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
Exercises
1 5
6. Left 1, down
1.1
1. 2
Right 2, up 3 y
y
(2, 3
x
x
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
12. —2(2) + (6) = —1 is false, so no the point is not
y
3
on the line
(–2, 1)
x
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, Chapter 1: Linear Equations and Straight Line ISM: Finite Math
1 24. 0 = 5
13 —2x + y = —1 Substitute the x and y no solution
. 3 x-
coordinates of the point into the equation: intercept: none
f1 ıh f1 h 1
' , 3 → —2' ı + (3)= —1 → —1+1 = —1 is
When x = 0, y =
y' I ' ı 5y-
intercept: (0, 5)
2 J y2J 3
a false statement. So no the point is not on 25. When y = 0, x = 7
theline. x-
f 1h f 1 h intercept: (7, 0)0
14 —2 ' ı +' ı(—1) = —1 is true so yes the point is =7
.
no solution
y-intercept: none
y'3ıJ 'y3ıJ
on the line. 26. 0 = –8x
15. m = 5, b = 8 x=0
x-intercept: (0, 0)
16. m = –2 and b = –6 y = –8(0)
y=0
17. y = 0x + 3; m = 0, b = y-intercept: (0, 0)
3
2 2 1
y= x + 0; m = , b =0 27 0= x –1
18 3
. 3 3 .
x=3
19. 14x + 7 y = 21 x-intercept: (3, 0)
1
7 y = —14x + 21 y = (0) – 1
y = —2x + 3 3
y = –1
y-intercept: (0, –1)
20 x— y =3 y
. —y = —x +A3
y = x —3
(3, 0)
21. 3x = 5 x
5 (0, –1)
x=
3
1 2 28. When x = 0, y = 0.
22 – 2 x + 3 y = 10 When x = 1, y = 2.
. 2 1 y
y = x +10
3 2
3
y = x +15 (1, 2)
4 x
(0, 0)
23. 0 = —4x + 8
4x = 8
x=2
x-intercept: (2, 0)
y = –4(0) + 8
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