to a feasible region
Optimal Solution: maximum or minimum of the solution set
Fundamental Theorem of Linear Programming: the
maximum or minimum always occurs at a vertex.
Ex.4 Mr. Jones is baking as many cakes & pies as possible
using the 18 eggs & 15 cups of milk he has on hand.
His cake recipe requires 2 eggs and 1 cup of milk and
his pie recipe requires 1 egg and 1.5 cups of milk.
a) Find the constraints and
graph the feasible region.
b) Write the objective function
and determine how many
cakes & pies should be made.
Ex.5
It takes 2 hrs to manufacture a pair of skis and 1 hr
for a snowboard. The finishing time for both is 1 hr.
The maximum time available for manufacturing is
40 hrs and for finishing is 32 hours each week. The
profit for a pair of skis is $70 & the profit for a
snowboard is $50. The manufacturer must produce
at least 8 snowboards every week because of a
contract with a sporting goods store.
a) Find the constraints and graph the feasible region.
b) Write the objective function
& find the maximum profit.
Summary: