Solution Manual for Finite Mathematics and Its Applications (13th Edition) – Larry J. Goldstein – Full Solutions (Chapters 1–12)

SOLUTION MANUAL

Finite Mathematics & Its Applications

13th Edition by Larry J. Goldstein,

Chapters 1 - 12, Complete

, Contents
Chapter 1: Linear Equations and Straight Lines
jd jd jd jd jd 1–1
Chapter 2: Matrices
jd 2–1

Chapter 3: Linear Programming, A Geometric Approach
jd jd jd jd jd 3–1

Chapter 4: The Simplex Method
jd jd jd 4–1

Chapter 5: Sets and Counting
jd jd jd 5–1

Chapter 6: Probability
jd 6–1

Chapter 7: Probability and Statistics
jd jd jd 7–1

Chapter 8: Markov Processes
jd jd 8–1

Chapter 9: The Theory of Games
jd jd jd jd 9–1

Chapter 10: The Mathematics of Finance
jd jd jd jd 10–1

Chapter 11: Logic
jd 11–1

Chapter 12: Difference Equations and Mathematical Models
jd jd jd jd jd 12–1

, Chapter 1
jd




Exercises 1.1 5
jd
6. Left 1, down
j d jd jd j d


2
1. Right 2, up 3 jd jd jd
y
y


(2, 3)
jd


x
x
( –1,j d –jd52jd )


7. Left 20, up 40
j d jd jd jd

2. Left 1, up 4
jd jd jd
y
y

(–20, 40)
(–1, 4)
jd
jd




x
x




8.j d Right 25, up 30 jd jd jd

3. Down 2
j d jd
y
y


(25, 30) jd




x
x
(0, –2)jd




9. Point Q is 2 units to the left and 2 units up or
4. Right 2
jd jd jd jd jd jd jd jd jd jd jd jd

jd


y (—2, 2). jd




10. Point P is 3 units to the right and 2 units down or
jd jd jd jd jd jd jd jd jd jd jd jd



(3,—2).
x
(2, 0)
jd
1 jd

11. —2(1) + (3) = —2 +1 = —1so yes the point is
jd jd jd jd jd jd jd jd jd jd jd


3
on the line.
jd jd




5. Left 2, up 1 1 jd


12. —2(2) + (6) = —1 is false, so no the point is not
jd jd jd
jd jd jd jd jd jd jd jd jd jd jd jd
y
3
on the linejd jd




(–2, 1) jd
x



1-1

, Chapter 1: Linear Equations and Straight Lines
jd jd jd jd jd jd ISM: Finite Math
jd jd




1 jd 24. 0 = 5
j d jd jd


13 —2x + y = —1 Substitute the x and y
jd jd jd jd j d jd jd jd jd
no solution jd

3
. x-
coordinates of the point into the equation: jd jd jd jd jd jd


f 1 hı f h intercept: none
' ,3 →—2 ' 1 ı +1(3)=—1→—1+1=—1 is
jd jd
jd jd jd

When x = 0, y = 5
y' ı ' ı
jd jd jd jd jd jd jd jd jd jd jd jd jd jd jd jd jd

jd
y-intercept: (0, 5) jd jd

2 J y2J 3 jdjdj d jd



a false statement. So no the point is not on t
jd jd jd jd jd jd jd jd jd jd 25. When y = 0, x = 7x
jd jd jd jd jd jd jd jd



heline. jd -
f 1h f1h intercept: (7, 0)0 jd jd jd jd



14 —2 ' ı + ' ı (—1) =—1 is true so yes the point is jd jd jd jd jd jd jd jd jd =7
. jd



no solution jd



'y3 ıJ 'y3 ıJ y-intercept: none jd

jdjdjd



on the line. jd jd
26. 0 = –8x
j d jd jd




15. m = 5, b = 8
j d jd jd jd jd jd
x=0 jd jd


x-intercept: (0, 0) jd jd




16. m = –2 and b = –6
j d jd jd jd jd jd jd
y = –8(0)
jd jd


y=0 jd jd



17. j d y = 0x + 3; m = 0, b = 3
jd jd jd jd jd jd jd jd jd jd
y-intercept: (0, 0) jd jd




2 2 1 jd


y = x +0; m = , b = 0
jd jd
27 0=jd x –1 jd jd jd

18 jd jd jd jd jd jd jd jd jd jd

3
3 3 .
. x=3 jd jd




19. 14x +7y = 21
j d jd jd jd jd jd
x-intercept: (3, 0) jd jd


1 jd

7 y = —14x +21
jd jd jd jd jd
y = (0) – 1
jd jd jd jd


3
y = —2x +3
jd jd jd jd
y = –1
jd jd


y-intercept: (0, –1) jd jd

20 x— y =3
jd jd jd jd y
. —y =—x +3 jd jd jd jd




y = x —3
jd jd jd jd




(3, 0) jd
x
21. 3x = 5
jdjdj d jd jd


5 (0, –1) jd

x= jd jd


3
1 2
28. When x = 0, y = 0. jd jd jd jd jd jd

22 – x+ jd y =1 0jd jd


. 2 3 When x = 1, y = 2. jd jd jd jd jd jd


2 jd 1 jd y
y= jd jd x +10 jd


3 2
3 jd

y= jd jd x +15 jd
(1, 2) jd

4 x
(0, 0) jd




23. 0 = —4x +8
jd jd jd jd




4x = 8 jd jd



x =2 jd jd


x-intercept: (2, 0) jd jd


y = –4(0) + 8
jd jd jd jd


y=8 jd jd



y-intercept: (0, 8) jd jd



1-2