Rational Inequalities
4t  5
#2:
3
t
3 3
4t  5
3  0
t
4t  5 3  t 
  0
t
1t 
4t  5 3t
 0
t
t
t 5
0
t
Find critical
values by
setting num and
denom = 0
Page 5
Steps:
CV for Numerator
Circle for this
CV depends
upon the
inequality!
t 5  0
t  5
CV for Denominator
1.
Get 0 on one side
of inequality.
2.
Get a common
denominator and
combine terms
3.
Find Critical
Values for
numerator and
denominator
4.
Graph solutions
on number line
5.
Test each interval
on number line
Circle for this
CV is always
open
t0
Plot CV on number line
6
1
5
1
0
-6: 3.17>3 - TRUE
-1: -1>3 - FALSE
Now we test each interval to
determine where to shade! Just pick
a number in the interval and test it in
the original equation.
1: 9>3 - TRUE
t  5
OR
t0
Page 5
#3:
3a  5
2
a 1
2 2
3a  5
20
a 1
3a  5 2  a  1 
 
0
a 1 1  a 1
3a  5 2a  1

0
a 1
a 1
3a  5 2a  2

0
a 1
a 1
3a  5 2a  2

0
a 1
a 1
Find critical
values by
setting num and
denom = 0
a7
0
a 1
8
0
1
a7  0
a7
CV for Denominator
a 1  0
a  1
Plot CV on number line
2
CV for Numerator
7
-2: 11  2 - FALSE
0: -5  2 - TRUE
8:2.1 2 - FALSE
Now we test each interval to
determine where to shade! Just pick
a number in the interval and test it in
the original equation.
1  a  7
Page 5
Plot CV on number line
#6:
x2  4
0
2
x  16
5
4
3
3
0
2
2
5
4
CV for Numerator
x2  4  0
x  2x  2  0
x  2 x  2
Circles for
these CV’s
depends upon
the inequality!
Now we test each interval to
determine where to shade! Just pick
a number in the interval and test it in
the original equation.
-5: 2.3 0 - FALSE
CV for Denominator
-3: -.7  0 - TRUE
x 2  16  0
x  4x  4  0
0: .25 0 - FALSE
x  4 x  4
Circles for
these CV’s are
always open
3: -.7  0 - TRUE
5: 2.3  0 - FALSE
 4  x  2
OR
2 x4
Page 5
#1:
2v
0
v3
CV for Numerator
2v  0
v0
Plot CV on number line
1
1
0
1: -1 < 0 - TRUE
4: 8 < 0 - FALSE
CV for Denominator
Always open circle
b/c it’s CV for denom
3
-1: .5 < 0 - FALSE
Open circle because
of inequality
v 3  0
v3
4
0v3
3x  7
#4:
2
x3
2 2
3x  7
20
x3
3x  7 2  x  3 
 
0
x 3 1  x 3
3x  7 2 x  6

0
x3
x3
x 1
0
x3
x 1
0
x3
CV for Numerator
x 1  0
x  1
Page 5
Plot CV on number line
4
2
3
0
1
-4: 5 > 2 - TRUE
Open circle because
of inequality
-2: 1 > 2 - FALSE
CV for Denominator
x3 0
x  3
Always open circle
b/c it’s CV for denom
0: 2.3 > 2 - TRUE
x  3
OR
x  1
#5:
x2
0
2
x  16
CV for Numerator
x20
x2
Closed circle because
of inequality
CV for Denominator
x 2  16  0
x  4x  4  0
x  4 x  4
Always open circle
b/c it’s CV for denom
Plot CV on number line
5
3
0
4
2
5
4
-5: -.8  0 - TRUE
0: .13  0 - FALSE
3: -.1
 0 - TRUE
5: .33  0 - FALSE
x  4
OR
2 x4
Page 5
Homework
• Page 6
#2,6,10
#2:
n2
3
n4
3 3
n2
3  0
n4
n2 3 n4
 
0
n4 1n4
 2n  14  0
 2n  14
n  7
Open circle because
of inequality
CV for Denominator
n  2 3n  12

0
n4
n4
 2n  14
0
n4
Page 6
CV for Numerator
n40
n  4
Always open circle
b/c it’s CV for denom
Plot CV on number line
8
5
7
0
4
 7  n  4
-8: 2.5 > 3 - FALSE
-5: 7 > 3 - TRUE
0: -.5 > 3 - FALSE
#6:
g  11
27
g 5
7 7
g  11
9  0
g 5
g  11 9  g  5 
  0
 
g 5 1 g 5
g  11 9 g  45

0
g 5
g 5
 8 g  56
0
g 5
Page 6
CV for Numerator
 8g  56  0
 8g  56
g  7
Open circle because
of inequality
CV for Denominator
g 5  0
g  5
Always open circle
b/c it’s CV for denom
Plot CV on number line
8
6
7
g  7
0
5
OR
g  5
-8: 4.3 < 7 - TRUE
-6: 15 < 7 - FALSE
0: -4.2 < 7 - TRUE
#10:
3d  2
1
d 4
1 1
3d  2
1  0
d 4
3d  2 1  d  4 
 
0
d  4 1 d  4 
3d  2 d  4

0
d 4 d 4
2d  2
0
d 4
Page 6
CV for Numerator
2d  2  0
2d  2
d 1
Closed circle because
of inequality
CV for Denominator
d 40
d  4
Always open circle
b/c it’s CV for denom
Plot CV on number line
5
0
4
d  4
-5: 13
2
1
OR
d 1
 1 - TRUE
0: .5  1 - FALSE
2: 1.3  1 - TRUE