FiniteMathematics & itsApplications
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 intercept:JnoneJ
' ,J3 →J—2 ' 1 ı +J1J(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.JSoJnoJtheJpointJisJnotJonJtheJ 25.JWhenJyJ=J0,JxJ=J7Jx-
line. intercept:J(7,J0)J0J
f 1h f1h =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) x
21.JJJ 3xJ=J5
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.