Download Section 3.1

Chapter 11
Systems of Equations
1
Review, Chapter 11.1
Two Equations, Two Unknowns
• Solve by
– Substitution
– Elimination/addition
– Graphing (inaccurate), where lines intersect
• Can have
– No solution (parallel lines)
– One solution
– Infinite solutions (identical equations)
• Check solution by substitution
2
Sample
3x + 4y = 12
2y = 6 – 3/2 x
3
Solution
3x + 4y = 12
2y = 6 – 3/2 x
• By substitution: make the first equation 2y = 6 – 3/2x, and
substitute for y in the second, giving 6 – 3/2x = 6 – 3/2 x; they
are identical equations.
• By addition: rewrite the 2nd to be 3/2x + 2y = 6 then multiply
the equation by – 2, yielding -3x – 4y = -12. Add the equations
to get an equation for y
4
Chapter 11.2
Gaussian Elimination
5
Fancy Name, Same Principle
• The bottom line, is what do we do when we have more than
two variables?
• Example:
3x +
2y –
z=
-3
5y -
2z =
2
5z =
20
• Start with z, find y, find x; This form is called UpperTriangular; If you can your equations in this form, there are
some techniques that make solutions easier
6
Can have Infinite or No Solutions
• Just like two equations and two unknowns
• Infinite Solutions:
– At least two equations are basically the same
– More variables than solutions
• No solutions
– Two of the lines are parallel (in parallel planes)
7
An Example
x + 2y + z = 3
2x + y + z = 16
x + y + 2z = 8
8
Solution
x + 2y + z = 3
2x + y + z = 16
x + y + 2z = 8
• There is more than one way to solve this! What follows is an
example method
Subtract the 2nd equation from the first
Subtract twice the 2nd from the third
Are left with two equations in x and y; solve as would two
equations then substitute to find z
9
Solution, Cont.
x + 2y + z = 3
2x + y + z = 16
x + y + 2z = 9
• Subtracting 2nd from 1st gives: -x + y = -13
• Subtracting twice 2nd from 3rd gives: -3x –y = -23
• Adding the resulting two equations gives -4x = -36, x = 9
• Substitute 9 for x in the first two: 2y + z = -6, y + z = -2.
Subtracting gives y = -4
• From 1st equation: 9 – 8 + z = 3, z = 2
• Check in 2nd and 3rd: 18 – 4 + 2 = 16, 9 – 4 + 4 = 9
10
Alternate Solution, resulting in upper triangle
x + 2y + z = 3
2x + y + z = 16
x + y + 2z = 8
• Remove x from 2nd and 3rd equations by adding multiples of
the third
x +2y + z = 3
-3y – z = 10
-y + z = 6
• Remove y from the 3rd by adding -1/3 times 2nd to 3rd
x +2y + z = 3
-y + z = 6
-4z = -8
11
Allowed Operations
• Just like with two equations
1. Multiply both sides of an equation by a non-zero constant
2. Interchange the order in which two equations of a system
are listed
3. Add a multiple of one equation to another in the system
12
Example
2x + 5y – 3z = 4
4x – 3y + 2z = 9
5x + 6y – 2z = 18
13
Solution
2x + 5y – 3z = 4
4x – 3y + 2z = 9
5x + 6y – 2z = 18
Let’s eliminate z from the 2nd two by adding: 9x + 3y = 27.
Then from the 1st two by mult the 1st by 2 and the 2nd by 3:
4x + 10y – 6z = 8 and 12x - 9y + 6z = 27 , and adding:
16x + y = 35. Multiply this last equation by -3, getting
-48x - 3y = -105 and add to 9x + 3y = 27, giving -39x = -78, x = 2
From 16x + y = 35, we have y = 3; 5x + 6y – 2z= 18 becomes
10 + 18 -2z = 18, z = 5
x, y, z = (2,3 5)
14
Example
3x + y – z = 10
8x – y - 6z = -3
5x – 2y – 5z = 1
15
Solution
3x + y – z = 10
8x – y - 6z = -3
5x – 2y – 5z = 1
Let’s find z: Eliminate y from 1st and 2nd by adding them:
11x – 7z = 7. Eliminate from 2nd and 3rd by mult 2nd by -2 and
adding: -11x + 7z = 7;
However, if we add these two equations, we get 0 = 7; there is
no solution
16
Chapter 11.3
Matrices
17
Matrix (Matrices)
• A Matrix is a rectangular array of numbers
235
3 5 19 8
345
4 5 2 0
3 1 9 21
• The numbers are called entries or elements
• Rows are horizontal, columns are vertical
A matrix with three rows and two columns is a 3 by 2 matrix
• A square matrix has the same number of rows as columns
18
Example
• How big is this matrix?
• What element is in the 3rd row, 1st column?
3
−8
−10
5
2
𝜋
3
19
Solution
• How big is this matrix? Three by two
• What element is in the 3rd row, 1st column? 2/3
3
−10
2
3
−8
5
𝜋
20
Equality
• Two matrices are equal if they
1. Are the same size (same number of rows and columns)
2. Have the same values in each position
21
Coefficient and Augmented Matrices
x + 2y – 3z = 4
3x
+ z=5
-x – 3y + 4z = 0
We call the matrix coefficients of the equations the coefficient
matrix :
1
2 −3
3
0
1
−1 −3 4
And the augmented matrix adds the right side
1
2 −3 4
3
0
1 5
−1 −3 4 0
22
Matrix Operations
• Just like systems of equations:
1. Multiply each entry in a row by a non-zero constant
2. Interchange two rows
3. Add a multiple of one row to another
• If we can get a matrix into triangular form, we can solve the
set of equations it represents
23
Matrix Addition
• To add (or subtract) two matrices, A and B add the value in
row 1, column 1 in the first matrix to the value in row 1 column
1 of the second matrix, etc.
1 9
3
6 7
5 −2 4
+
=
4 −3 −8
7 5
3 8 3
7
−5
• Clearly A + B = B + A, provided A and B are the same size
• A + (B+C) = (A+B) + C
24
Scalar Multiplication
• A scalar is a number, any number
• To multiply a matrix by a scalar, multiply every element
by the scalar
10 4 −2
5 2 −1
2 0 4 4 = 0
8
8
6 16 −4
3 8 −2
• If A is a matrix and c and k are scalars, then c(k)A = k(c)A
• If B is also a matrix, then c(A+B) = cA + cB
25
Zero Matrix
0 0
0 0
0 0
0
0 = 0, is the zero matrix. It can have any dimension.
0
It is also the additive identity: A + 0 = A, A – A = 0
26
Vectors
• A vector (row vector) is a row of a matrix; it can have any
length, depending on the size of the matrix
(2 3 5), (-10 5 4 8), (0 0 2 0)
• A column vector is a column of a matrix
3
2 ,
5
−10
,
8
5
−2
1/2
27
Inner Product
• An inner product is the multiple of a row and a column vector.
The row must have the same length as the vector
3 4
5 ∙
1
6 = 3 ∙ 1 + 4 ∙ 6 + 5 ∙ 8 = 3 + 24 + 40 = 67
8
The value of an inner product is just a number (a scalar)
28
Multiplying Matrices
• If A and B are two matrices, and the number of columns in A
is equal to the number of rows in B, then the value of the
element in the ith column, jth row is the inner product of the ith
row of A and the jth row of B
1 0
2 1
2 4
3 0
1
2 4
=
1
7 8
1
3
29
Examples
• A=
1
3
2
5 6
and B =
4
7 8
• Find AB and BA
30
Solution
• A=
1
3
2
5 6
and B =
4
7 8
• AB =
19 22
43 50
• BA =
23 34
21 46
31
Why do we use Matrices?
• Matrices were invented to provide an endless supply of
homework for the poor students
• Used in every branch of physics, including classical
mechanics, optics, electromagnetics, and quantum
• Used to solve linear equations; first example Chinese 12th
century BC.
• Make problems very amenable to computer solutions
32
Chapter 11.4
The Inverse of a Square Matrix
33
Identity Matrix
1
I= 0
0
0 0
1 0
0 1
For all 3 x 3 matrices A, A I = I A = A
34
Inverse
• If AB = I and BA = I, then A is invertible, and B is its inverse
35
Finding an Inverse
Find A-1 = B, the inverse of A
AB=I
AB =
1 2
3 4
𝑎
𝑐
1
𝑏
=
0
𝑑
0
1
36
Solution
• AB =
1 2
3 4
𝑎
𝑐
1
𝑏
=
0
𝑑
0
1
• Need:
a + 2c = 1
b + 2d = 0
3a + 4c = 0
3b + 4d = 1
• Four equations in four unknowns
• Combining 1st and 3rd, gives a = -2, c = 3/2
• Combining 2nd and 4th gives b = 1, d = -1/2
B = A -1 =
−2
1
3/2 −1/2
37
Checking Inverses
• If A and A-1 are inverse matrices, then A A-1 = I, the identity
−2
1
1 2
-1
• In the previous example, A=
,A =
3/2 −1/2
3 4
−2 + 3
1−1
1 0
• Multiplying gives:
=
−6 + 12/2 3 − 4/2
0 1
38
Alternate Approach to Inverse
• Row operations
5 0 2 1 0
2 2 2 0 1
−3 1 −1 0 0
0
0
1
Let R1 = -2 R2 + R1; Do the same to each matrix pair
1 −4 −2 1 −2 0
2
2
2 0 1 0 ;
−3 1 −1 0 0 1
Try to get the identity matrix on the left
39
Continued
• R1 = -2 R2 + R1
•
1 −4 0 1 −2 0
2
2
2 0 1 0 ;
−3 1 −1 0 0 1
• Let R3 = 3R1 + R3
•
1 −4
0 1 −2 0
2
2
2 0 1 0 ;
0 −11 −1 3 −6 1
40
Continued
1 −4
0 1 −2 0
• 2
2
2 0 1 0
0 −11 −1 3 −6 1
• R2 = R2 – 2R1
•
1 −4
0 1 −2 0
2
2
2 0 1 0
0 −11 −1 3 −6 1
41
Keep going until have:
•
1 0
0 1
0 0
1
0 1 −1/2
0 1 −1/4 3/2
1 −2 5/4 −5/2
1 −1/2
1
5 0
• A -1 = 1 −1/4 3/2 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 𝐴 2 2
−2 5/4 −5/2
−3 1
1 0
= 0 1
0 0
2
2
−1
0
0
1
42
An Alternate Method for Inverses
𝑎 𝑏
𝑑
, the inverse is
𝑐 𝑑
−𝑐
is the determinant of the original matrix.
• Given a matrix
• Example:
𝑎
𝑐
−𝑏
/D, where D
𝑎
𝑏
5 −7
=
𝑑
−4 6
• The determinant of this matrix is 30 – 28 = 2=D
•
𝑑
−𝑐
−𝑏
becomes
𝑎
• Dividing by 2 gives:
6 7
4 5
3 7/2
as the inverse
2 5/2
43
Recap
1. Matrices are a way of keeping track of a lot of numbers
a) Rows and columns
b) Any kind of numbers
2. We define rules for arithmetic operations on a matrix
a) Multiplication by a scalar (any number); just multiply each
term by the same number
b) Adding matrices; add the values that are in the same
place in the matrix
c) Zero matrix; all terms zero – adding it doesn’t change the
matrix
44
Recap 2
3. Dot Product (scalar product); multiply the elements of a row by
a column and add together
1
(3, 2, 2) 4 = 3 + 8 + 10 = 21
5
4. Matrix multiplication; multiply(using the dot product) the ith row
by the jth column and put the result in the ith jth place
2 1 4 5
8 12
=
; a 2x3 matrix can multiply a 3x2
3 0 0 2
12 15
matrix, but a 2x3 matrix cannot multiply a 2x3 matrix
45
Recap 3
5. Matrix multiplication is not commutative: AB ≠ BA
6. There is an identity matrix I for matrix multiplication:
1 0 0
0 1 0 , 1 on the diagonal, zero elsewhere
0 0 1
7. A matrix times an identity matrix, or an identiy matrix times a
matrix results in the matrix: IA = AI = A
8. Most matrices are invertible (although the inverse my be
difficult to find). Given the inverse of A, A-1, we have
A A-1 = A-1A=I
46
Writing a System of Equations as a Matrix
• Suppose we have the following system
x + 2y = 8
3x + 4y = 6
1
• Let A =
3
𝑥
2
8
,X= 𝑦 ,B=
4
6
• Then AX = B
47
Using a Matrix to Solve a System
• If AX = B, then A-1 A X = I X = X = A-1 B
1
• A=
3
𝑥
2
8
,X= 𝑦 ,B=
4
6
• We found A -1 =
• X = A -1
−2 3/2
1 −1/2
−2
1
8
=
3/2 −1/2
6
8
−10
=
6
9
48
What are you going to have to know?
