August 28-30, 2024
Solving
Rational Inequalities
Ms. Lynde Amor P. Casayas
:
:
−∞ +∞
Get a representative for each interval and substitute it to each of the factors to determine the sign.
−∞ +∞
Get a representative for each interval and substitute it to each of the factors to determine the sign.
−∞ +∞
For example,
you can use −𝟐 from the
interval (-∞, -1) and
substitute it to
(x+1) and (x-5)
Get a representative for each interval and substitute it to each of the factors to determine the sign.
−∞ +∞
For example, For example,
you can use −𝟐 from the you can use 𝟏 from the
interval (-∞, -1) and interval (-1, 5) and
substitute it to substitute it to
(x+1) and (x-5) (x+1) and (x-5)
Get a representative for each interval and substitute it to each of the factors to determine the sign.
−∞ +∞
For example, For example, For example,
you can use −𝟐 from the you can use 𝟏 from the you can use 𝟕 from the
interval (-∞, -1) and interval (-1, 5) and interval (5, +∞) and
substitute it to substitute it to substitute it to
(x+1) and (x-5) (x+1) and (x-5) (x+1) and (x-5)
𝒙 𝟏
Since 𝒙 𝟓
> 𝟎 should be positive (numbers greater than zero are positive numbers), the solution must be (−∞, −1) ∪ ( 5, +∞) .
Note:
The use of a square bracket indicates that it is part of the solution, while an open bracket (parenthesis) denotes that it’s not.
The parentheses around the infinity symbols are necessary to indicate that the intervals are open, meaning they do not include endpoints.
Solution Set:
SOLVING RATIONAL INEQUALITIES
𝒙+𝟏
1) > 𝟎 2)
𝒙+𝟏
≥𝟎
𝒙−𝟓 𝒙−𝟓
Critical Values: numerator: denominator: Critical Values: numerator: denominator:
𝑥+1 =0 𝑥−5 = 0 𝑥+1 =0 𝑥−5 = 0
𝒙 = −𝟏 𝒙=𝟓 𝒙 = −𝟏 𝒙=𝟓
Solution set: Solution set:
SOLVING RATIONAL INEQUALITIES
𝒙+𝟏
3) < 𝟎 4)
𝒙+𝟏
≤𝟎
𝒙−𝟓 𝒙−𝟓
Critical Values: numerator: denominator: Critical Values: numerator: denominator:
𝑥+1 =0 𝑥−5 = 0 𝑥+1 =0 𝑥−5 = 0
𝒙 = −𝟏 𝒙=𝟓 𝒙 = −𝟏 𝒙=𝟓
Solution set: Solution set:
SOLVING RATIONAL INEQUALITIES
𝒙+𝟒
5) >𝟑
𝒙−𝟐
𝑥+4
−3 >0
𝑥−2
𝑥+4 − 3(𝑥−2)
>0
𝑥−2
𝑥+4 − 3𝑥+6
>0
𝑥−2
−𝟐𝒙 + 𝟏𝟎
>𝟎
𝒙−𝟐
Critical Values:
numerator:
−2𝑥 + 10 = 0
−2𝑥 = −10
−2𝑥
−2
= −10
−2
𝒙=𝟓
Solution set:
denominator:
𝑥−2=0
𝒙=𝟐
SOLVING RATIONAL INEQUALITIES
𝒙+𝟒
6) ≥𝟑
𝒙−𝟐
𝑥+4
−3 ≥0
𝑥−2
𝑥+4−3(𝑥−2)
≥0
𝑥−2
𝑥+4−3𝑥+6
≥0
𝑥−2
−𝟐𝒙 + 𝟏𝟎
≥𝟎
𝒙−𝟐
Critical Values:
numerator:
−2𝑥 + 10 = 0
