Regents Review #4
Inequalities
and
Systems
Simple Inequalities
1) Solve inequalities like you would solve an
equation (use inverse operations to isolate
the variable)
2) When multiplying or dividing both sides of an
inequality by a negative number, flip the
inequality sign
3) Graph the solution set on a number line
Simple Inequalities
3(2x – 1) + 3x 4(2x + 1)
-3x – 4 > 8
6x – 3 + 3x 8x + 4
-3x > 12
9x – 3 8x + 4
x<-4
x 7
-7 -6 -5 -4 -3
-2 -1
4 5 6 7 8 9 10
Simple Inequalities
Words to Symbols
Example

Minimum 
In order to go to the movies, Connie and Stan
decide to put all their money together. Connie has
three times as much as Stan. Together, they have
more than $17. What is the least amount of
money each of them can have?
At Least
Cannot Exceed

Maximum 
At Most

Let x = Stan’s money
Let 3x = Connie’s money
x + 3x > 17
4x > 17
x > 4.25
Since Stan has to have more than $4.25, the least
amount of money he can have is $4.26.
Since Connie has three times as much as Stan,
she has $12.78.
Compound Inequalities
A compound inequality is a sentence with two
inequality statements joined either by the word
“OR” or by the word “AND”
“AND”  Graph the solutions that both
inequalities have in common
“OR”  Graph the combination of both
solutions sets
Compound Inequalities
“AND”
-3 < x 
4
-12 2x < -8
x > -3 and x 
4
2x -12 and 2x < -8
x -6 and x < -4
-6 x < - 4
-4 -3 -2 -1 0 1 2 3 4 5 6
-9 -8 -7 -6 -5 -4 -3 -2 -1 0
Compound Inequalities
“OR”
x < -4 or x 
6
-10 -8 -6 -4 -2 0 2 4 6 8 10
2x + 5 < 11 or 3x > 15
2x < 6
or x > 5
x < 3
x < 3 or x > 5
0 1 2 3 4 5 6 7 8 9 10 11
Linear Inequalities
Graph Linear Inequalities the same way you graph
Linear Equations but…
1) Use a dashed line (----) if the signs are < or >
2) Use a solid line (
) if the signs are  or 
3) Shade above the line if the signs are > or 
4) Shade below the line if the signs are < or 
Linear Inequalities
Graph -2y > 2x – 4
-2y > 2x – 4
y<-x+2
m=
1 1
or
1
1
b = 2 (0,2)
Test point (0,0)
-2y > 2x – 4
-2(0) > 2(0) – 4
0 > 0–4
0 > - 4 True
Systems
A "system" of equations is a collection of equations
in the same variable
When solving Linear Systems, there are three types
of outcomes…
No Solution
One Solution
y = 2x + 5
y = 2x – 4
y = -2x + 4
y = 3x - 2
Infinite
Solutions
y = 2x + 3
3y = 6x + 9
Systems
There are two ways to solve a Linear System
1) Graphically-graph both lines and determine
the common solution (point of intersection)
2) Algebraically
-Substitution Method
-Elimination Method
Solve the system y = 4x – 1
and 3x + 2y = 20 graphically
y = 4x – 1
m=
4
1
b = -1 (0,-1)
3x + 2y = 20
2y = -3x + 20
3
y = - 2 + 10
m=- 3
2
b = 10 (0,10)
Check (2, 7)
y = 4x – 1
7= 4(2) – 1
7=8–1
7=7
3x + 2y = 20
3(2) + 2(7) = 20
6 + 14 = 20
20 = 20
Systems
Solution
(2,7)
Systems
Solving Linear Systems Algebraically (Substitution)
Finding y
x=7–y
x+y=7 
3x = 17 + y
3x = 17 + y
Check
x+y=7
6+1=7
7=7
3x = 17 + y
3(6) = 17 + 1
18 = 18
Finding x
x+y=7
3(7 – y) = 17 + y
x+1=7
21 – 3y = 17 + y
x=6
-4y = -4
Solution (6,1)
y=1
Systems
Solving Linear Systems Algebraically (Elimination)
5x – 2y = 10 5x – 2y = 10  5x – 2y = 10
2x + y = 31  2[2x + y = 31] + 4x + 2y = 62
Finding y
2x + y = 31
2(8) + y = 31
16 + y = 31
y = 15
9x + 0y = 72
9x = 72
x=8
Solution (8, 15)
Check
5x – 2y = 10
5(8) – 2(15) = 10
40 – 30 = 10
10 = 10
4x + 2y = 62
4(8) + 2(15) = 62
32 + 30 = 62
62 = 62
Systems
Using Systems to Solve Word Problems
A discount movie theater charges $5 for an adult ticket and $2 for a
child’s ticket. One Saturday, the theater sold 785 tickets for $3280.
How many children’s tickets were sold?
Finding x
Let x = the number of adult tickets
Let y = the number of children tickets
5x + 2y = 3280
x + y = 785
5x + 2y = 3280
-5[x + y = 785]
+
5x + 2y = 3280
-5x – 5y = -3925
0x – 3y = -645
-3y = -645
y = 215
x + y = 785
x + 215 = 785
x = 570
570 adult tickets
215 children tickets
Systems
Solving Linear-Quadratic Systems Graphically
Two Solutions
One Solution
No Solution
Systems
Solving Linear-Quadratic Systems Graphically
y = x2 – 4x – 2
y=x–2
y = x2 – 4x – 2
x=
y=x–2
b
 ( 4)
4


2
2a
2(1)
2
1
m=
1
x
y
-1
3
b = -2 (0,-2)
0
-2
1
-5
2
-6
3
-5
4
-2
5
3
y = x2 – 4x – 2
y=x–2
Solutions (0,-2) and (5,3)
Systems
Solving Systems of Linear Inequalities
Solve the system:
y < 3x
y
-2x + 3
y < 3x
m = 3/1

y < 3x
b = 0 (0,0)
y 
-2x + 3 m = -2/1 b = 3 (0,3)
1) Graph each inequality
2) Label each inequality
3) Label the solution region with S
S
Now it’s your turn to review on your own!
Using the information presented today and your review packet,
complete the practice problems in the packet.
Regents Review #5
is
FRIDAY, May 31st
BE THERE!!!!