DIP1003 Algebra Assignment: Semester 1, 2019 - Expressions, Equations

This algebra assignment focuses on simplifying expressions, solving equations, and working with inequalities. It includes problems such as expanding and simplifying algebraic expressions using identities, finding the values of x in equations using the zero-product property, solving equations for a sp...Read More

DIP1003 SEMESTER 1,2019
ASSIGNMENT 2
MARCH 26, 2019
Submitted by:
Timo Weiss

1. Expand and simplify the following expressions:
a) (𝟒𝒙 + 𝒚)𝟐 − 𝟑𝒙(𝒚 + 𝟐𝒙)
Sol.
Using (𝑎 + 𝑏)2 = 𝑎2 + 𝑏2 + 2𝑎𝑏
(4𝑥 + 𝑦)2 − 3𝑥(𝑦 + 2𝑥)
= ((4𝑥)2 + 𝑦2 + 2(4𝑥)(𝑦)) − 3𝑥(𝑦 + 2𝑥)
= 16𝑥 2 + 𝑦2 + 8𝑥𝑦 − 3𝑥𝑦 − 6𝑥 2
= 10𝑥 2 + 𝑦2 + 5𝑥𝑦
b)
(𝒙𝟒𝒚𝟎𝒛𝟔)
𝟏
𝟐(𝒙
−𝟑
𝟐 𝒚
𝟔
𝟐)
−𝟐
𝒙𝟒𝒚𝟐𝒛𝟑
𝒙−𝟒𝒚𝒛𝟑
Sol.
(𝑥4𝑦0𝑧6)1
2(𝑥−3
2 𝑦6
2)−2𝑥4𝑦2𝑧3
𝑥−4𝑦𝑧3
= (𝑥2𝑧3)𝑥4𝑦2𝑧3
(𝑥−3𝑦6)(𝑥−4𝑦𝑧3)
= 𝑥6𝑦2𝑧6
𝑥−7𝑦7𝑧3
= 𝑥13 𝑦−5𝑧3
= 𝑥13 𝑧3
𝑦5
2. Fid the values of x in the following equation using the rule that if 𝒂 ∗ 𝒃 = 𝟎 then 𝒂 =
𝟎 𝒐𝒓 𝒃 = 𝟎 . Do not expand the equation or use the quadratic formula.
(𝟓𝒙 + 𝟔) (𝟑𝒙 − 𝟐
𝟑) = 𝟎
Sol.
𝐸𝑖𝑡ℎ𝑒𝑟 (5𝑥 + 6) = 0 𝑜𝑟 (3𝑥 − 2
3) = 0
5𝑥 + 6 = 0
𝑥 = −6
5
(3𝑥 − 2
3) = 0

3𝑥 = 2
3
𝑥 = 2
9
𝑥 = −6
5 𝑜𝑟 2
9
3. Solve the following equation for 𝒎.
Hint: 𝒂𝟐 − 𝒃𝟐 = (𝒂 − 𝒃)(𝒂 + 𝒃)
(𝟐𝒎)𝟐 − 𝟒
𝒎𝟐 + 𝟏 = 𝟐𝒎 + 𝟐
𝒎 + 𝟐
Sol.
(2𝑚)2 − 4
𝑚2 + 1 = 2𝑚 + 2
𝑚 + 2
= (2𝑚)2 − 22
𝑚2 + 1 = 2𝑚 + 2
𝑚 + 2
= (2𝑚 + 2)(2𝑚 − 2)
𝑚2 + 1 = 2𝑚 + 2
𝑚 + 2
= 4(𝑚 + 1)(𝑚 − 1)
𝑚2 + 1 = 2(𝑚 + 1)
𝑚 + 2
= 2(𝑚 − 1)
𝑚2 + 1 = 1
𝑚 + 2
2(𝑚 − 1)(𝑚 + 2) = 𝑚2 + 1
2(𝑚2 + 2𝑚 − 𝑚 − 2) = 𝑚2 + 1
2𝑚2 + 𝑚 − 2 = 𝑚 2 + 1
= 𝑚2 + 𝑚 − 3
4. Factorise the following expressions:
a) 𝟑𝟔𝒙𝟐 − 𝟖𝟏
Sol.
36𝑥2 − 81
= (6𝑥) 2 − 92
𝑈𝑠𝑖𝑛𝑔 𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)
= (6𝑥 + 9)(6𝑥 − 9)

= 9(2𝑥 + 3)(2𝑥 − 3)
b) 𝟐𝟕𝒙𝟖𝒚𝟗 + 𝟒𝟓𝒙𝟗𝒚𝟓
Sol.
27𝑥8𝑦9 + 45𝑥9𝑦5
= 9𝑥 8 𝑦5(3𝑦4 + 5𝑥)
c) 𝒙𝟐 + 𝟐𝒙 − 𝟏𝟓
Sol.
𝑥2 + 2𝑥 − 15
= 𝑥 2 + 5𝑥 − 3𝑥 − 15
= (𝑥 + 5)(𝑥 − 3)
5. Solve each inequality and write your answer in interval notation.
a) −𝟐 < 𝟕𝒙−𝟓
𝟑 ≤ 𝟑
Sol.
−2 < 7𝑥 − 5
3
−6 < 7𝑥 − 5
6 > 5 − 7𝑥
1 > −7𝑥
−1 < 7𝑥
𝑥 > −1
7
7𝑥 − 5
3 ≤ 3
7𝑥 − 5 ≤ 9
7𝑥 ≤ 14
𝑥 ≤ 2
−1
7 < 𝑥 ≤ 2
b) |𝟕𝒙 + 𝟓| > 𝟐

Sol.
|7𝑥 + 5| > 2
−2 < 7𝑥 + 5 < 2
7𝑥 + 5 > −2
7𝑥 > −7
𝑥 > −1
7𝑥 + 5 < 2
7𝑥 < −3
𝑥 < −3
7
−1 < 𝑥 < −3
7
c) 𝟑𝒙 + 𝟓 < 𝟖𝒙 ≤ 𝟐𝟓
Sol.
8𝑥 > 3𝑥 + 5
5𝑥 > 5
𝑥 > 1
8𝑥 ≤ 25
𝑥 ≤ 25
8
1 < 𝑥 ≤ 25
8
6. The area of the front of a book is 𝟔𝟎 𝒄𝒎𝟐. The length of the book is 𝟕 𝒄𝒎 longer than
the width.
a) If the length of the book is 𝒙 𝒄𝒎, find the quadratic equation for the area of the
book 𝑨 using only the variables 𝒙 and 𝑨.
Sol.
𝐴 = 𝑥 ∗ 𝑤𝑖𝑑𝑡ℎ … … … (1)

𝑥 = 𝑤𝑖𝑑𝑡ℎ + 7
𝑤𝑖𝑑𝑡ℎ = 7 − 𝑥 … … … (2)
𝐹𝑟𝑜𝑚 𝐸𝑞(1) 𝑎𝑛𝑑 𝐸𝑞(2)
𝐴 = 𝑥 (7 − 𝑥)
𝐴 = 7𝑥 − 𝑥 2
𝑥2 − 7𝑥 + 𝐴 = 0
b) What are the dimensions of the book?
Sol.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑜𝑘 𝑏𝑒 𝑙 𝑐𝑚 𝑎𝑛𝑑 𝑏 𝑐𝑚 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦
𝐴𝑐𝑐 𝑡𝑜 𝑡ℎ𝑒 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛,
𝑙𝑏 = 60 … … … (1)
𝑙 = 𝑏 + 7 … … … (2)
𝐹𝑟𝑜𝑚 𝐸𝑞(1) 𝑎𝑛𝑑 𝐸𝑞(2)
(𝑏 + 7)(𝑏) = 60
𝑏2 + 7𝑏 − 60 = 0
𝑏2 + 12𝑏 − 5𝑏 − 60 = 0
𝑏(𝑏 + 12) − 5(𝑏 + 12) = 0
(𝑏 + 12)(𝑏 − 5) = 0
𝑏 = −12 𝑜𝑟 𝑏 = 5
∵ 𝑤𝑖𝑑𝑡ℎ 𝑐𝑎𝑛 𝑛𝑜𝑡 𝑏𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒, 𝑤𝑒 𝑖𝑔𝑛𝑜𝑟𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑣𝑎𝑙𝑢𝑒
𝑏 = 5
𝐹𝑟𝑜𝑚 𝐸𝑞(2)
𝑙 = 5 + 7 = 12
𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠: 𝑙𝑒𝑛𝑔𝑡ℎ = 12𝑐𝑚, 𝑤𝑖𝑑𝑡ℎ = 5𝑐𝑚