Coverage:
Equalities
and
Inequalities
in
One
Variable, Partial Fraction Decomposition
5. Solve the quadratic inequality:
𝑥 − 4𝑥 − 5 < 0
a. (−1, 5)
b. [−1,5]
I. Multiple Choice
1. Solve for the solution set of
3(𝑥 − 4) − 2(𝑥 + 1) = 4 − 5𝑥.
a. 𝑥 = 1
b. 𝑥 = −
c. [-5, 1]
d. (-5, 1)
II. Problem Solving
1. A bus leaves Bulacan City and travels north at 60
c. 𝑥 = 3
d. 𝑥 = −
km/h. Three hours later, another bus leaves the
same station, heading in the same direction at 90
2. The roots of the equation 3𝑥 − 3𝑥 − 5 = 0 are:
a. 𝑥 = +
√
, −
c. 𝑥 =
km/h. How long will it take for the second train
to catch up to the first?
2. Baduya needs to prepare 50 milliliters of a 20%
b. 𝑥 = , −1
√
− + −
√
,
√
− − −
d. 𝑥 = − , 1
3. Solve for the solution set of the inequality
2𝑥 − 5
≤1
𝑥+3
a. (−∞, 8]
b. (−∞, −2]
c. (−2, 8)
d. (−3, 8]
4. Find the solution set of the linear equations
2𝑥 + 3𝑦 = 7
5𝑥 − 𝑦 = 1
a. 𝑥 =
,𝑦 =
b. 𝑥 =
,𝑦 = −
c. 𝑥 =
,𝑦 =
d. 𝑥 =
,𝑦 =
methamphetamine solution. He has on hand a
10%
caffeine
solution
and
a
30%
methamphetamine solution. How many milliliters
of each solution should he mix to obtain the
desired concentration?
3. Mrs. Williams has 4 children, and 3 of them are
gymnasts: Heliza, Barry, and Raiden. The
combined age of Heliza and Barry is equal to the
combined age of Mrs. Willians and Barry five
years ago. If Heliza’s age is 24 and Barry was 14
years old two years ago, how old is Mrs. Williams
now? Note: Raiden’s age is 15.
4. Decompose the following using the case where
one of the factors is repeated:
5𝑥 + 4
(𝑥 − 1) (𝑥 + 2)
5. Decompose the following using Case 3:
3𝑥 + 7
(𝑥 + 1)(𝑥 − 2)
BONUS | Solve the quadratic equation by completing
the square: 3𝑥 + 12𝑥 + 7 = 0
1
ANSWERS
I. Multiple Choice
3. D
1. C
3(𝑥 − 4) − 2(𝑥 + 1) = 4 − 5𝑥
GIVEN
The constant multiplied to (𝑥 + 1) is negative 2.
(3𝑥 − 12) + (−2𝑥 − 2) = 4 − 5𝑥
DISTRIBUTE.
NOTE: Transposing the terms will change its signs.
3𝑥 − 2𝑥 + 5𝑥 = 12 + 2 + 4
6𝑥 = 18
6𝑥 18
=
3
6
𝒙=𝟑
LIKE TERMS.
DIVIDE BOTH SIDES.
2𝑥 − 5
≤1
𝑥+3
2𝑥 − 5
−1 ≤0
𝑥+3
2𝑥 − 5 𝑥 + 3
−
≤0
𝑥+3 𝑥+3
(2𝑥 − 5) − (𝑥 + 3)
≤0
𝑥+3
2𝑥 − 5 − 𝑥 − 3
≤0
𝑥+3
2𝑥 − 𝑥 − 5 − 3
≤0
𝑥+3
𝑥−8
≤0
𝑥+3
2. A
The critical 𝑥-values are 8 and 3. The primary
3𝑥 − 3𝑥 − 5 = 0
goal is to let the expression
The standard form of a quadratic equation is 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0.
or a negative value.
Hence, 𝑎 = 3, 𝑏 = −3, 𝑐 = −5.
−
+
be equal to zero
8−8
≤0
8+3
0
≤0
11
NOTE: The quadratic formula is
√
−𝑏 ± 𝑏 − 4𝑎𝑐
𝑥=
2𝑎
NOTE: The product of two terms with similar signs is positive. The
product of two terms with different signs is negative.
𝑥=
−(−3) ±
(−3) − 4(3)(−5)
2(3)
3 and −5 are multiplied.
𝑥=
−1(−3) ±
(−3)(−3) − 4(−15)
6
Multiply the terms −1 and −3, −3 and −3, −4 and −15.
3±
√
9 + 60
𝑥=
√6
3 ± 69
𝑥=
6
At 𝑥 = 8, the expression
−
+
will be equal to 0.
Thus, 𝑥 = 8.
9−8
≤0
9+3
1
≤0
12
Meanwhile, substituting the values more than 8
will result in positive values. Therefore, 𝑥 < 8.
2
3−8
≤0
3+3
−5
≤0
0
At 𝑥 = 3, the expression will be undefined. Thus,
𝑥 ≠ 3.
