MAT150 Homework for Test # 3 Solutions
Day 40 Homework: Topics: Graphing Rational Functions: Vertical Asymptotes, Horizontal Asymptotes, Y-intercept, X-intercept,
crossing the horizontal Asymptote.
Supplemental Reading:
1. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut40_ratgraph.htm (skip content on oblique/slant
asymptotes and example 6.)
2. http://www.purplemath.com/modules/grphrtnl.htm
3. http://www.coolmath.com/algebra/23-graphing-rational-functions/index.html
x 2  5x  6  x  6  x  1
x2  5x  6
1. y 
Find
each
of
the
following…..
y


x2  4
x2  4
 x  2 x  2
A) The equation(s) for the vertical asymptote(s)
From the factored form of the function
We can see that the vertical asymptotes will occur at x  2 and
x2
B) The equation(s) for the horizontal asymptote(s)
From the original form of the function we can see that the degree of the numerator equals the degree of the
denominator, therefore the horizontal asymptote will occur at y = “the ratio of lead coefficients” so……
1
y   y 1
1
C) The y-intercept(s)
Let x = 0 in the original form
6 3
and you can see that y 

4 2
 3
so the y intercept is  0, 
 2
D) The x-intercept(s)
From the factored form of the function we can see that
the x intercepts will occur at .  6,0 and
E) Does this function cross it’s horizontal asymptote? If so find out where.
x2  5x  6
1
 1 x 2  4   x 2  5 x  6  x 2  4  x 2  5 x  6  4  5 x  6
x2  4
2
 2 
5 x  6  4  5 x  2  x  
so, this function crosses it ' s H . A. at   ,1
5
 5 
F) Sketch the graph of this function on the axes below.
 1,0
(Day 40 Homework continued) :
x 2  3x  4
Find each of the following…..
x3  2 x 2  4 x  8
 x  4  x  1
 x  4  x  1
 x  4  x  1
 x  4  x  1
x 2  3x  4
y 3
 2



2
2
x  2 x  4 x  8 x  x  2   4  x  2   x  2   x  4   x  2  x  2  x  2   x  2 2  x  2 
2. y 
A) The equation(s) for the vertical asymptote(s)
From the factored form of the function
we can see that the vertical asymptotes will occur at x  2 and
x  2
B) The equation(s) for the horizontal asymptote(s)
Since the degree of the numerator is less than the degree of the numerator we know that
Y = 0 must be our H.A.
C) The y-intercept(s)
Let x = 0 in the original form
4
1

and you can see that y 
8
2
1

so the y intercept is  0,  
2

D) The x-intercept(s)
From the factored form of the function we can see that
the x intercepts will occur at .  4,0 and
E) Does this function cross it’s horizontal asymptote? If so find out where.
Yes, we already found this in part D)
F) Sketch the graph of this function on the axes below.
Note: We need to plot more points in order to sketch in the “outside” portions of this graph.
1 4  4

 25 1 25
 7  2  14
x 3 y 

1 5 5
x  3  y 
1,0
Day 41 Homework: Topics: Graphing Rational Functions:
Vertical Asymptotes, Horizontal Asymptotes,
Y-intercept, X-intercept, crossing the horizontal Asymptote.
Supplemental Reading: (The same as yesterday!)
1. y 
 x  3 x  3
x2  9

2 x  3x  5  2 x  5 x  1
2
A) The equation(s) for the vertical asymptote(s)
From the factored form of the function
we can see that the vertical asymptotes will occur at x  
5
2
and
x 1
B) The equation(s) for the horizontal asymptote(s)
From the original form of the function we can see that the degree of the numerator equals the degree of the
denominator, therefore the horizontal asymptote will occur at y = “the ratio of lead coefficients” so……
1
y
2
C) The y-intercept(s)
D) The x-intercept(s)
Let x = 0 in the original form
and you can see that y 
From the factored form of the function we can see that
9 9

5 5
the x intercepts will occur at .  3,0 and
 9
so the y intercept is  0, 
 5
E) Does this function cross it’s horizontal asymptote? If so find out where.
1
x2  9
 2
 1 2 x 2  3x  5   2  x 2  9   2 x 2  3x  5  2 x 2  18
2 2 x  3x  5
13
 13 1 
3x  5  18  3x  13  x  
so, this function crosses it ' s H . A. at   , 
3
 3 2
F) Sketch the graph of this function on the axes below.
3,0
(Day 41 Homework continued) :
2. y 
 x  3 x  2 
 x  3 x  2 
 x  3 x  2 
 x  2
x2  5x  6
 2



3
2
2
x  3x  x  3 x  x  3  1 x  3  x  3  x  1  x  3 x  1 x  1  x  1 x  1
A) The equation(s) for the vertical asymptote(s)
From the factored form of the function
we can see that the vertical asymptotes will occur at x  1 and
ALSO note that we have a “hole” when x  3 y 
3  2
1

 3  13  1 8
x 1
 1
 3, 
 8
B) The equation(s) for the horizontal asymptote(s)
Since the degree of the numerator is less than the degree of the numerator we know that
Y = 0 must be our H.A.
C) The y-intercept(s)
Let x = 0 in the original form
6
and you can see that y   2
3
so the y intercept is  0, 2
D) The x-intercept(s)
From the factored form of the function we can see that
the only x intercept will occur at .  2,0
E) Does this function cross it’s horizontal asymptote? If so find out where.
Yes, we already found this in part D)
F) Sketch the graph of this function on the axes below.
HOLE
Note: We need to plot one more point in order to sketch in the far left “outside” portions of this graph.
x  2
y
 4
4

 1 3 3
4

 2,  
3

x3
Day 42 Homework: Topics: Exponential Functions and their graphs,
compound interest, Logarithmic Functions and their graphs,
the definition of logarithm.
Supplemental Reading:
1. http://www.coolmath.com/algebra/17-exponentials-logarithms/index.html (check out lessons 1, 3, 4, 7 and 8)
2. http://www.purplemath.com/modules/expofcns.htm
3. http://www.purplemath.com/modules/graphexp.htm
4. http://www.purplemath.com/modules/graphlog.htm
5. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut42_expfun.htm
6. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut43_logfun.htm
1. Use your calculator and evaluate each of the following. Round your answer to three decimal places.
A) f  x   3.4x
3.4
1.3
x  1.3
at
10
 4.908
C) f  x   e x
B) f  x   10x
at
x  3.2
e3.2  24.533
.24 
3  2
x  .24
 .004
1
D) f  x    
3
1
 
3
at
x2
at
x  3
1
1
  3
3
2. Graph the following two functions on the same set of axes… (Fill out the tables!!!)
f  x   3x
x
0
1
2
-1
-2
x
0
1
2
-1
-2
1
g  x   
3
x
f (x)
1
3
9
1/3
1/9
g (x)
1
1/3
1/9
3
9
3. $20,000 is placed in a savings account that earns 8% interest. Find the amount of money after 5 years IF…
A) interest is compounded quarterly
 .08 
A  20000 1 

4 

B) interest is compounded monthly
12 5
4 5
 $29,718.95
 .08 
A  20000 1 

 12 
C) interest is compounded continuously
A  20000e.08 5  $29,836.49
 $29,796.91
(Day 42 Homework continued) :
4. Repeat problem 3. Is the interest rate is 10%.
A)
B)
 .10 
A  20000 1 

4 

4 5
C)
12 5
 $32,772.33
 .10 
A  20000 1 

 12 
 $32,906.18
A  20000e.10 5  $32,974.43
5. Graph y  log 2 x  x  2 y
x
y
0
1
2
-1
-2
1
2
4
1/2
1/4
6. Use the definition of logarithm to convert the following logarithmic equations into their equivalent exponential form.
A) 3  log 2 8
8  23
 1 
B)  2  log5  
 25 
1
 52
25
C ) log10 1  0
1  100
D) ln e  1
e  e1
7. Use the definition of logarithm to convert the following exponential equations into their equivalent logarithmic form.
A) 53 
1
125
 1 
3  log5  
 25 
B) 64  43
C ) 104  10, 000
3  log 4 64
4  log10 10000
D ) 50  1
0  log5 1
Day 43 Homework: Topics: Properties of logarithms.
Supplemental Reading:
1. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut44_logprop.htm
2. http://www.coolmath.com/algebra/17-exponentials-logarithms/10-inverses-tricks-01.htm
3. http://www.purplemath.com/modules/logs.htm
4. http://www.purplemath.com/modules/logrules.htm
1. Evaluate the following WITHOUT using a calculator.
A) log 3 81
B) log 6 1
log 3 34
0
4
 1 
C ) log 4  
 64 
1
log 4 3
4
log 4 43
D) log 2 213
13
E ) ln e13
F ) log100
log e e13
log10 102
13
2
3
2. Use the change of base formula, and your calculator, to evaluate the following. Round your answer to three decimal places in A
and B.
 1 
B) log3  
 75 
1
log
75
log 3
A) log 2 28
ln 28
ln 2
C ) log100 1000
log10 1000
log10 100
 4.807
 1.5
 3.930
3. Find he exact value of log3 5 9 WITHOUT using a calculator.
1
2
1
log3 5 9  log3 9 5  log3  32  5  log3 35 
4. Expand logb
4
b3 x 2
y6
1
log b
4
2
5
3
1
3 1
3
3
1
3
 b3 x 2  4
b3 x 2
b4 x2
3 1
3
 log b  6   log b 3  log b b 4 x 2  log b y 2  log b b 4  log b x 2  log b y 2   log b x  log b y
6
4 2
2
y
 y 
y2
5. Condense 3log x  2log y  5log z
3log x  2 log y  5log z  log x 3  log y 2  log z 5  log
x3
x3 5
x3 z 5
5

log
z

log

z

log
y2
y2
y2
Day 44 Homework: Topics: More with properties of logarithms.
Supplemental Reading: Same as yesterday!
1. Find the value for each of the following without using calculator.
C) log4  16
B) ln 3 e2
A) log100
Does not exist!
(You cannot feed
a negative number to
a logarithm!)
2
log10 100
ln e 3
log10 10 2
2
log e e 3
2
D) log5 75  log5 3
log 5
2
3
E) log6 2  log6 18
F) 3ln e6  2ln e5
log 6  2 18 
ln  e 6   ln  e5 
log 6 36  log 6 62  2
ln e 18  ln e10
75
3
log 5 25  log 5 5  2
2
3
ln  e 18  e10 
2. Expand ln
x5 y 2
z4
ln  e 18 10 
1
 52 
 x5 y 2  2
x5 y 2
x y
ln
 ln  4   ln  2 
4
z
 z 
 z 


5
ln e 8  8
5
 ln x 2 y  ln z 2  ln x 2  ln y  ln z 2

5
ln x  ln y  2 ln z
2
3. Condense 5log 2 y  log 2 z  3log 2 x
5log 2 y  log 2 z  3log 2 x  log 2 y 5  log 2 z  log 2 x3  log 2
4. Write 4 3 
3  log 4
1
in logarithmic form.
64
1
64
5. Write 7  log 2 128 in exponential form.
128  27
y5
y5
 log 2 x3  log 2 3
z
xz
2
Day 45 Homework: Topics: Solving Exponential Equations
Supplemental Reading:
1. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut45_expeq.htm
2. http://www.purplemath.com/modules/solvexpo.htm
3. http://www.coolmath.com/algebra/17-exponentials-logarithms/11-solving-exponential-equations-01.htm
1. Solve the following exponential equations by “making the bases the same”.
A. 32 x 1  27
B. 2  4x1  8
32 x 1  33
2 x  1  3
2 x  2
x  1
2  4 x 1 8

2
2
4 x 1  4
x 1  1
x0
2
D.  
3
C. 3x  9 x  27
2
3x   32   33
2 x2  x  3  0
3x  32 x  33
 2 x  3 x  1  0
x2
2
3x  2 x  33
2
x
3
2
x 1
x  2x2  3
x 1

9
4
2
 
3
x 1
3
 
2
2
 
3
x 1
  2  1 
   
 3  


x 1
2
2
   
3
3
x  1  2
E. 2  32 x1   1  17
2
2
2
x  1
2  32 x 1   18
32 x 1  9
32 x 1  32
2x  1  2
x
1
2
2. Solve the following exponential equations. Give an exact answer AND an approximate answer rounded to three decimal places.
A. 2x1  3
B. 2  32 x1   1  15
ln 2 x 1  ln 3
2  32 x 1   16
 x  1 ln 2  ln 3
x 1 
ln 3
ln 3
 x  1
 2.585
ln 2
ln 2
32 x 1  8
ln 32 x 1  ln 8
 2 x  1 ln 3  ln 8
C. 32 x 1  5x 2
ln 32 x 1  ln 5x  2
 2 x  1 ln 3   x  2  ln 5
2 x ln 3  ln 3  x ln 5  2 ln 5
ln 3  2 ln 5  x ln 5  2 x ln 3
ln 8
ln 3
ln 8
2 x  1 
ln 3
1 ln 8
x 
2 2 ln 3
x  .446
2x 1 
D. e2 x  3ex  4  0
ln 3  2 ln 5  x  ln 5  2 ln 3
ln 3  2 ln 5
x
ln 5  2 ln 3
 7.345  x
 e   3e
 e  4  e
x 2
x
4  0
x
x
 1  0
e 4  0
x
ex  4
x  ln 4
x  1.386
ex  1  0
e x  1
no solution
Day 46 Homework: Topics: Solving Logarithmic Equations
Supplemental Reading:
1. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut46_logeq.htm
2. http://www.purplemath.com/modules/solvelog.htm
3. http://www.coolmath.com/algebra/17-exponentials-logarithms/15-solving-logarithmic-equations-01.htm
4. http://www.coolmath.com/algebra/17-exponentials-logarithms/14-tricks-to-help-with-calculus-01.htm
1. Solve the following logarithmic equations.
A. ln x  ln 4  0
B. ln  x  3  1  2
ln x  ln 4
x4
ln  x  3   1
C. log  3  x   2
D. 2log3 1  x   3  11
3  x  10
3  x  100
 x  97
x  97
2 log 3 1  x   8
x  3  e1
x  3 e
2
log 3 1  x   4
1  x  34
1  x  81
 x  80
x  80
E. log4  x   log4  x  12  3
F. log3  x   log3  x  6  3
log 4 x  x  12   3
log 3 x  x  6   3
x  x  12   4
x  x  6   33
3
x 2  12 x  64
x 2  6 x  27
x 2  12 x  64  0
x 2  6 x  27  0
 x  16  x  4   0
 x  9  x  3  0
x  16
x  9 x  3
x4
Extraneous
Extraneous
G. log2  x   log2  x  6  2
H. log3  x  5  log3  x  3  2
log 2
x
2
x6
x
 22
x6
x
4
x6
x  4  x  6
x5
 2
x3
x5
 32
x3
x5 1

x3 9
9  x  5   1 x  3
log 3
x  4 x  24
9 x  45  x  3
3x  24
8 x  48
x8
x6
Day 47 Homework: Topics: Applications involving exponential equations
Supplemental Reading:
1. http://www.coolmath.com/algebra/17-exponentials-logarithms/12-solving-exponential-equations-rate-time-01.htm
2. http://www.coolmath.com/algebra/17-exponentials-logarithms/13-radioactive-decay-decibel-levels-01.htm
3. http://www.purplemath.com/modules/expoprob2.htm
4. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut47_growth.htm
1. Complete the table for a savings account in which interest is compounded continuously.
Initial Investment
$10,000
$5000
$165,298.89
A)
B)
C)
A) T2 : 20, 000  10, 000e.04t  2  e.04t  ln 2  .04t  t 
Annual % Rate
4%
6.9%
12%
Time to Double
17.33 yrs
10 yrs
5.78 yrs
Amount after 15 yrs
$18,221.19
$14,075.53
$1,000,000
ln 2
 17.33 years
.04
A 15  : A 15   10, 000e.0415  $18, 221.19
B) % : 10, 000  5000e r 10  2  e10 r  ln 2  10r  r 
ln 2
 .069  6.9%
10
A 15  : A 15   5000e.06915  $14, 075.53
C ) T2 : 2 P  Pe.12t  2  e.12t  ln 2  .12t  t 
P : 1, 000, 000  Pe.1215  P 
1, 000, 000
e.1215
ln 2
 5.78 years
.12
 $165, 298.89
2. Determine how long it would take for $1000 to double if it is invested at an interest rate of 7% compounded…
Round your answer to the nearest hundredth of a year.
A) Annually
B) Monthly
C) Continuously
12 t
1t
12
t
 .07 
t
 .07 
2000  1000e.07t  2  e.07t
2000  1000 1 
  2  1.07  2000  1000 1  12   2  1.00583
1




ln 2  ln e.07t  ln 2  .07t
12 t
t
ln 2  ln 1.07   ln 2  t ln 1.07 
ln 2  ln 1.00583  ln 2  12t ln 1.00583 ln 2
 t  t  9.90 yrs
ln 2
.07
ln 2
t
 10.24 yrs
t


9.93
yrs
ln 1.07 
12 ln 1.00583

3. The half life of radioactive radium

226







Ra  is 1620 years. What percent of a present amount of radioactive radium will remain
after 100 years? (Round your answer to the nearest tenth of a percent)
1
ln
1
1
1
k 1620
1620 k
A0  A0 e
 e
 ln  1620k  k  2
2
2
2
1620
A  A0 e
1
ln
2 100 
1620
 A  .958115578 A0
So about 95.8% will remain after 100 years.
4. The population P of a city (in thousands) is P  548ekt where t = 0 represents the year 2001. In 1969 the population was 238,000.
Find the value of k and use the result to predict the population in the year 2025.
Round your answer to the nearest thousand.
(note: if 2001 = year zero then 1969 = year # -32)
238
ln
238
238
k  32 
238  548e

 e 32 k  ln
 32k  k  548
548
548
32
238
548  24 
32
ln
so the amount in 2025  year # 24  A  548e
So the population in 2025 will be about 1,024,000
 1024.314356
Day 48 Homework: Topics: Solving Linear systems of equations in two variables by graphing and substitution.
Supplemental Reading:
1. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut49_systwo.htm (Read everything up through
example 6)
2. http://www.purplemath.com/modules/systlin1.htm (read the first four pages only)
3. http://www.coolmath.com/algebra/12-2x2-systems-of-equations/index.html (read items 1, 2 and 4 only)
1. Solve each of the following systems of equations by Graphing.
3

y   x  2
B) 
2
 2 x  3 y  7
 x y  3
A) 
 2 x  3 y   1
 x  3y  1
C) 
 2 x  3 y  1
 2,1
 2,1
 2, 1
2. Solve each of the following systems of equations by the Substitution method.
3

 x y  3
y   x  2
A) 
B) 
2
 2 x  3 y   1
 2 x  3 y  7
x  y  3  x  3 y
 3

2x  3  x  2   7
2 x  3 y  1  2  3  y   3 y  1
 2

9
 6  2 y  3 y  1
2x  x  6  7
2
 6  5 y  1
4
x

9
x  12  14
5y  5  y  1
x  3  y  x  3 1  x  2
 2,1
 3x  6 y  7
D) 
 2 x  4 y  2
2 x  4 y  2  2 x  4 y  2
x  2 y 1
3 x  6 y  7  3  2 y  1  6 y  7
6y  3 6y  7
13 x  26  x  2
3
3
y   x  2  y    2   2  1
2
2
 2, 1
 x  3y  1
C) 
 2 x  3 y  1
x  3y  1  x  1 3y
2 x  3 y  1
2 1  3 y   3 y  1
2  6 y  3 y  1
3 y  3  y  1
x  1  3 y  x  1  3 1
x  2
 2,1
 5 x  15 y  10
E) 
 3 x  9 y  6
3 x  9 y  6  3 x  9 y  6  x  3 y  2
5 x  15 y  10  5  3 y  2   15 y  10
 15 y  10  15 y  10
 10  10
3  7
No Solution !
An infinite # of solutions of the form  3 y  2, y 
Day 49 Homework: Topics: Solving Linear systems of equations in two variables by elimination.
Supplemental Reading:
1. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut49_systwo.htm (start reading after example 6)
2. http://www.purplemath.com/modules/systlin5.htm (read this page only)
3. http://www.coolmath.com/algebra/12-2x2-systems-of-equations/index.html (read item 3 only)
1. Solve each of the following systems of equations using the Elimination method!!!
A)  x  y  3

2 x  3 y  1
*2  
2x  2 y  6
 2 x  3 y  1
5y  5  y  1
x y 3
x 1  3  x  2
B) 
3
y   x  2
2


2 x  3 y  7
*2   2 y  3x  4  3x  2 y  4 *3  9 x  6 y  12


2 x  3 y  7

2 x  3 y  7 *2   4 x  6 y  14
13x
3
3
y   x  2  y    2   2  y  3  2  y  1
2
2
C)  x  3 y  1



2 x  3 y  1
 2,1
 26  x  2
 2, 1
*  2   2 x  6 y  2
 2 x  3 y  1
 3 y  3  y  1
x  3 y  1  x  3 1  1  x  3  1  x  2
D)  3x  6 y  7



2 x  4 y  2
 2,1
*2   6 x  12 y  14
*3  6 x  12 y  6
0  20  No Solution !
E)  5 x  15 y  10



3x  9 y  6
*3  15 x  45 y  30
*5  15 x  45 y  30
0  0  find infinite generating set
5 x  15 y  10  5 x  15 y  10  x  3 y  2
So the infinite generating set will be
 3 y  2, y 
PART II (Still Day 49)
1. Solve each of the following systems of equations using the Elimination method!!!
*2  
A) 
 x  2y  3


2 x  4 y  1
2x  4 y  6
 2 x  4 y  1
0  7 No Solution!
B) 2 x  y  1



4 x  2 y  2
*  2   4 x  2 y  2
 4x  2 y  2
0 = 0 So we must find the infinite generating set!
 x, 2 x  1
2 x  y  1   y  2 x  1  y  2 x  1
C)  3 x  2 y  5


2 x  5 y  7
*2   6 x  4 y  10
*3  6 x  15 y  21
19 y  11  y  
 3x  2 y  5

2 x  5 y  7
*5   15 x  10 y  25
*  2   4 x  10 y  14
19 x
D)  1
2
 2 x  3 y  4

  x  3 y  7

5
*6  
 39  x 
3 x  4 y  24
11
19
 39 11 
 , 
 19 19 
39
19
*3 
9 x  12 y 
*5  5 x  3 y  35 *4   20 x  12 y  140
 11x
3 x  4 y  24

5 x  3 y  35
*5   15 x  20 y  120
*3  15 x  9 y  105
 11y  15  y  
 68 15 
 , 
 11 11 
72
15
11
 68  x 
68
11
Day 50 Homework: Topics: Gaussian Elimination with Back Substitution for 3x3 systems
Supplemental Reading: There are resources out there on this BUT everyone seems to do it slightly differently (particularly notation
wise) SO, I recommend that you just refer heavily to your class notes!
1. Solve the following systems of equations using Gaussian Elimination with back substitution. Be sure to SHOW your row
operations between steps (on top and bottom of your arrows, just like I did in class). If you find that your system has an infinite
number of solutions be sure to find the infinite generating set.
A)
 x  9 y  2 z  50

 6 x  2 y  9 z  14
5 x  6 y  4 z  43

 x  9 y  2 z  50

 
56 y  3z  286
R3  5 R1
  39 y  6 z  207

R2  6 R1
 x  9 y  2 z  50


2184 y  117 z  11154
56 R3
  2184 y  336 z  11592

39 R2
 x  9 y  2 z  50

  2184 y  117 z  11154

 219 z  438

R3  R2
So from the last "row" we can see that  219 z  438  z  2 now if we "back substitute" this answer for z into an "earlier"
(easier) version of "row" two , 56 y  3 z  286  56 y  3  2   286  56 y  6  286  56 y  280  y  5.
Finally, substituting BOTH of these results for z and y into the first "row" we obtain...
x  9 y  2 z  50  x  9  5   2  2   50  x  45  4  50  x  49  50  x  1
So the answer to our system is 1, 5, 2 
B)
4 x  5 y  6 z  26

5 x  3 y  7 z  14
 x  6 y  z  40

 x  6 y  z  40

 5 x  3 y  7 z  14
4 x  5 y  6 z  26

1  x  6 y  z  40
 x  6 y  z  40
R2
3 

  27 y  12 z  186  
9 y  4 z  62
R3  4 R1
 19 y  10 z  134
 19 y  10 z  134


x  6 y  z  40
x  6 y  z  40


R3  R2
19 R2


   171y  76 z  1178    171y  76 z  1178
9 R3
 171y  90 z  1206

 14 z  28


So from the last "row" we can see that  14 z  28  z  2. Substituting that into an "earlier" version of "row" two
R1  R3
R2  5 R1
27 y  12 z  186  27 y  12  2   186  27 y  24  186  27 y  162  y  6. Substituting BOTH of these
into the first "row" we obtain x  6 y  z  40  x  6  6   2  40  x  36  2  40  x  38  40  x  2
So the answer to our system is  2, 6, 2 
 x  2y  z  3

 1
C)  x  4 y
2 x
 4 z  12

D)  x  y  z  7

 3x  2 y  z  3
 x  6 y  3 z  25

x  2 y  z  3

  2y  z  2
R3  2 R1
 4 y  2z  6

R2  R1
x  y  z  7

   5 y  2 z  18
R3  R1
  5 y  2 z  18

R2  3 R1
x  2 y  z  3

  2y  z  2
NO SOLUTION

02

R3  2 R2
x  y  z  7

   5 y  2 z  18

00

R3  R2
Infinite generating set is...
2
18
5 y  2 z  18  5 y  2 z  18  y   z 
5
5
2
18
3
17
x  y  z  7  x   z   z  7  5 x  2 z  18  5 z  35  5 x  3 z  17  x   z 
5
5
5
5
17 2
18 
 3
 z  , z  , z
5 5
5 
 5
Day 51 Homework: Topics: Gaussian Elimination with Back Substitution AND Gauss-Jordan Elimination
Supplemental Reading: There are resources out there on this BUT everyone seems to do it slightly differently (particularly notation
wise) SO, I recommend that you just refer heavily to your class notes!
1. Solve problems A and B below using the Gauss – Jordan method and use EITHER Gaussian Elimination OR the Gauss – Jordan
method for problems C and D. Be sure to SHOW your row operations between steps (on top and bottom of your arrows, just like I did
in class). If you find that your system has an infinite number of solutions be sure to find the infinite generating set.
A)
 3x  y  z  3

 2 x  3 y  z  8
2 x  y  2 z  2


x

R1  2 R2

 
R3  5 R2




 x  2 y  2z  5

  2 x  3 y  z  8
2 x  y  2 z  2

R1  R2

x
7
R3 
17 





4
1
z
7
7
5
18
y z 
7
7
17
34
z
7
7

x  2 y  2z  5
 7 y  5 z  18
5 y  6z  8


 
R3  2 R1


R2  2 R1
4
1
z
7
7
5
18
y z 
7
7
z  2

x

 
5
R2  R3 
7

4
R1  R3
7
y
1
4
So the solution to our system is 1, 4, 2 
z  2
B) 
x  y  2z  7
3z  6
x  y  2z  7 R R x 
R2  R1
1
2



y  z  1
  x  2 y  z   8 R
 y  z  1 R 

6 R
3  2 R1
 2x  4 y  z  1
 6 y  3z  15 3 2 
 9 z  9



3
x
R1  3 R3

 
y
 2 So the answer to our system is  3, 2,1
R2  R3

z
 1

C)  x  2 y  z  4

4 x  3 y  z  8
5 x  y
 12

x  2 y  z  4

   11 y  5 z  8
R3  5 R1
  11 y  5 z  8

R2  4 R1
 x  2 y  2z  5

5
18


y z 
7
7

5 y  6z  8

1
 R2
7
3z  6
x 


y  z  1

z 1

1
 R3
9
x  2 y  z  4

   11 y  5 z  8 must find infinite generating set

00

R3  R2
5
8
z
11 11
8
10
16
5
x  2 y  z  4  x  2  z    z  4  x  z   z  4  11x  10 z  16  11z  44
11
11
11
11


11 y  5 z  8  11 y  5 z  8  y 
 11x  z  28  x 
D)  x  3 y  4 z  1

 3x  4 y  5 z  3
 x  8 y  11z  2

1
28
z
so our answer is
11
11
x  3y  4z  1

   5y  7z  0
R3  R1

5y  7 y  1

R2  3 R1
28 5
8 
1
 z  , z  ,z
11 11 11 
 11
x  3y  4z  1

   5y  7z  0
NO SOLUTION!

0 1

R3  R2
PART II (Still Day 51)
1. Solve problem B using the Gauss – Jordan method and do the others using EITHER Gaussian Elimination with Back Substitution
OR the Gauss – Jordan method. Be sure to SHOW your row operations between steps (on top and bottom of your arrows, just like I
did in class). If you find that your system has an infinite number of solutions be sure to find the infinite generating set.
A) 5 x  y  3z  2

9 x  2 y  5 z   7
 2x  y  z  3

 11
 11
 11
 x  4y
 x  4y
 x  4y
R2  5 R3
R2  9 R1



 9 x  2 y  5 z   7   34 y  5 z  92    y
 3
R3  2 R1
 2x  y  z  3

  7 y  z  19

  7 y  z  19

This problem is unusual as we can now solve the system. Take the second "row" and solve for y  y  3
Now substitute that into the first and third rows to find x and z respectively.
R1  3 R3
x  4y  11  x  4  3 =11  x  1
 7 y  z  19  7  3  z  19  21  z  19  z  2
and
So the solution to our system is  1,3, 2 
B)  3x  y  z   5
 x  2y  4 z   6 R  2 R  x  2y  4 z   6  R  x  2y  4 z   6
R1  R3
2
1
2




y  9 z  23
 2 x  3 y  z  11   2 x  3 y  z  11 R 
  y  9 z  23  

2
R
3
1
2 x  y  3 z   1
2 x  y  3z   1


5 y  11z  13
5 y  11z  13




 14 z  40 1 R3  x
 14 z  40
 2
x
x
R1 14 R3
R1  2 R2
34 


 
y  9 z  23  
y  9 z  23  
y
4
R3  5 R2
R2  9 R3



34 z  102
z 3
z 3



So the solution to our system is  2, 4,3
 x  2y  z  8

C) 2 x  y  z  4
8 x  y  z  2

D)
 2 x  y  3z  1

 x  4y  z  6
 4 x  7 y  z 13

x  2 y  z   8

   5 y  3z  20
R3 8 R1
  15 y  9 z  66

R2  2 R1
 x  4y  z  6

 2 x  y  3z  1
4 x  7 y  z  13

R1  R2
x  2 y  z   8

   5 y  3z  20 No Solution!

06

R3  3 R2
 x  4 y  z  6
1
x  4 y  z  6
R2  
9
5
11


  9 y  5 z  11  
y z 
R3  4 R1
9
9
 9 y  5 z  11


 9 y  5 z  11
R2  2 R1
11
10

 z
x
9
9

R1  4 R2
5
11

 
y z 
So we can see that we have an infinite number of solutions since we got 0 = 0
R3  9 R2
9
9

00



in the bottom "row". Note that I chose to do Gauss-Jordan here to demonstrate that when you end up with
an infinite number of solutions Gauss-Jordan allows you to almost immediately come up with the infinite
11
10
11 10
z
 x z
9
9
9
9
11 
 11 10 5
So our solution is  z  , z  , z 
9
9
9
9 

generating set. x

5
11
5
11
y z   y  z
9
9
9
9
Day 52 Homework: Topics: Graphing linear inequalities and Systems of linear inequalities.
Supplemental Reading:
1. http://www.purplemath.com/modules/ineqgrph.htm
2. http://www.purplemath.com/modules/syslneq.htm
1. Sketch the graph of each individual inequality.
A. y  2 x  4
B) 2x  3 y  6
C)
 x  1
2
2. Sketch the graph for the following systems of inequalities.
3x  2 y  6

A.  x  4 y  2
 2x  y  3

x  2 y  4
x  y  1

B. 
x  0
 y  0
3. List the “vertices” in each of your graphs in 2A and 2B above.
 2,0  0,3  2, 1
A. ______________________________________
 0,0  0, 2  2,1 1,0
B. ___________________________________
  y  2  4
2
Day 53 Homework: Topics : Linear Programming
Supplemental Reading:
1. http://www.purplemath.com/modules/linprog.htm
1. Sketch the region determined by the constraints. Then find the minimum and maximum values of the objective function and where
they occur, subject to the indicated constraints.
A) Objective function: z = 6x + 10y
Constraints:
x  0

y  0
2 x  5 y  10

z  6 x  10 y
 0, 0   z  6  0   10  0   0  Minimum value!
 0, 2   z  6  0   10  2   20
 5, 0   z  6  5  10  0   30  Maximum value!
B) Objective function: z = 4x + 5y
Constraints:
x  0
y  0


x  y  8
3 x  5 y  30
z  4x  5 y
8, 0   z  4 8  5  0   32  Minimum value!
 5,3  z  4  5  5  3  35
10, 0   z  4 10   5  0   40  Maximum value!
(Day 53 Homework continued)
2. A manufacturer wants to maximize the profit for two types of nuts that he sells. A box of Peanuts yields a profit of $1.50 per box
and a box of Cashews yields a profit of $2.00 per box. Market tests and available resources have indicated the following
constraints……
*** The combined production level should not exceed 1200 boxes per month
***The demand for a box of Cashews is no more than one half the demand for a box of Peanuts.
***The production level of a box of Peanuts should be less than or equal to 600 boxes plus three times the production level of a box of
Cashews.
Find the number of boxes of Peanuts and the number of boxes of Cashews that will maximize the profit. Let x = the number of boxes
of Peanuts and y = the number of boxes of Cashews. Also, let each marking on the x and y axes of the graph below represent 100
boxes.
Objective Function (Profit) P  1.5 x  2 y
Constraints

x  0

y  0

 x  y  1200

1
y  x
2

 x  600  3 y
1200
600
800, 400
1050,150
600
1200
P  1.5 x  2 y
 0, 0   P  1.5  0   2  0   $0
800, 400   P  1.5 800   2  400   $2000 
1050,150   P  1.5 1050   2 150   $1875
 600, 0   P  1.5  600   2  0   $900
Max Profit!!!!
The profit will be maximized with 800 boxes of Peanuts and 400 boxes of Cashews
Day 54 Homework: Topics: Matrix Size (or order), Augmented Matrices, Gauss-Jordan with Matrices
Supplemental Reading:
1. http://www.purplemath.com/modules/matrices.htm (first two pages only)
2. http://www.purplemath.com/modules/mtrxrows.htm (you can read this BUT realize their notation is slightly different than what
we have been doing)
1. Write the size (or order) of the following matrices at the lower right corner of each matrix.
1 3 5 
 2 7 1

 2 x3
1 3 
5 2 


7 1  3 x 2
3 
 2
 
1  3 x1
0
 2
 1  2

2 x2
1
2 31x 3
2. Write the following system as an augmented matrix…
 x  2y  z  7
 1 2 1


 3x  y  2 z  11   3 1  2
2 x  y  7 z  10
 2 1  7

7
 11
10 
3. Write the following augmented matrix as a system of equations…
3
 1 2
 0 3 2

 1 4
0
 5   x  2 y  3x  5

1   
3y  2z  1

3  x  4 y
 3
4. Use matrices and the Gauss-Jordan method to solve the following systems of equations.
A)  2 x  3 y  3
3  R1  R2  1  1
 2 3

 

2
 1
2
 x  2 y  1  1
 1
R1  R2
0
3
 1
 
So our solution is  3,1
1
1 
 0
2  R2  R1  1
 
 1 
 0
1
1
2
1
B)
3
 x  2 y  3z  11
 1 2

 2 3

2
x

3
y

5
z

1

5


 3x  3z  z  4
 3 3  1

0
 19
 35  1 R3 1
1
R1  2 R2
89

 0
1
 11
 23  0
R3  9 R2
0
0
0
89
178 
11
1
4 
0
1
0
1
 0
R3  3 R1
0
R2  2 R1
 19
 11
1
2
3
1
11
9  10
 35 
 23 
2 
11 
1
 R2
23   0
 29 
0
0
0
1
R1 19 R3

 0
1
0
R2 11R3 
0
0
1
2
3
1  11
9  10
11 
 23 
 29 
3
 1 So our solution is  3, 1, 2 
2 
Day 55 Homework: Topics: Solving systems using Matrices.
Supplemental Reading: (None) We will be following “similar” steps to what we did in our notes (and homework) on days 50 – 51
BUT we will create an augmented matrix first, do the work without writing all of the variables during every step and re-insert the
variables at the end.
1. Use matrices and the Gauss-Jordan method to solve the following systems of equations.
A)
 4 x  2 y  3z  5

8 x  y  z   5
 2x  y  2z  5

 2
R1  R2
  0
R3  4 R2
 0
 2 1 2
  8  1 1
 4  2  3
0 1
1
3
0
5
5
 5 
5 
R1  R3
0
5 
15 
 2
  0
 0
1
R3
5
 2 1 2
  0 3 9
R3  2 R1 
 0  4  7
0 1
1
3
0
1
5  1 R2  2 1 2
3
15    0 1 3
 0  4  7
 5 
R2  4 R1
0
5 
3 
 2
R1  R3
  0
R2  3 R3
 0
0
0
1
0
0
1
5
5 
 5

1
3  1 R1 
2 
 4    0
 0
3 


0
0
1
0
0
1
3

So our solution is  , 4,3 
2

B)
3
 x  2 y  3z  11
 1 2


2 3
 4 x  2 y  3z  4   4
3 x  3 y  z  4

1

 3  3

 1

R1  2 R2
  0
R3  3 R2 

 0

7

1
3 2 
 R3 
5
20  5 
1
 0

2
3


5
 0
0
9


2
 73 7 18 
So our solution is  ,  , 
 15 3 5 
C)
0
2

2 1
 x  2 y  z  8
 1

 2 1
2
x

y

z

4

1


8 x  y  z  2
 8  1  1


 1

R1  2 R2
  0
R3 15 R2 

 0

1
5
3
1 
5
0 0
0
11
3
 1 2
R2  4 R1


4   0  6  15
R3  3 R1
 0 3
 4 
10
0
2
1
5
2
0
1
 8
4 
 2 
7
3

20 
3

18 
5 

11
 40 
29 

 1

R1  2 R3
  0
5 
R2  R3
2 
 0

2 1
 1
  0  5
3
R3  8 R1
 0 15  9
R2  2 R1
 1 2

 0 1

 0 3

73 
0
15 

7
0 
3

18 
1
5 
1
 R2
6
0
1
0
8 
20 
 66 
 1

 0

 0

1
 R2
5
11 
20 
3
29 
3
5
2
10
2 1
3
1 
5
15  9
8 

4 

 66 

 2

 4  So our system has NO SOLUTION !


6

D) 3x  4 y  z  6
6
3  4  1


1
 1 
2 x  y  z   1   2  1
4 x  7 y  3z  18
 4  7  3
18

1 1  3
2
7
0
1
1
R2
R1  3 R2
5
 0 1
1
 3   0 1
1
R3  5 R2
0 5
0 0
5  10 
0
1  3  2
  2  1
1
 4  7  3
R1  R2
 2
 2 
5 
No Solution
7
 1 
18
1  3  2
 0 5
5
R3  4 R1 
0 5
5
R2  2 R1
7
 15 
 10 
3
2

 4
3


Day 56 Homework: Topics: Equal Matrices, Addition and Subtraction of Matrices, Multiplication of Matrices.
Supplemental Reading: 1. http://www.purplemath.com/modules/matrices3.htm
2. http://www.purplemath.com/modules/mtrxadd.htm
3. http://www.purplemath.com/modules/mtrxmult.htm
4. http://www.coolmath.com/algebra/24-matrices/02-adding-subtracting--01.htm
5. http://www.coolmath.com/algebra/24-matrices/03-scalar-multiplication-01.htm
6. http://www.coolmath.com/algebra/24-matrices/04-multiplying-matrices-01.htm
1. Find x, y and z
3
0

a b 
2. M  

c d 
2y 5
7  3

1
3 x  2   4  2 z
e f
N 
h i
7
1
7 
 1 
g
j 
A) Find M  N
B) Find N  M
 a b  e f g 
c d    h i j 

 

 ae  bh af  bi ag  bj 
ce  dh cf  di cg  dj 


e f g   a b 
 h i j   c d 

 

Can't do it! (# of columns in N  # of rows in M)
 1 2 
3. A  

 3 5
 2  3
B
4 
0
0 2
1
C
2

3
1 

 1 3 
D   2 5
 1 2 
I. Find 2A – 3B
4  6  9   8 13
 1 2 
 2  3  2
2
 3





4   6 10  0 12   6  2 
 3 5
0
II. Find A  B
 1 2   2  3  2 11

 3 5   0
4   6 11

 
III. Find B  A
 2  3  1 2   11  11


0
4   3 5  12
20 

IV. Find B  C
0 2   4
9 1
 2  3 1
0
   2  3 1   8  12 4 
4

 
 

V. Find C  B
0 2  2  3
1
 2  3 1   0
4 

 
Can't do it! (# of columns in C  # of rows in B)
VI. Find C  D
 1 3 
0 2 
7
1
  1
 2  3 1    2 5   7  7 

  1 2 



VII.
 1
 2

 1
Find D  C
3
 5  9 1
0 2 
1

5  
 12  15 9
2  3 1  

 5  6 4 
2 
Day 57 Homework: Topics: Inverse Matrices
Supplemental Reading:
1. http://www.purplemath.com/modules/mtrxinvr.htm (first page only) (remember, their notation is slightly different than what we
have been using)
2. http://www.coolmath.com/algebra/24-matrices/06-inverse-matrices-01.htm
3. http://www.coolmath.com/algebra/24-matrices/07-solving-systems-using-matrices-01.htm
1 3 
1
1. A  
 Find A
2
4


1 3
2 4

1

0

3

2

2
A1  

 1 1

2 
1 0  R2  2 R1 1
3
 
0 1
0

2


1
 
0

3
2

1
1  
2 
2
0
R1  3 R2
1
1
 R2
2
1 0
 2 1
0
1
1  
2
3
1
1
2. Use your result from problem 1 to solve the following system.
 x  3y  9

 2 x  4 y  14
1 3  x 
 9
 2 4    y   14

  
 
AX  B
A1 AX  A1 B
IX  A1 B
3

2
x  
2  9
  
 y  
   1  1  14 

2 
 x  3 
 y    2  So our answer is  3, 2 
   
2
5
3. M  
3

1 


1
 
0

5
R3
11
4.
1

0
5

3
5

11
11 
2
5
1
 6  9
N 
6 
 4
 6 9
 4
6

M 1
1
2
R 
1 0  5 1 1

5

0 1 
3  1
2
5
3  1


Find
Find
1

0
5

0 1 

1
 
0

2
R1  R2
5
0
1
2

1
5
 
11
0 

5
1
2
11
11 

3
5
 
11
11 
R2  3 R1
1

0
5

3

1

5
M 1
2
1
11
11 


3  5 
11
11 
N 1
1
3
R 
1 0 6 1  1 

2

0 1
6
 4
N 1 DOES NOT EXIST !
X  A 1 B
1

0
6

0 1 
3

 1 2
 
 0
0

R2  4 R1
1

0
6

2 
1
3 
PART II (still Day 57)
1  1 0 
1. A  1
0  1  Find
6  2  3
1  1 0
1
0 1

6  2  3
 1
  0
R3  4 R2 
 0
0
R1  R2
 1 1 0
  0
1 1
R3  6 R1 
 0
4 3
1 0
 1
R1  R3

1
1 0    0
R2  R3
 0
 2  4 1 
1 0 0
0 1 0 
0 0 1
1
0
1 1
0
1
A1
R2  R1
1 0 0
 1 1 0 
 6 0 1
0 0
 2  3 1
1 0
 3  3 1
0 1
 2  4 1
 2  3 1
A   3  3 1
 2  4 1 
1
2
0
 1

2. A   3  1
2  Find
 2 3  2 
A1
2
0
 1
 3 1
2

 2
3 2
2
0
 1
  0  7
2
R3  2 R1
 0
7 2
 1
1
 R2 
7
 0

 0

A1
2
1 0 0
0 1 0 
0 0 1 
0
1
2
7
7 2
3
7
2
1 
DOES NOT EXIST !
R2  3 R1
0 0

1

0
7

0 1

 1

R1  2 R2
  0
R3  7 R2 

 0

1 0 0
 3 1 0 
2 0 1 
4
1
2
0
7
7
7
2
3
1
1 

7
7
7
0
0
1 1

0

0


1

Day 58 Homework: Topics: Finding Determinants for 2x2 and 3x3 matrices.
Supplemental Reading:
1. http://www.purplemath.com/modules/determs.htm (Note: on the second page they show a “short cut” specifically for 3x3
determinants BUT that method does not work for 4x4 and bigger SO, please do the 3x3 problems using the “expanding by cofactors”
method that I taught…see the next reading!)
2. http://www.purplemath.com/modules/minors.htm
1. Find the determinant of each matrix below…
 2 3
A) 

 1 5 
6
 2
B) 

 1  3 
0 0
C) 

0 0
1 0 
D) 

 0 1
 5 2    3 1
 2  3   6  1
 0  0    0  0 
11   0  0 
10  3
6  6
00
1 0
7
0
0
1
2. Find the determinant of the matrix as indicated in each part.
2 1
0
3  1 2 


 4 0 1
A) Find the determinant by expanding about row 2
3  C21  1  C22  2  C23
3  1
2 1
3  1
3
M 21  1 1
2 2
M 22  2  1
23
M 23
2 1
2
4 0 1
5 0
 1 1
 2  1
0 1
4 1
4 0
3  1 2   11 4   2  1 8 
6  4  16
14
B) Find the determinant by expanding about column 2
2  C12  1  C22  0  C32
2  1
1 2
2  1
3
M 12  1 1
2 2
3 2
4 0 1
 1 1
4 1
4 1
2  1 5   11 4 
10  4
14
M 22  0
Day 59 Homework: Topics: Cramer’s Rule (using Determinants to solve systems)
Supplemental Reading:
1. http://www.purplemath.com/modules/cramers.htm
2 x  3 y  1
1. Solve 
by using Cramer’s Rule if possible (if not then use another method)
 x  3y  5
D
2 3
  2  3   31  6  3  9
1
3
Dx 
1 3
 1 3   3 5   3  15  18
5
3
Dy 
2 1
  2  5   11  10  1  9
1 5
x
Dx 18

2
D
9
y
Dy
D

9
1
9
So our solution is  2,1
3 x  2 y  7
2. Solve 
by using Cramer’s Rule if possible (if not then use another method)
 2 x  3 y  4
3 2
  3 3   2  2   9  4  13
2
3
D
Dx 
7 2
  7  3   2  4   21  8  13
4
3
Dy 
3 7
  3 4    7  2   12  14  26
2 4
x
Dx 13
 1
D 13
y
Dy
D

26
 2
13
So our solution is 1, 2 
2 x  2 y  7
3. Solve 
by using Cramer’s Rule if possible (if not then use another method)
3 x  3 y  4
D
2 2
  2  3   2  3  6  6  0
3 3
So Cramer's Rule won't work!!!!
* 3

2 x  2 y  7  6 x  6 y  21


*2
6 x  6 y  8
3x  3 y  4  

0  29 No Solution !
 6x  2 y  8
4. Solve 
by using Cramer’s Rule if possible (if not then use another method)
9 x  3 y  12
6 2
D
  6  3   2  9   18  18  0 So Cramer's Rule won't work!!!!
9
3

 6x  2 y  8

9 x  3 y  12
*3
  18 x  6 y  24

*2
18 x  6 y  24
 
0  0 we must find the infinite generating set !
6x  2 y  8  6x  2 y  8  x 
1
4
y
3
3
4 
1
 y  , y
3 
3