SOLUTION MANUAL Finite Mathematics & Its Applications 15th Edition by Larry J. Goldstein, Chapters 1 - 12, Complete

SOLUTION MANUAL
Finite Mathematics & Its Applications
15th Edition by Goldstein Chapters 1 - 12

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

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

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

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

Chapter 6: Probability
xp 6–1

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

Chapter 8: Markov Processes
xp xp 8–1

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

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

Chapter 11: Logic
xp 11–1

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

, Chapter 1
xp




Exercises 1.1
xp 5
6. Left 1, down
x p xp xp x p




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


(2, 3)
xp


x
x




7. Left 20, up 40
2. Left 1, up 4
x p xp xp xp

xp xp xp
y
y


(–20, 40)
(–1, 4)
xp
xp




x
x




8. x p Right 25, up 30xp xp xp

3. Down 2
x p xp
y
y



(25, 30)xp




x
x
(0, –2)xp




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


y (—2, 2). xp




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



(3,—2).
x
(2, 0)
xp



1 xp

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


3
on the line. xp xp




5. Left 2, up 1 1 xp

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


y xp xp xp xp xp xp xp xp xp xp xp xp


3
on the line xp xp




(–2, 1)xp
x

1-1

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




1 xp
24. 0 = 5
x p xp xp


13. —2x + xp xp y = —1 Substitute the x and y
xp xp x p xp xp xp xp
no solution xp

3 x-
coordinates of the point into the equation: xp xp xp xp xp xp
intercept: none W
f 1 ıh f h
xp xp


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

y' ı
xp xp xp xp xp xp xp xp xp xp


intercept: (0, 5) xp xp


ı
2 'J y2J 3 xpxpx p
xp
xp


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



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



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


no solution xp




'y3 ıJ 'y3 ıJ xpxpxp
y-intercept: none xp




on the line. xp xp
26. 0 = –8x
x p xp xp




15. m = 5, b = 8
x p xp xp xp xp xp
x=0 xp xp



x-intercept: (0, 0) xp xp



16. m = –2 and b = –6
x p xp xp xp xp xp xp
y = –8(0)
xp xp


y=0 xp xp



17. x p y = 0x + 3; m = 0, b =
xp xp xp xp xp xp xp xp xp xp y-intercept: (0, 0) xp xp



3

2 2 1 xp


18. y= xp xp
xp

x +0; m = , b = 0
xp xp xp xp xp
xp

xp xp xp
27. 0 = xp xp x –1 xp xp



3 3 3
x=3 xp xp




19. 14x+7y = 21
x p xp xp xp xp xp
x-intercept: (3, 0) xp xp


1 xp

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


3
y = —2x +3
xp xp xp xp
y = –1
xp xp




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

xp xp xp xp y
—y =—x +3 xp xp xp xp




y = x —3
xp xp xp xp




(3, 0) xp


21. xpxpx p 3x = 5 xp xp
x
5 (0, –1) xp


x= xp xp


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




22. – 2 x + 3 y =10 xp xp xp

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


2 1 xp xp y
y = x +10 xp xp xp


3 2
3 xp

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

4 x
(0, 0) xp




23. 0 = —4x +8
xp xp xp xp



4x = 8 xp xp




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



.