Finite Mathematics & Its Applications | 13th Edition by Larry J. Goldstein | Complete Solution Manual for Chapters 1–12

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

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

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

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

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

Chapter 6: Probability
of 6–1

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

Chapter 8: Markov Processes
of of 8–1

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

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

Chapter 11: Logic
of 11–1

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

, Chapter 1
of




Exercises 1.1 5
of

6. Left 1, down
o f of of o f




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


(2, 3)
of



x
x
( –1, – 52 o f of )
of




7. Left 20, up 40
o f of of of


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

(–20, 40) of

(–1, 4) of




x
x




8. Right 25, up 30
o f of of of


3. Down 2
o f of
y
y


(25, 30) of




x
x
(0, –2) of




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


4. Right 2 of



y (—2,2). fo




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




(3,—2).
x
(2, 0)
of
1 of



11. —2(1) + (3) =—2+1= —1so yes the point isof of of fo of fo of of of of of




3
on the line. of of




5. Left 2, up 1 1 of




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

y
3
on the line of of




(–2, 1) of

x



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

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




1 o f
24. 0 = 5
o f of of




13 —2x + y = —1 Substitute the x and y of of of of o f of of of of
no solution of


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



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


When x = 0, y = 5y
y' ı ' ı
fo fo fo fo fo fo fo fo fo o f
of of of of of of fo


of

-intercept: (0, 5) of of



2 J y2J 3 ofofo f fo




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




line. intercept: (7, 0)0 of of fo of




f 1h f1h =7 of




14 —2 ' ı + ' ı (—1) =—1 is true so yes the point is no solution
.
of fo o f of of of of of of of




'y3 ıJ 'y3 ıJ y-intercept: none of

ofofof




on the line. of of
26. 0 = –8x
o f of of




15. m = 5, b = 8
o f of of of of of
x=0 of of




x-intercept: (0, 0) of of




16. m = –2 and b = –6
o f of of of of of of
y = –8(0)
of of




y=0 of of




17. y = 0x + 3; m = 0, b = 3
o f of of of of of of of of of of
y-intercept: (0, 0) of of




2 2 1 of




y = x+0; m = , b =0
of of

27 0 = x –1
of of of fo


18 of of fo fo of of of of of of

3
3 3 .
. x=3 of of




19. 14x+7y =21
o f of fo fo of of
x-intercept: (3, 0) of of




1 of



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




3
y =—2x +3
y = –1
of of of fo

of of




y-intercept: (0, –1) of of


20 x— y =3
fo of of of
y
. —y =—x +3 of fo of fo




y = x —3 of of fo fo




(3, 0) of
x
21. 3x =5
ofofo f of fo




5 (0, –1) of


x= of of




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



22 2
of


3
of fo


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




2 o f 1 o f
y
y= x +10
of of of




3 2
3 of


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


4 x
(0, 0) of




23. 0 =—4x +8 of of of fo




4x = 8 of of




x =2 of of




x-intercept: (2, 0) of of




y = –4(0) + 8
of of of of




y=8 of of




y-intercept: (0, 8) of of




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