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

SOLUTION MANUAL as




Finite Mathematics & Its Applications
as as as as




13th Edition by Larry J. Goldstein,
as as as as as




Chapters 1 - 12, Complete
as as as as

, Contents
Chapter 1: Linear Equations and Straight Lines
as as as as as 1–1

Chapter 2: Matrices
as 2–1

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

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

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

Chapter 6: Probability
as 6–1

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

Chapter 8: Markov Processes
as as 8–1

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

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

Chapter 11: Logic
as 11–1

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

, Chapter 1
as




Exercises 1.1 5
as
6. Left 1, down
a s as as a s


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



(2, 3)
as


x
x
( –1, a s –2as5

)
as




7. a s Left 20, up 40 as as as

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

(–20, 40) as
(–1, 4) as




x
x




8. a s Right 25, up 30 as as as

3. Down 2
a s as
y
y


(25, 30) as




x
x
(0, –2) as




9. Point Q is 2 units to the left and 2 units up or
as as as as as as as as as as as as

4. Right 2 as


y (—2, 2). as




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



(3,—2).
x
(2, 0)
as
1 as

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


3
on the line. as as




5. Left 2, up 1 1 as


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




(–2, 1) as
x



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

, Chapter 1: Linear Equations and Straight as as as as as ISM: Finite
as


Lines
as Math
as




1 as 24. 0 = 5
a s as as


13 —2x + as as y = —1 Substitute the x and y
as as a s as as as as
no solution as

3
. x-intercept: none as

coordinates of the point into the equation: as as as as as as


f 1 hı f h When x = 0, y =
' ,3 →—2 ' 1 ı +1(3)=—1→—1+1=—1 is
as as as as as as
as sa sa

5y-intercept: (0,
y' ı '
sa sa sa sa sa sa sa sa sa as as sa as



5)
ı
as

as
2 J y2J 3 as as a s sa



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



theline.
as sa 7 x-intercept: (7,
as sa as




f 1h f1 h 0)0 = 7
as sa as as



14 —2 ' ı + ' ı (—1) =—1 is true so yes the point is as sa as as as as as as as no solution
. as




'y3 ıJ 'y3 ıJ y-intercept: none as

as as as



on the line. as as
26. 0 = –8x
a s as as




15. m = 5, b = 8
a s as as as as as
x=0 as as


x-intercept: (0, 0) as as




16. m = –2 and b = –6
a s as as as as as as
y = –8(0) as as



y=0 as as



17. a s y = 0x + 3; m = 0, b
as as as as as as as as
y-intercept: (0, 0) as as



=3
as as




2 2 1 as


y=
as

x +0; m = , b = 0
as
27 0= x –1
as as as as

18 as as as sa as as as as as as

3
3 3 .
. x=3 as as




19. a s 14x +7y = 21 as sa as as as
x-intercept: (3, 0) as as


1 as

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


3
y = —2x +3as as as sa
y = –1
as as


y-intercept: (0, –1) as as

20 x— y =3as as as as
y
. —y =—x +3 as sa as sa




y = x —3 as as as sa




(3, 0) as

21. as as a s 3x = 5 as as
x
5 (0, –1) as

x= as as


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

22 – x+ as y =1 0 as sa


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


2 as 1 as y
y = as as x +10 as


3 2
3 as

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

4 x
(0, 0) as




23. 0 = —4x +8as as as sa




4x = 8 as as



x =2 as as


x-intercept: (2, 0) as as


y = –4(0) + 8
as as as as


y=8 as as




1-2 Copyright © 2023 Pearson Education, as as as as



Inc.
as