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

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

, Contents
Chapter 1: Linear Equations and Straight Lines 1–1

Chapter 2: Matrices 2–1

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

Chapter 4: The Simplex Method 4–1

Chapter 5: Sets and Counting 5–1

Chapter 6: Probability 6–1

Chapter 7: Probability and Statistics 7–1

Chapter 8: Markov Processes 8–1

Chapter 9: The Theory of Games 9–1

Chapter 10: The Mathematics of Finance 10–1

Chapter 11: Logic 11–1

Chapter 12: Difference Equations and Mathematical Models 12–1

, Chapter 1
Exercises 1.1 5
6. Left 1, down
2
1. Right 2, up 3 y
y



(2, 3)
x
x
( )
–1, – 52




7. Left 20, up 40
2. Left 1, up 4 y
y

(–20, 40)
(–1, 4)
x
x




8. Right 25, up 30
3. Down 2 y
y


(25, 30)
x
x
(0, –2)




9. Point Q is 2 units to the left and 2 units up or
4. Right 2
y (—2, 2).

10. Point P is 3 units to the right and 2 units down or
(3,—2).
x
(2, 0) 1
11. —2(1) + (3) = —2 +1 = —1so yes the point is
3
on the line.

5. Left 2, up 1 1
y 12. —2(2) + (6) = —1 is false, so no the point is not
3
on the line

(–2, 1)
x



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


1v 24. v 0 v= v5
13 —2x v+ v y v = v—1 v Substitute v the v x v and v y no vsolution
3
. x-intercept:
coordinates vof vthe vpoint vinto vthe vequation:
f 1 v ıhv f h vnone vWhen vx
' ,v3 → v—2 ' 1 ı + v1 v(3)v=v—1 v→ v—1+1 v=v—1 v is v= v0, vy v= v5vy-
'y ı ' ı
intercept: v(0, v5)
2 v v vJ yv2J 3
a vfalse vstatement. vSo vno vthe vpoint vis 25. vWhen vy v= v0, vx
vnot von vthevline. v= v7 vx-intercept:
v(7, v0)v0 v= v7
f 1h f1h
14 —2 ' ı + ' ı (—1) v=v—1 v is vtrue vso vyes vthe no vsolution
.
vpoint vis

'y3 ıJ v v v'y3 ıJ y-intercept: vnone
on vthe vline. 26. v 0 v= v–8x
15. v m v= v5, vb v= x v= v0
v8
x-intercept: v(0, v0)
y v= v–8(0)
16. v m v= v–2 vand vb v= v–6 y v= v0
y-intercept: v(0, v0)
17. v y v= v0x v+ v3; vm v=
v0, vb v= v3

2v 2v 1v
y v= v x v+v0; v m v= v , v b 27 0 v= v x v– v1
18 3
. .
v= v0 x v= v3
3 3
19. v 14x v+v7 vy v=
x-intercept: v(3, v0)
1v
v21 y v = v (0) v– v1
3
7 vy v= v—14x v+v21
y v= v–1
y v = v—2x v+v3
y-intercept: v(0, v–1)
20 x v— vy v= v3 y
. —y v= v—x v+v3
y v= vx v—v3
(3, 0)
21. v v v 3x v= v5 x
5 (0, –1)
x v= v
3
1 2
28. When vx v= v0, vy v= v0.
22 – x v+ y v =v10
. 2 3 When vx v= v1, vy v= v2.
2v 1v y
y v= v x
v+10
3 2 (1, 2)
3v x
y v= v x (0, 0)

v+15
4
23. 0 v= v—4x v+v8
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