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Math 103 – Sections 3.3 & 3.4
Interval Notation
(1,5)
Interval Notation
Inequality
1<X<5
(-,7)
X7
[-4,5)
-4≤ X < 5
(0, 5]
0x5
(2,)
X>2
(-,1]
X ≤ -1
[6,)
X≥6
[-2,8]
-2≤ X ≤8
(-, )
All real numbers
(-,8]
X≤8
(-1, 5)
-1  x  5
[-5, 4)
-5 ≤ X < 4
Graph
2) Properties of Inequalities – Due the day we do this section in class
Fill in the blanks with appropriate numbers and inequalities:
Add 3 to both sides and get
7

10 > 6
Subtract 8 from both sides
_2 > -2
Multiply both sides by 5
_10 > -10
Divide both sides by (-1)
_-10 < 10
Multiply both sides by (- 3)
_30 > -30
Divide both sides by 6
_5 > -5
3
The only time we switch the sense of the inequality is when we __MULTIPLY_ or _DIVIDE_ both sides
of the inequality by a __NEGATIVE_ number.
1
3) Solve inequalities algebraically.
a. Linear inequalities
Solve as if you were solving a linear equation
In the last step, when you divide by the coefficient of the variable, switch the inequality
only if you are dividing by a negative number
(1) 3(x – 5) + 2 > 5x - 10
3X – 15 + 2 > 5X – 10
3X – 13 > 5X – 10
-2X – 13 > -10
-2X > 3
x < -3/2 which is the same as (-,-3/2)
(You graph it)
(2) Now do problem number 22 on page 175 (rewrite without denominators!!! – then, solve)
5  2x 2x 1

2
4
Multiply by 4 both sides of the equation – after you simplify the denominator, you get
2(5  2 x)  2 x  1
10  4 x  2 x  1
6 x  9
x
Interval notation is (-,3/2]
3
2
b. Compound inequalities (section 3.4)
x ≤ -10 or x > 8
(3) Graph
_____●________o_________
-10
8
Interval notation: (-,-10] U (8, )
(4) Solve and graph
3x < -15 or 2x > 22
X < -5 or x > 11
_____o ________o_________
-5
11
Interval notation: (-,-5] U (11, )
x ≤ 10 and x > - 8
(5) Graph
________ o ________ ● ____________
-8
10
Interval notation: (-8, 10]
(6) Solve and graph
2x ≥ -10 and x < 8
X ≥ - 5 and x < 8
Interval notation: [-5, 8)
________●________ o ____________
-5
8
2
(7) Graph
-1 < x < 2 (all numbers between -1 and 2)
_____ o _______ o _______
-1
2
Interval notation: (-1, 2)
-15 < 3x – 7 ≤ 8
(8) Solve and graph
Add 7
-8 < 3x ≤ 15
Divide by 3
-8/3 < x ≤ 5
Interval notation
(-8/3, 5]
_____ o _______ o _______
-8/3
5
(9) Now do problem number 54 on page 188 (rewrite without denominators!!! – then, solve)
1
3 1
 y 
2
2 2
Multiply by 2
-1 < 2y – 3 < 1
Add 3
2 < 2y < 4
Divide by 2
1<y<2
Interval notation: (1, 2)
____o _______o__________
1
2
3