SOLUTION MANUAL
Finite Mathematics & Its Applications
13th Edition by Larry J. Goldstein,
Chapters 1 - 12, Complete
, Contents
Chapterj 1: Linearj Equationsj andj Straightj Lines 1–1
Chapterj 2: Matrices 2–1
Chapterj 3: Linearj Programming,j Aj Geometricj Approach 3–1
Chapterj 4: Thej Simplexj Method 4–1
Chapterj 5: Setsj andj Counting 5–1
Chapterj 6: Probability 6–1
Chapterj 7: Probabilityj andj Statistics 7–1
Chapterj 8: Markovj Processes 8–1
Chapterj 9: Thej Theoryj ofj Games 9–1
Chapterj 10: Thej Mathematicsj ofj Finance 10–1
Chapterj 11: Logic 11–1
Chapterj 12: Differencej Equationsj andj Mathematicalj Models 12–1
, Chapterj 1
Exercisesj 1.1 5
6.j Leftj 1,j downj
2
1. Rightj 2,j upj 3 y
y
(2, 3)
x
x
( )
–1, – 52
7.j Leftj 20,j upj 40
2. Leftj 1,j upj y
4
y
(–20, 40)
(–1, 4)
x
x
8.j Rightj 25,j upj 30
3.j Downj 2 y
y
(25, 30)
x
x
(0, –2)
9. Pointj Qj isj 2j unitsj toj thej leftj andj 2j unitsj upj or
4. Rightj 2
y (—2,j2).
10. Pointj Pj isj 3j unitsj toj thej rightj andj 2j unitsj downj or
(3,—2).
x
(2, 0) 1j
11. —2(1)j +j (3)j =j—2j+1j=j—1soj yesj thej pointj is
3
onj thej line.
5. Leftj 2,j upj 1j
1 12. —2(2)j +j (6)j =j—1jisj false,j soj noj thej pointj isj not
3
y
onj thej line
(–2, 1)
x
Copyright © 2023 Pearson Education, Inc. 1-1
, Chapter 1: Linear Equations and Straight Lines ISM: Finite Math
1j 24.j 0j =j 5
13 —2xj +j yj =j—1j Substitutej thej xj andj y noj solution
3
. x-intercept:j
coordinatesj ofj thej pointj intoj thej equation:
f 1j hıj f h nonej Whenj xj =j
' ,j3 →j—2 ' 1 ı +j1j(3)j=j—1j→j—1+1j=j—1j is 0,j yj =j 5jy-
y' ı ' ı
intercept:j (0,j 5)
2j j j J yj2J 3
aj falsej statement.j Soj noj thej pointj isj notj 25.j Whenj yj =j 0,j xj =j
onj thejline. 7jx-intercept:j (7,j
f 1h f1 h 0)j0j =j 7
14 —2 ' ı + ' ı (—1)j =j—1j isj truej soj yesj thej pointj is noj solution
.
'y3 ıJj j j 'y3 ıJ y-intercept:j none
onj thej line. 26.j 0j =j –8x
15.j mj =j 5,j bj =j 8 xj =j 0
x-intercept:j (0,j 0)
16.j mj =j –2j andj bj =j –6 yj =j –8(0)
yj =j 0
17.j yj =j 0xj +j 3;j mj =j 0,j bj y-intercept:j (0,j 0)
=j 3
2j 2j 1j
yj =j xj+j0;j mj =j ,j bj =j0 27 0j =j xj –j1
18 3
3 3 .
. xj =j 3
19.j 14xj+j7jyj =j21 x-intercept:j (3,j 0)
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—j yj =j3 y
. —yj =j—xj +j3
yj =j xj—j3
(3, 0)
21.j j j 3xj =j5 x
5 (0, –1)
xj =j
3
1 2
28. Whenj xj =j 0,j yj =j 0.
22 – xj + yj =j10
. 2 3 Whenj xj =j 1,j yj =j 2.
2j 1j y
yj =j xj +10
3 2
3j
yj =j xj +15 (1, 2)
4 x
(0, 0)
23. 0j =j—4xj +j8
4xj =j 8
xj =j 2
x-intercept:j (2,j 0)
yj =j –4(0)j +j 8
yj =j 8
1-2 Copyright © 2023 Pearson Education, Inc.