• Find the inverse of a 2x2 matrix
• Given the inverse of a 3x3 matrix, use it for problem solving
• Know when two matrices can be multiplied and multiply 2x2
and 3x3 matrices
49
Practice
Find:
2 1
1 6
+
=
3 2
4 9
3
4 6
=
2 0
1 4
0 2
0
3
2 5
=
4 0
50
Solution
2 1
1 6
3 7
+
=
3 2
4 9
7 11
3
4 6
12 18
=
2 0
6
0
1 4
0 2
0
3
2 5
12
=
4 0
6
18 5
8 0
51
Practice
• Show AI = AI if A =
4
1
3
2
52
Solution
• Show AI = IA=A if A =
•
4 3
1 2
1 0
4 3
=
0 1
1 2
•
1 0
0 1
4 3
4 3
=
1 2
1 2
4 3
1 2
53
Example
• If A =
3
4
−7
−9 7
and A-1 =
use this this to solve
−9
−4 3
3x – 7y = 30
4x – 9y = 39
54
Solution
• If A =
3
4
−7
−9 7
and A-1 =
use this this to solve
−9
−4 3
3x – 7y = 30
4x – 9y = 39
• X = A-1 B =
−9 7
−4 3
30
= (-270+273, -120+117)= (3, -3)
39
55
Example
•
The inverse of the Matrix A is given below. Verify the matrices are inverses, then use A and its inverse to solve the
following system of equations
x –y + z = 5
2x – 3y + 2z = -15
-4x + 6y + z = 25
3
1 −1 1
A= 2 −3 2 , A-1 = 2
0
−4 6 1
−7/5 −1/5
−1
0
product is
2/5
1/5
3−2
0
0
7
− 5 + 1 + 2/5
−
14
5
0
+ 3 + 4/5
0
0
4
5
, the identity matrix
+ 1/5
5
(A-1 )(−15) = (15+21-5, 10+15, -6+5) = 31, 25, -1
25
Check: x-y+z = 31 +25 – 1=5 OK, 2x-3y+2z = 62 -75 -2 = -15 OK, -4x+6y+z = -124+150-1 = 25 OK
56
Solution
3
1 −1 1
• A= 2 −3 2 , A-1 = 2
0
−4 6 1
•
1 −1 1
2 −3 2
−4 6 1
3
2
0
−7/5 −1/5
−1
0
2/5
1/5
−7/5 −1/5
1 0
−1
0
= 0 1
2/5
1/5
0 0
0
0
1
57
Solution Cont
3 −7/5 −1/5
−1
0
• A-1B = 2
0 2/5
1/5
5
−15 = (15+21-1, 10+15, -6+1)
5
= (35, 25, -5) = (x, y, z)
x –y + z = 5 ; 35-25+5=5 OK
2x – 3y + 2z = -15; 70 – 75 – 10 = -15 OK
-4x + 6y + z = 25; -140 + 150 – 5 = 5
58
Example
2x – 3y = 5
-x + 4y = 8
Use matrix techniques to solve this system of equations
59
Solution
2x – 3y = 5
-x + 4y = 8
Use matrix techniques to solve this system of equations
A=
2 −3
. Find its inverse
−1 4
AA-1 =
1
0
0
𝑎
. Let A-1=
0
𝑐
𝑏
𝑑
2a –3c = 1
2b – 3d = 0
-a + 4c = 0
-b + 4d = 1
5c = 1, c =1/5
5d = 2, d = 2/5
a = 4/5
b = 3/5
60
Solution Continued
4/5 3/5
4/5 3/5
-1
A
,A B=X=
1/5 2/5
1/5 2/5
= 44/5, 21/5
-1 =
5
= 4+24/5, 1+16/5
8
2x – 3y = 5; 88/5 – 63/5 = 5
-x + 4y = 8; -44/5+ 84/5 = 50/5 = 8
61
Chapter 11.5
Determinants and Cramer’s Rule
62
Determinant
𝑎 𝑏
= ad –bc, the product
𝑐 𝑑
of the downhill minus the uphill diagonal
• The determinant of a 2x2 matrix
63
Examples
• Find the determinant:
3 8
−1 4
3
2
1/2 4
2 0
0 3
64
Solution
• Find the determinant:
3 8
= 12 + 8 = 20
−1 4
3
2
= 12 – 1 = 11
1/2 4
2 0
=6 – 0 = 6
0 3
65
Three by Three Determinant
8
2
9
1.
3 5
4 6
1 7
Pick an entry, and remove the rows and columns containing
8 3 5
that entry 2 4 6 . Here I picked 8.
