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

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

, TABLE OF CONTENTS wj wj




Chapter 1:
wj Linear Equations and Straight Lines
wj wj wj wj 1–1


Chapter 2:
wj Matrices 2–1


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


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


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


Chapter 6:
wj Probability 6–1


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


Chapter 8:
wj Markov Processes wj 8–1


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


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


Chapter 11:
wj Logic 11–1


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

, Chapter 1
wj




Exercises 1.1 5
wj
6. Left 1, down
w j wj wj w j


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


(2, 3)
wj


x
x
(–1, – 2 wj



5
wj) wj




7. w j Left 20, up 40 wj wj wj

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

(–20, 40)
(–1, 4)
wj
wj




x
x




8. w j Right 25, up 30 wj wj wj

3. Down 2
w j wj
y
y


(25, 30) wj




x
x
(0, –2)
wj




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

wj


y (—2, 2). wj




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



(3,—2).
x
(2, 0)
wj
1 wj

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


3
on the line. wj wj




5. Left 2, up 1 1 wj


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




(–2, 1) wj
x



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

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




1 wj 24. 0 = 5
w j wj wj


13 —2x + y = —1 Substitute the x and y
wj wj wj wj w j wj wj wj wj
no solution wj

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


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

When x = 0, y =
y' ı
w
j w
j wj jw w
j wj w
j wj w
j w j wj wj wj wj wj wj



5y-
' ı
jw

wj
intercept: (0, 5) wj wj

2 J y2J 3
wjwjw j jw



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



theline. jw x-
f 1h f1h intercept: (7, 0)0 wj wj jw w



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



no solution wj



'y3 ıJ 'y3 ıJ y-intercept: none wj

wjwjwj



on the line.
wj wj
26. 0 = –8x
w j wj wj




15. m = 5, b = 8
w j wj wj wj wj wj
x=0 wj wj


x-intercept: (0, 0) wj wj




16. m = –2 and b = –6
w j wj wj wj wj wj wj
y = –8(0) wj wj


y=0 wj wj



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



3

2 2 1wj


y=
wj

x + 0; m = , b = 0
wj
27 0=
wjx –1 wj wj wj

18 wj wj wj w
j wj wj wj wj wj wj

3
3 3 .
. x=3 wj wj




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


1 wj

7 y = —14x + 21
wj wj wj wj w
j
y = (0) – 1
wj wj wj wj


3
y = —2x + 3
wj wj wj wj
y = –1
wj wj


y-intercept: (0, –1) wj wj

20 x— y =3
wj wj wj wj y
. —y = —x + 3 wj wj wj w
j




y = x —3
wj wj wj jw




(3, 0) wj
x
21. wjwjw j 3x = 5 wj wj


5 (0, –1) wj

x= wj wj


3
1 2
28. When x = 0, y = 0.
– x+ y = 10
wj wj wj wj wj wj

22 2
wj

3
wj w
j

When x = 1, y = 2.
. wj wj wj wj wj wj


2 wj 1 wj y
y= wj wj x +10wj


3 2
3 wj

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

4 x
(0, 0) wj




23. 0 = —4x + 8
wj wj wj wj




4x = 8 wj wj



x =2 wj wj


x-intercept: (2, 0) wj wj


y = –4(0) + 8
wj wj wj wj



1-2 Copyright © 2023 Pearson Education, Inc wj wj wj wj wj



.