SOLUTION MANUAL
Finite Mathematics & Its Applications
13th Edition by Larry J. Goldstein,
Chapters 1 - 12, Complete
, Contents
Chapter 1: Linear Equations and Straight Lines
j j j j j 1–1
Chapter 2: Matrices
j 2–1
Chapter 3: Linear Programming, A Geometric Approach
j j j j j 3–1
Chapter 4: The Simplex Method
j j j 4–1
Chapter 5: Sets and Counting
j j j 5–1
Chapter 6: Probability
j 6–1
Chapter 7: Probability and Statistics
j j j 7–1
Chapter 8: Markov Processes
j j 8–1
Chapter 9: The Theory of Games
j j j j 9–1
Chapter 10: The Mathematics of Finance
j j j j 10–1
Chapter 11: Logic
j 11–1
Chapter 12: Difference Equations and Mathematical Models
j j j j j 12–1
, Chapter 1 j
Exercisesj1.1 5
6.j Leftj1,jdownj
2
1. Rightj2,jupj3 y
y
(2,j3)
x
x
( )
–1,j –j52j
7.j Leftj20,jupj40
2. Leftj1,jupj4 y
y
(–20,j40)
(–1,j4)
x
x
8.j Rightj25,jupj30
3.j Downj2 y
y
(25,j30)
x
x
(0,j–2)
9. PointjQjisj2junitsjtojthejleftjandj2junitsjupjor
4. Rightj2
y (—2,j2).
10. PointjPjisj3junitsjtojthejrightjandj2junitsjdownjor
(3,—2).
x
(2,j0) 1j
11. —2(1)j+j (3)j=j—2j+1j=j—1soj yesj thej pointj is
3
onjthejline.
5. Leftj2,jupj1 1j
y 12. —2(2)j+j (6)j=j—1jisj false,j soj noj thej pointj isj not
3
onjthejline
(–2,j1)
x
Copyrightj©j2023jPearsonjEducation,jInc. 1-1
, Chapterj1:jLinearjEquationsjandjStraightjLines ISM:jFinitejMath
1j 24.j 0j=j5
13 —2xj+j yj =j—1j Substitutej thej xj andj y nojsolution
3
. x-
coordinatesjofjthejpointjintojthejequation:
f 1j hıj f h j intercept:jnonej
' ,j3 →j—2 ' 1 ı +1 j (3)j=j—1j→j—1+1j=j—1j is Whenjxj=j0,jyj=j5jy-
y' ı 'j ı
intercept:j(0,j5)
2jjj J yj2J 3
ajfalsejstatement.jSojnojthejpointjisjnotjonjthejli 25.jWhenjyj=j0,jxj=j7jx-
ne. intercept:j(7,j0)j0j
f 1h f1 h =j7
14 —2 ' ı + ' ı (—1)j=j—1j isjtruejsojyesjthejpointjis nojsolution
.
'y3 ıJjjj'y3 ıJ y-intercept:jnone
onjthejline. 26.j 0j=j–8x
15.j mj=j5,jbj=j8 xj=j0
x-intercept:j(0,j0)
16.j mj=j–2jandjbj=j–6 yj=j–8(0)
yj=j0
17.j yj=j0xj+j3;jmj=j0,jbj=j3 y-intercept:j(0,j0)
2j 2j 1j
yj=j xj+j0;j mj=j ,j bj=j0 27 0j=j xj–j1
18 3
3 3 .
. xj=j3
19.j 14xj+j7jyj=j21 x-intercept:j(3,j0)
1j
7jyj=j—14xj+j21 yj =j (0)j–j1
3
yj =j—2xj+j3
yj=j–1
y-intercept:j(0,j–1)
20 xj—jyj =j3 y
. —yj =j—xj+j3
yj =jxj—j3
(3,j0)
21.jjj 3xj=j5 x
5 (0,j–1)
xj=j
3
1 2
28. Whenjxj=j0,jyj=j0.
22 – xj+ yj =j10
. 2 3 Whenjxj=j1,jyj=j2.
2j 1j y
yj =j xj+10
3 2
3j
yj =j xj+15 (1,j2)
4 x
(0,j0)
23. 0j=j—4xj+j8
4xj =j8
xj=j2
x-intercept:j(2,j0)
yj=j–4(0)j+j8
yj=j8
y-intercept:j(0,j8)
1-2 Copyrightj©j2023jPearsonjEducation,jInc.
