Show Interval(s) On Real Number Line

[nycmathguy]

Set 1.2

Problems: 50, 52, 56, 60

I will use your work on these 4 to do the rest. Raining all day in NYC. Ruined my day off. Stayed home all day. I made the best of it by answering math problems and packing for the big move on 9/9/21. Thank God!!

 

[MathLover1]

50

abs(x)<2

Apply absolute rule: If abs(u) <=a, a > 0 then -a<=u<=a

-2<x<2

Interval notation
(-2, 2)

Integer solutions
x = ± 1, x = 0


52
abs(x)>0
sqrt(x^2)>0
solutions:
x>0
x<0

Interval notation
(0, ∞)
(-∞, 0)

56

abs(x-1)<=1/2

Apply absolute rule: If abs(u) <=a, a > 0 then -a<=u<=a
-1/2<=x -1 <=1/2
-1/2+1<=x -1+1 <=1/2+1
1/2<=x<=3/2

Integer solution: x=1

interval notation: [1/2, 3/2]

60

abs(x+5)>=2
Apply absolute rule:
If abs(u)>=a, a > 0 then u<=-a or u>=a

x+5<=-2 or x+5>=2
solutions:
x+5<=-2=>x<=-2-5=>x<=-7
x+5>=2=>x>=2-5=>x>=-3

Interval notation:
[-3, ∞)
(-∞, -7]

 

[nycmathguy]

50

abs(x)<2

Apply absolute rule: If abs(u) <=a, a > 0 then -a<=u<=a

-2<x<2
View attachment 308
Interval notation
(-2, 2)
View attachment 309
Integer solutions
x = ± 1, x = 0


52
abs(x)>0
sqrt(x^2)>0
solutions:
x>0
x<0
View attachment 310
Interval notation
(0, ∞)
(-∞, 0)

View attachment 311

56

abs(x-1)<=1/2

Apply absolute rule: If abs(u) <=a, a > 0 then -a<=u<=a
-1/2<=x -1 <=1/2
-1/2+1<=x -1+1 <=1/2+1
1/2<=x<=3/2

Integer solution: x=1
View attachment 312

interval notation: [1/2, 3/2]

60

abs(x+5)>=2
Apply absolute rule:
If abs(u)>=a, a > 0 then u<=-a or u>=a

x+5<=-2 or x+5>=2
solutions:
x+5<=-2=>x<=-2-5=>x<=-7
x+5>=2=>x>=2-5=>x>=-3
View attachment 313
Interval notation:
[-3, ∞)
(-∞, -7]
View attachment 314

Perfect.