−2𝑥 = −10
−2𝑥
−2
= −10
−2
𝒙=𝟓
Solution set:
denominator:
𝑥−2=0
𝒙=𝟐
SOLVING RATIONAL INEQUALITIES
𝒙+𝟒 𝒙+𝟒
5) 𝒙−𝟐
>𝟑 6) 𝒙−𝟐
≥𝟑
Critical Values: numerator: denominator: Critical Values: numerator: denominator:
𝑥=5 𝑥=2 𝑥=5 𝑥=2
Solution set: Solution set:
SOLVING RATIONAL INEQUALITIES
𝒙+𝟒
7) <𝟑
𝒙−𝟐
𝑥+4
−3 <0
𝑥−2
𝑥+4 − 3(𝑥−2)
<0
𝑥−2
𝑥+4 − 3𝑥+6
<0
𝑥−2
−𝟐𝒙 + 𝟏𝟎
<𝟎
𝒙−𝟐
Critical Values:
numerator:
−2𝑥 + 10 = 0
−2𝑥 = −10
−2𝑥 −10
−2
= −2
𝒙=𝟓
denominator: Solution set:
𝑥−2=0
𝒙=𝟐
SOLVING RATIONAL INEQUALITIES
𝒙+𝟒
8) ≤𝟑
𝒙−𝟐
𝑥+4
−3 ≤0
𝑥−2
𝑥+4 − 3(𝑥−2)
≤0
𝑥−2
𝑥+4 − 3𝑥+6
≤0
𝑥−2
−𝟐𝒙 + 𝟏𝟎
≤𝟎
𝒙−𝟐
Critical Values:
numerator:
−2𝑥 + 10 = 0
−2𝑥 = −10
−2𝑥 −10
−2
= −2
𝒙=𝟓
denominator: Solution set:
𝑥−2=0
𝒙=𝟐
SOLVING RATIONAL INEQUALITIES
𝒙+𝟒 𝒙+𝟒
7) 𝒙−𝟐
<𝟑 8) 𝒙−𝟐
≤𝟑
Critical Values: numerator: denominator: Critical Values: numerator: denominator:
𝑥=5 𝑥=2 𝑥=5 𝑥=2
Solution set: Solution set:
SOLVING RATIONAL INEQUALITIES
𝟐 𝟑
9) >
𝒙+𝟑 𝒙+𝟐
2 3
− >0
𝑥+3 𝑥+2
2 𝑥+2 − 3(𝑥+3)
>0
𝑥+3 (𝑥+2)
2𝑥+4 − 3𝑥−9
>0
𝑥+3 (𝑥+2)
−𝒙−𝟓
𝒙+𝟑 (𝒙+𝟐)
Critical Values:
numerator:
−𝑥 − 5 = 0
−𝑥 5
=
−1 −1
𝒙 = −𝟓
denominator: Solution set:
𝑥+3=0 𝑥+2 = 0
𝒙 = −𝟑 𝒙 = −𝟐
SOLVING RATIONAL INEQUALITIES
𝟐 𝟑
10) ≥
𝒙+𝟑 𝒙+𝟐
2 3
− ≥0
𝑥+3 𝑥+2
2 𝑥+2 − 3(𝑥+3)
≥0
𝑥+3 (𝑥+2)
2𝑥+4 − 3𝑥−9
≥0
𝑥+3 (𝑥+2)
−𝒙−𝟓
𝒙+𝟑 (𝒙+𝟐)
Critical Values:
numerator:
−𝑥 − 5 = 0
−𝑥 5
=
−1 −1
𝒙 = −𝟓
denominator: Solution set:
𝑥+3=0 𝑥+2 = 0
𝒙 = −𝟑 𝒙 = −𝟐
SOLVING RATIONAL INEQUALITIES
𝟐 𝟑 𝟐 𝟑
9) >
𝒙+𝟑 𝒙+𝟐 10) ≥
𝒙+𝟑 𝒙+𝟐
Critical Values: numerator: denominator: Critical Values: numerator: denominator:
𝑥 = −5 𝑥 = −3 𝑥 = −5 𝑥 = −3
𝑥 = −2 𝑥 = −2
Solution set: Solution set:
SOLVING RATIONAL INEQUALITIES
𝟐 𝟑
11) <
𝒙+𝟑 𝒙+𝟐
2 3
− <0
𝑥+3 𝑥+2
2 𝑥+2 − 3(𝑥+3)
<0
𝑥+3 (𝑥+2)
2𝑥+4 − 3𝑥−9
<0
𝑥+3 (𝑥+2)
−𝒙−𝟓
𝒙+𝟑 (𝒙+𝟐)
Critical Values:
numerator:
−𝑥 − 5 = 0
−𝑥 5
=
−1 −1
𝒙 = −𝟓
denominator: Solution set:
𝑥+3=0 𝑥+2 = 0
𝒙 = −𝟑 𝒙 = −𝟐
SOLVING RATIONAL INEQUALITIES
𝟐 𝟑
12) ≤
𝒙+𝟑 𝒙+𝟐
2 3
− ≤0
𝑥+3 𝑥+2
2 𝑥+2 − 3(𝑥+3)
≤0
𝑥+3 (𝑥+2)
2𝑥+4 − 3𝑥−9
≤0
𝑥+3 (𝑥+2)
−𝒙−𝟓
𝒙+𝟑 (𝒙+𝟐)
Critical Values:
numerator:
−𝑥 − 5 = 0
−𝑥 5
=
−1 −1
𝒙 = −𝟓
denominator: Solution set:
𝑥+3=0 𝑥+2 = 0
𝒙 = −𝟑 𝒙 = −𝟐
SOLVING RATIONAL INEQUALITIES
𝟐 𝟑 𝟐 𝟑
11) <
𝒙+𝟑 𝒙+𝟐 12) ≤
𝒙+𝟑 𝒙+𝟐
Critical Values: numerator: denominator: Critical Values: numerator: denominator:
𝑥 = −5 𝑥 = −3 𝑥 = −5 𝑥 = −3
𝑥 = −2 𝑥 = −2
Solution set: Solution set:
