SOLUTION MANUAL Finite Mathematics & Its Applications 13th Edition by Larry J. Goldstein, Chapters 1 - 12, Complete

SOLUTION MANUAL
Finite Mathematics & Its Applications
13th Edition by Goldstein Chapters 1 - 12

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

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

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

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

Chapter 6: Probability
th 6–1

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

Chapter 8:
th Markov Processes th 8–1

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

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

Chapter 11: Logic
th 11–1

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

, Chapter 1
th




Exercises 1.1
th 5
6. Left 1, down
t h th th t h


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


(2, 3)
th


x
x
(–1, – 52)
th th
th




7. t h Left 20, up 40
th th th

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

(–20, 40)
th
(–1, 4) th




x
x




8. t h Right 25, up 30
th th th

3. Down 2
t h th
y
y


(25, 30) th




x
x
(0, –2) th




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


y (—2,2). th




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



(3,—2).
x
(2, 0)
th 1 th

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


3
on the line. th th




5. Left 2, up 1 1 th

12. —2(2) + (6) = —1 is false, so no the point is not
th th th


y th th th th th th th th th th th th


3
on the line th th




(–2, 1)
x
th




Copyright © 2023 Pearson Education, Inc.
th th th th th 1-1

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




1 th
24. 0 = 5
t h th th


13. —2x + y = —1 Substitute the x and y th th th th t h th th th th
no solution th

3 x-
coordinates of the point into the equation: th th th th th th
intercept: none Wh
f 1 hı f h
th th


' ,3 →—2 ' 1 ı +1(3)=—1→—1+1=—1 is
th
th th
en x = 0, y = 5y- th th th th th th th

y' ı ' ı
th th th th th th th th th th

th
intercept: (0, 5) th th


2 J y2J 3 ththt h th


a false statement. So no the point is not on theline.
th th th th th th th th th th th 25. When y = 0, x = 7x-
th th th th th th th th




f 1h f1h intercept: (7, 0)0 = th th th th



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



14. no solution th


'y3 ıJ 'y3 Jı ththth
y-intercept: none th




on the line. th th
26. 0 = –8x
t h th th




15. m = 5, b = 8
t h th th th th th
x=0 th th



x-intercept: (0, 0) th th



16. m = –2 and b = –6
t h th th th th th th
y = –8(0) th th


y=0 th th



17. t h y = 0x + 3; m = 0, b = 3
th th th th th th th th th th y-intercept: (0, 0) th th




2 2 1 th


18. y = x+0; m = , b = 0
th th
th

th th th th th
th

th th th
27. 0 = x –1 th th th th



3 3 3
x=3 th th




19. 14x+7y = 21
t h th th th th th
x-intercept: (3, 0) th th


1 th

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


3
y = —2x +3
th th th th
y = –1 th th



y-intercept: (0, –1)
20. x— y = 3
th th

th th t h th y
—y =—x +3 th th th th




y = x —3
th th th th


(3, 0) th

21. 3x =5
ththt h th th
x
5 (0, –1) th


x= th th


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


22. – x + y = 10 th th th

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


2 1 th th y
y = x +10 th th th


3 2
3 th

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

4 x
(0, 0) th



23. 0 = —4x +8th th th th




4x = 8 th th



x =2 th th


x-intercept: (2, 0) th th


y = –4(0) + 8
th th th th


y=8
th th


y-intercept: (0, 8) th th




1-2 Copyright © 2023 Pearson Education, Inc. th th th th th