9 1 7
2. Find the determinant of the remaining matrix, called the
minor; 4x7 - 1x6 = 22.
3. Find the cofactor = (-1)i +j (minor), where i and j are the row
and column of the matrix. If i and j are 1, (-1)2=1, the
cofactor is 22
4. Multiply each entry in the first row by its cofactor and add to
get the determinant; 8x22 -3x(14-54)+5x(2-36) =
176 +120-170 = 126
66
Example
8 4 2
3 9 3
−2 8 6
67
Solution
8 4 2
3 9 3
−2 8 6
For 8, minor is 54 – 24 = 30, cofactor is 30
For 4, minor is 18 + 6 = 24, cofactor is -24
For 2, minor is 24 + 16 = 40, cofactor is 40
8x30 – 4x24 + 2x40 = 240 – 96 + 80=228
68
Example
9 9
4 4
7 7
12
6
5
69
Solution
9 9 12
4 4 6
7 7 5
For 9, minor is 20 – 42 = -22, cofactor is -22
For the 2nd 9, minor is 20 – 42 = -22, cofactor is 22
For the 12, minor is 28 – 28 = 0, cofactor is 0
9 x22 – 9 x 22 = 0, the determinant is zero
70
Did we need to use the first row?
• No, can expand on any row.
8 4 2
3 9 3 Expand on 2nd row
−2 8 6
71
Solution
8 4 2
3 9 3 Expand on 2nd row
−2 8 6
For 3, have 4x6-8x2 = 24 – 16 = 8, cofactor is -8
For 9, have 8x6+2x2= 48 + 4 = 52, cofactor is 52
For 3 have 8x8 +2x4 = 64 + 8 = 72, cofactor is -72
3(-8) + 9(52) – 3(72) = 228
72
An Alternate Method
9 9 12
4 4 6
7 7 5
The determinant is the sum of the products of the downhills
minus the sum of the products of the uphills:
9x4x5+9x6x7 + 4x7x12 – (7x4x12 + 4x9x5 + 7x6x9)
180+378+336-(336+180+378) = 0
73
Note on 3x3
• If you are having trouble with the up and down hills, try the
following
𝑎
Original Matrix 𝑑
𝑔
𝑐 𝑎
Augment it 𝑓 𝑑
𝑖 𝑔
𝑏
𝑒
ℎ
𝑏
𝑒
ℎ
𝑐
𝑓
𝑖
𝑐 𝑎
𝑐 𝑎
𝑓 𝑏 the diagonals are visible 𝑓 𝑑
𝑖 𝑔
𝑖 𝑐
𝑏
𝑒
ℎ
𝑐 𝑎
𝑓 𝑏
𝑖 𝑐
74
Manipulating Determinants
1. If each entry in a row is multiplied by a constant, k, then the
determinant is multiplied by a constant k.
2. If a multiple of one row is added to another row, the value of
the determinant is not changed. The same holds for columns
3. If two rows are interchanged, the determinant is multiplied by
-1. The same holds for columns
These rules may make calculating determinants easier.
75
Example
15 14 26
18 17 32
21 20 42
Remove a 3 from the 1st column and 2 from the third column
5 14 13
6 17 16 (3x2)
7 20 21
Subtract the third column from the 2nd
5 1 13
6 1 16 (3x2)
7 −1 21
76
Example Continued
5 1 13
6 1 16 (6) Add the third row to the 1st and 2nd
7 −1 21
12 0 34
13 0 37 (6)
7 −1 21
Now expand along the 2nd column; only one term is left
=6(-1)(-1)
12
1
34
= 6(36-34) = 12
3
• Strategy – get zeroes and small numbers
77
Zero Determinants
9
4
13 17
9
4
−3
5
−3
9x17x(-3) + 4x5x9 + 13x4x(-3) – [9x17x(-3) + 13x4x(-3) + 4x5x9]
=0
If two rows are equal, or two columns are equal then the
determinant is zero.
