Solving Absolute-Value, Compound,
and Quadratic Inequalities
Reminder: Compound Inequalities
The following are examples of how to
algebraically write the following graphs:
0≤x<4
0
4
x<-1 or x>2
-1
2
REMINDER
How to solve a one variable
inequality.
Solving a 1 Variable Inequality
Represent the solutions to the following inequality
algebraically and on a number line.
Closed or Open Dot(s)?
2 x  7  12
Find the Boundary
Graphical
Solution
Test Every Region
x
Change inequality to equality
Solve
2 x  7  12
2x  5
x  2.5
Plot Boundary Point(s)
0
Pick a point in
each region
x=0
Substitute
into Original
x=3
2  0   7  12
2  3  7  12
7 < 12
True
13 < 12
False
Shade True
Region(s)
x  2.5
Algebraic
Solution
Apply this method to more
complicated Ineqaulities
Solving an Absolute Value Inequality
Represent the solutions to the following inequality
algebraically and on a number line.
Closed or Open Dot(s)?
x2 3
Find the Boundary
Test Every Region
Graphical
Solution
x
Change inequality to equality
Solve
x2 3
x  2  3
x  2  3 x  2  3
x  5 or x  1
Plot Boundary Point(s)
Pick a point in
each region
x = -2
2  2  3
Substitute
into Original
4>3
True
Shade True
Region(s)
0
x=0
02 3
2>3
False
x=6
62 3
4>3
True
x  1 or x  5
Algebraic
Solution
Solving a Compound Inequality
Represent the solutions to the following inequality
algebraically and on a number line.
Closed or Open Dot(s)?
12  2 x  8  2
Find the Boundary
Test Every Region
Graphical
Solution
x
Change inequality to two equalities
12  2 x  8 2 x  8  2
Solve Both
4  2x
2x  6
0
Pick a point in
each region
x=0
x = -3
x=4
12  2  3  8  2
12  2  4   8  2
12  2  0   8  2
Substitute into Original
x  2 or x  3
Plot Boundary Point(s)
-12<-14≤-2
False
Shade True
Region(s)
-12<-8≤-2
True
2  x  3
-12<0≤-2
False
Algebraic
Solution
Solving a Quadratic Inequality
Represent the solutions to the following inequality
algebraically and on a number line.
Closed or Open Dot(s)?
x  2x 1  4
2
Find the Boundary
Graphical
Solution
Test Every Region
x
Change inequality to equality
x  2x 1  4
x=2
x=0
x = -4
2
x  2 x  3  0 Substitute
 4  2  4  1  4
 2  2  2  1  4
 0  2  0  1  4
into
Original
 x  3 x  1  0
1<4
9≤4
9<4
2
Solve
0
Pick a point in
each region
2
2
2
x  3 or x  1
Plot Boundary Point(s)
False
Shade True
Region(s)
True
3  x  1
False
Algebraic
Solution