2−8
≤0
2+3
−6
≤0
5
However, substituting the values less than 3 to 𝑥,
the expression will become positive as it would
result negative values on both numerator and
denominator. Hence, 𝑥 > 3.
(−3, 8]
4. C
2𝑥 + 3𝑦 = 7
5𝑥 − 𝑦 = 1
50
−𝑦 =1
17
50
−1=𝑦
17
50 17
−
=𝑦
17 17
33
=𝑦
17
5. A
𝑥 − 4𝑥 − 5 < 0
(𝑥 − 5)(𝑥 + 1) < 0
The critical 𝑥-values are 5 and −1.
(6 − 5)(6 + 1) < 0
(1)(7) < 0
Let 𝑥 = 6, which will cause the result of
(𝑥 − 5)(𝑥 + 1) a positive value. Therefore, 𝑥 < 5.
(−2 − 5)(−2 + 1) < 0
(−7)(−1) < 0
Using Substitution Method:
5𝑥 − 𝑦 = 1  5𝑥 − 1 = 𝑦
Let 𝑥 = −2, which will cause the result of the
same expression a positive value. Therefore, 𝑥 >
2𝑥 + 3𝑦 = 7
Substitute the value of 𝑦:
2𝑥 + 3(5𝑥 − 1) = 7
2𝑥 + 15𝑥 − 3 = 7
17𝑥 = 10
10
𝑥=
17
Substitute the value of 𝑥:
5𝑥 − 𝑦 = 1
10
5
−𝑦 =1
17
−1.
(−1, 5)
II. Problem Solving
1. A bus leaves Bulacan City and travels north at 60
km/h. Three hours later, another bus leaves the
same station, heading in the same direction at 90
km/h. How long will it take for the second bus to
catch up to the first?
3
GIVEN
Therefore, 0.1𝑐 + 0.3𝑚 = 10
𝑡=0
𝑡=1
𝑡=2
𝑡=3
𝑡=4
0
60
120
180
240
0
0
0
0
90
Bus 1
(60 kph)
Bus 2
(90 kph)
Solve by substitution method:
Distance of First Bus = 60𝑡 + 180
𝑐 + 𝑚 = 50
𝑚 = 50 − 𝑐
180 signifies the head start of the first bus.
Distance of Second Bus = 90𝑡
0.1𝑐 + 0.3(50 − 𝑐) = 10
0.1𝑐 + 15 − 0.3𝑐 = 10
−0.2𝑐 = 10 − 15
−0.2𝑐 = −5
𝑐 = 25
60𝑡 + 180 = 90𝑡
180 = 90𝑡 − 60𝑡
180 = 30𝑡
6=𝑡
Hence, it will take 6 hours for the second bus to
catch up to the first.
2. Baduya needs to prepare 50 milliliters of a 20%
methamphetamine solution. He has on hand a
10% caffeine solution and a 30% marijuana
solution. How many milliliters of each solution
should
he
mix
to
obtain
Hence, we have a system of linear equations.
𝑐 + 𝑚 = 50
0.1𝑐 + 0.3𝑚 = 10
the
desired
concentration?
Let 𝑐 = caffeine and 𝑚 = methamphetamine.
Baduya needs to prepare 50 milliliters of a 20%
methamphetamine solution.
Therefore, 𝑐 + 𝑚 = 50
The concentration of 𝑚 is 50 × 20% = 10.
𝑐 + 𝑚 = 50
25 + 𝑚 = 50
𝑚 = 50 − 25
𝑚 = 25
Therefore, Baduya should mix 25 milliliters of the
10% caffeine solution and another 25 milliliters
of 30% methamphetamine .
3. Mrs. Williams has 4 children, and 3 of them are
gymnasts: Heliza, Barry, and Raiden. The
combined age of Heliza and Barry is equal to the
combined age of Mrs. Williams and Barry five
years ago. If Heliza’s age is 24 and Barry was 14
years old two years ago, how old is Mrs. Williams
now? Note: Raiden’s age is 15.
Let 𝑊 = Mrs. Williams, 𝐻 = Heliza, 𝐵 = Barry,
He has on hand a 10% caffeine solution and a 30%
and 𝑅 = Raiden.
marijuana solution.
4
The combined age of Heliza and Barry…
𝐻 +𝐵
is equal to the combined age of Mrs. Williams and
Barry five years ago.
(𝑊 − 5) + (𝐵 − 5)
Hence, 𝐻 + 𝐵 = (𝑊 − 5) + (𝐵 − 5)
If Heliza’s age is 24 and Barry was 14 years old
two years ago
𝐻 = 24; 𝐵 = 14 + 2
Note: Raiden’s age is 15.
𝑅 = 15
(24) + (14 + 2) = (𝑊 − 5) + (14 − 5)
40 = 𝑊 − 5 + 9
40 + 5 − 9 = 𝑊
36 = 𝑊
Hence, Mrs. William’s age is 36 years old .
4. Decompose the following using the case where
one of the factors is repeated:
5𝑥 + 4
(𝑥 − 1) (𝑥 + 2)
5𝑥 + 4
𝐴
𝐵
𝐶
=
+
+
(𝑥 − 1) (𝑥 + 2) 𝑥 − 1 (𝑥 − 1)
𝑥+2
Multiply (𝑥 − 1) (𝑥 + 2) on both sides:
5𝑥 + 4 = 𝐴(𝑥 − 1)(𝑥 + 2) + 𝐵(𝑥 + 2) + 𝐶(𝑥 − 1)
Expand the terms.
Hence, a system of linear equations is formed.
𝐴+𝐶 =5
𝐴 − 2𝐶 + 𝐵 = 0
−2𝐴 + 2𝐵 + 𝐶 = 4
Using substitution method:
𝐴 =5−𝐶
𝐴 − 2𝐶 + 𝐵 = 0
(5 − 𝐶) − 2𝐶 + 𝐵 = 0
5 − 3𝐶 + 𝐵 = 0
𝐵 = 3𝐶 − 5
−2𝐴 + 2𝐵 + 𝐶 = 0
−2(5 − 𝐶) + 2(3𝐶 − 5) + 𝐶 = 4
−10 + 2𝐶 + 6𝐶 − 10 + 𝐶 = 4
2𝐶 + 6𝐶 + 𝐶 = 4 + 10 + 10
9𝐶 = 24
8
𝐶=
3
𝐴 =5−𝐶
8
𝐴=5−
3
15 8
𝐴=
−
3 3
7
𝐴=
3
𝐴 − 2𝐶 + 𝐵 = 0
7
8
−2
+𝐵 =0
3
3
7 16
− +𝐵 =0
3 3
9
− +𝐵 =0
3
9
𝐵=
3
𝐵=3
5𝑥 + 4 = 𝐴(𝑥 + 𝑥 − 2) + 𝐵(𝑥 + 2) + 𝐶(𝑥 − 2𝑥 + 1)
Group the terms by powers of 𝑥.
5𝑥 + 4 = (𝐴 + 𝐶)𝑥 + (𝐴 − 2𝐶 + 𝐵)𝑥 + (−2𝐴 + 2𝐵 + 𝐶)
5
Hence,
7
8
5𝑥 + 4
3
= 3 +
+ 3
(𝑥 − 1) (𝑥 + 2) 𝑥 − 1 (𝑥 − 1)
𝑥+2
or
5𝑥 + 4
7
3
8
=
+
+
(𝑥 − 1) (𝑥 + 2) 3(𝑥 − 1) (𝑥 − 1)
3(𝑥 + 2)
5. Decompose the following using Case 3:
3𝑥 + 7
(𝑥 + 1)(𝑥 − 2)
3𝑥 + 7
𝐴
𝐵
=
+
(𝑥 + 1)(𝑥 − 2) (𝑥 − 1) (𝑥 − 2)
Multiply (𝑥 − 1)(𝑥 + 2) on both sides:
3𝑥 + 7 = 𝐴(𝑥 − 2) + 𝐵(𝑥 + 1)
Expand the terms.
3𝑥 + 7 = 𝐴𝑥 − 2𝐴 + 𝐵𝑥 + 𝐵
13
3
9 13
𝐴= −
3 3
4
𝐴=−
3
𝐴=3−
Hence,
4
13
3𝑥 + 7
3
3
=
+
(𝑥 + 1)(𝑥 − 2) (𝑥 − 1) (𝑥 − 2)
or
3𝑥 + 7
4
13
=
+
(𝑥 + 1)(𝑥 − 2) 3(𝑥 − 1) 3(𝑥 − 2)
BONUS | Solve the quadratic equation by completing
the square: 3𝑥 + 12𝑥 + 7 = 0
3𝑥 + 12𝑥 + 7 = 0
3(𝑥 + 4𝑥) = −7
Group the terms by powers of 𝑥.
3𝑥 + 7 = (𝐴 + 𝐵)𝑥 − (2𝐴 + 𝐵)
Hence, a system of linear equations is formed.
𝐴+𝐵 =3
−2𝐴 + 𝐵 = 7
3(𝑥 + 4𝑥 + 4 − 4) = −7
3[(𝑥 + 2) − 4] = −7
3(𝑥 + 2) − 12 = −7
3(𝑥 + 2) = −7 + 12
3(𝑥 + 2) = 5
Using substitution method:
𝐴=3−𝐵
−2𝐴 + 𝐵 = 7
−2(3 − 𝐵) + 𝐵 = 7
−6 + 2𝐵 + 𝐵 = 7
2𝐵 + 𝐵 = 7 + 6
3𝐵 = 13
13
𝐵=
3
𝐴+𝐵 =3
13
𝐴+ =3
3
(𝑥 + 2) =
5
3
𝑥+2=±
5
3
𝑥=±
5
−2
3
Rationalize the roots.
√
√
5× 3
√ −2
𝑥 = ±√
3× 3
√
15
𝑥=±
−2
3
6
Reviewer is made by:
Isaiah Anthony Thomas Barredo
BS IE 123
Jethro Josh Aragon
BSMS CE 123
7