However, a determinant can be zero even if neither applies
What is the implication of a zero determinant
78
Cramer’s Rule
• Suppose we have
a 1x + b 1y + c 1z = d 1
a 2x + b 2y + c 2z = d 2
a 3x + b 3y + c 3z = d 3
Calculate the determinants if D0, x = Dx/D, y= Dy/D, z= Dz/D
𝑎1
D= 𝑎2
𝑎3
𝑎1
Dy= 𝑎2
𝑎3
𝑏1
𝑏2
𝑏3
𝑏1
𝑏2
𝑏3
𝑐1
𝑑1
𝑐2 , Dx= 𝑑2
𝑐3
𝑑3
𝑐1
𝑎1
𝑐2 , Dz= 𝑎2
𝑐3
𝑎3
𝑏1
𝑏2
𝑏3
𝑏1
𝑏2
𝑏3
𝑐1
𝑐2 ,
𝑐3
𝑐1
𝑐2
𝑐3
79
Example
2x + 2y – 3z = -20
x – 4y + z = 6
4x – y + 2z = -1
80
Solution
2x + 2y – 3z = -20
x – 4y + z = 6
4x – y + 2z = -1
2
D= 1
4
2 −1
−20 2 −1
−4 1 = -55, Dx= 6
−4 1 = 144, etc.
−1 2
−1 −1 2
Dy = 61, Dz= -230
x= - 144/55, y = -61/55, z = 46/11
81
Example
• Solve:
5x – 3y – z = 16
2x + y – 3z = 5
3x – 2y + 2z = 5
82
Solution
• Solve:
5x – 3y – z = 16
2x + y – 3z = 5
3x – 2y + 2z = 5
• D = 10 + 27 + 4 – (-3 -12 + 30) = 26
• Dx= 32 + 45+ 10 – ( 5 -30 +96) = 26
• x=1, y= -3, z= -2
83
Higher Order Determinants
• Every square matrix has a determinant
• However, in order to find an nth order determinant, we need
to be able to find an (n-1)th order determinant, etc.
84
Chapter 11.6
Non-Linear Systems of Equations
85
Non-Linear Equations
• For example:
2x + y = 1
y = 4 – x2
• If we have one linear equation, we can solve by substitution
86
Solution
• 2x + y = 1
y = 4 – x2
subst for y:
2x + 4-x2 = 1;
find x and subst to get y.
x = 3 or x = -1,
(-1, 3) (3, -5)
87
Intersecting Graphs
• y = x2
• x2 + y2 = 1
88
Solution
• y = x2
• x2 + y2 = 1
• Again by substitution, y + y2 = 1, y =
substituted to find x =± (−1 +
−1 ± 5
2
and we can
5 )/2
89
Example
• xy = 1
• y = 3x + 1
90
Solution
• xy = 1
• y = 3x + 1
the first becomes x(3x + 1) = 1, a quadratic, and
x=
−1 ± 13
, leaving
6
y=
−1 ± 13
+
2
1,
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There is not always a solution
y= 𝑥
(x+2)2 + y2 = 1
Substituting gives (x+2)2 + x = 1, and x =
−5± 13
.
2
However, that
is negative, and y = 𝑥 cannot be computed over the real
numbers
92
Creating a Linear System
•
2
𝑥2
•
3
𝑥2
-
3
𝑦2
= -6
-
4
𝑦2
= 59
• Simply substitute for x and y and solve
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Solution
•
2
𝑥2
•
3
𝑥2
-
3
𝑦2
= -6
-
4
𝑦2
= 59
1/x2 = u, 1/y2 = v
2u-3v=-6 becomes 6u-9v = -18
3u-4v=59 becomes -6u +8v = -118
Adding gives -v = -136, v=136, y = 
2u = -6 + 3v = -6 + 408 = 402, x=
1
136
1
201
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More Examples
• y = 3x, y = 32x -2
• y = - 𝑥 − 6 , (x-3)2 + y2 = 4
• y = log(x+1), y = 5 – log(x-3)
• a 2x + a 2y = 10, a x+y = 4 (a>0)
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What did all the preceding have in common?
• Either an x or y was isolated
• We could substitute, e.g., replace x2 by u
• In log/exp problems, we could remove coefficients on
variables (a2x = ax ax ) and manipulate the equations
• ALWAYS LOOK FOR A WAY TO SUBSTITUTE
96
Approach to Solution
• y = 3x, y = 32x -2
2nd becomes y = y2 - 2
• y = - 𝑥 − 6 , (x-3)2 + y2 = 4
substitute for y in the 2nd; (x-3) 2 + x-6 = 4
• y = log(x+1), y = 5 – log(x-3)
log(x+1) = log(x-3) + 5; add logs and exponentiate
• a 2x + a 2y = 10, a x+y = 4 (a>0)
let ax = u, ay = v; u2 + v2 = 10, uv=4 or v = u/4
97