MATH1021 Mathematics for Engineers Preliminary
Lecture 2: Algebraic Skills Continue - Equations
1. Addition/Subtraction of algebra fractions
As in arithmetic algebraic fractions may be added or subtracted by expressing them in
terms of a common denominator.
Examples: Express as a single fraction in its lowest terms:
(2−𝑥)(𝑥−2)
𝑥
2−𝑥
𝑥(𝑥+3)
+ 𝑥+3 = (𝑥−2)(𝑥+3) + (𝑥−2)(𝑥+3)
𝑥−2
𝑥(𝑥+3)+(2−𝑥)(𝑥−2)
𝑥 2 +3𝑥+2𝑥−𝑥 2 −4+2𝑥
a.
=
b.
c.
1
(𝑥−2)(𝑥+3)
(𝑥−2)(𝑥+3)
7𝑥−4
= (𝑥−2)(𝑥+3)
2𝑚
𝑚−1
− 𝑚2 −1
1
2
𝑥+1
=
+ 𝑥 2 −1
2. Formulae and Change of Subject
A formula is an expression, which describes the relationship between two or more
quantities. For example, the volume of a pyramid with square base is given by the formula
1
𝑉 = 𝑎2 ℎ
3
where 𝑎 is the side of base and ℎ is the perpendicular height.
Given the values of 𝑎 and ℎ then the value of 𝑉 may be found by simple arithmetic after
substituting the given values. For example, given that 𝑎 = 20 𝑚 and ℎ = 30 𝑚 then the
corresponding volume is
𝑉 =
1
× 202 × 30 = 4000 𝑚3
3
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Suppose that for the pyramid we know the values of 𝑉 and ℎ then it is possible to calculate
the value of 𝑎. Algebraically we can write
𝑎2 =
3𝑉
ℎ
to obtain
𝑎 =√
3𝑉
ℎ
this process of re-writing the formula is called ‘changing the subject’, in this case from 𝑉
to 𝑎.
Examples:
a.
Make 𝑢 the subject of the formula
1 2
𝑎𝑡
2
Bring all the terms with 𝑢 to one side and other terms to another side
1
𝑠 − 𝑎𝑡 2 = 𝑢𝑡
2
Factorize (if required) to bring 𝑢 outside (not in this question) then bring all terms to
another side so that only 𝑢 in one side.
In this question, we divide by 𝑡 both side to get rid of 𝑡 on the right hand side.
1
𝑠 − 2 𝑎𝑡 2
=𝑢
𝑡
1
1
Therefore, 𝑢 = 𝑡 (𝑠 − 2 𝑎𝑡 2 )
𝑠 = 𝑢𝑡 +
b. Make u the subject of the formula
𝑓=
𝑢𝑣
𝑢+𝑣
c. Make ℎ the subject of the formula
𝐴 = 𝜋𝑟 √𝑟 2 + ℎ2
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3. Surds and Rationalisation
a. Surds
Surds are numbers in root form, e.g √2, √10, √45 = 3√5 (simplest form)
Examples: Express the following in simplest surd form
a. √27 − 4√3 + √48 = √9 × 3 − 4√3 + √16 × 3 = √32 × √3 − 4√3 + √42 × √3
= 3√3 − 4√3 + 4√3 = 3√3
b. √12 + √3 − 2√48
c. √6(√3 − √2)
b. Rationalisation:
The process by which a fraction involving surds is rewritten so that the denominator
contains only rational numbers (no more surds in the denominator). To rationalise a
fraction, we multiply the numerator and denominator with the denominator’s conjugate.
Examples:
1. Find the conjugate of the following terms:
a. √3 + √2. Conjugate is √3 − √2
b.
√7 − 2√5
2. Rationalise the following fractions:
a.
b.
c.
1
√3−√2
=(
1×(√3+√2)
=
)
√3−√2)(√3+√2
√3+√2
2
2
(√3) −(√2)
=
√3+√2
3−2
= √3 + √2
1
2√5+3√2
3√2
√5−√3
3. Simplify:
1
2√7 + 6
−
2
3√2 − 2
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4. Division
a. Polynomial Expression
An expression of the form:
𝑎𝑥 + 𝑏 is said to be a linear in 𝑥
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 is said to be a
is said to be a cubic in 𝑥
is said to be a quartic in 𝑥, etc
These are all examples of polynomial expressions in 𝑥 and such expressions may be
divided by algebraic division (long division).
b. Long Division
See Refresher Unit 2, page 9 and 10 for description of steps.
Examples:
a. Divide 2𝑥 2 – 3𝑥 + 4 by 𝑥 + 1
Dividend →
2𝑥 2 − 3𝑥 + 4
𝑥+1
 divisor
2𝑥 2 + 2𝑥
2𝑥 − 5
 quotient
− 5𝑥 + 4
−5𝑥 − 5
9  remainder
Therefore, 2𝑥 2 − 3𝑥 + 4 = (𝑥 + 1)(2𝑥 − 5) + 9
b. Divide 𝑥 3 – 3𝑥 2 + 6𝑥 – 4 by 𝑥 – 1
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c. Divide 4𝑥 2 − 3𝑥 + 2 by 𝑥 – 2
5. Algebraic Equations
5.1Linear Equation:
An equation contains only the first power of a single unknown 𝑥 is said to be a linear
equation. General form after simplified: 𝑎𝑥 + 𝑏 = 0
Eg: 3𝑥 − 2 = 10 − 𝑥
To solve linear equation, we bring all terms with 𝑥 to one side and remaining terms to
another side.
Eg: 3𝑥 − 2 = 10 − 𝑥
3𝑥 + 𝑥 = 10 + 2
4𝑥 = 12
𝑥=
12
=3
4
Examples: Solve the following equations:
a. 7𝑥 + 5 = 4𝑥 − 6
b.
𝑦−1
7
=
𝑦+1
4
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1
c.
2+3𝑝
+
1
3−𝑝
=0
1
1
=−
2 + 3𝑝
3−𝑝
3 − 𝑝 = −(2 + 3𝑝)
3 − 𝑝 = −2 − 3𝑝
−𝑝 + 3𝑝 = −2 − 3
2𝑝 = −5
𝑝=−
d.
3
𝑞+5
=
5
2
4
𝑞−2
5.2 Simultaneous linear equations:
Consider the two equations
3𝑥 + 2𝑦 = 21 (1)
2𝑥 + 5𝑦 = 3 (2)
each of which contains the two unknown quantities 𝑥 and 𝑦. Such a pair of equations are
called simultaneous equations. Their solution is the values of 𝑥 and 𝑦 which satisfy both
equations.
Method 1: Elimination
- multiplying each equation by appropriate numbers so that the coefficient of 𝑥 or 𝑦
is the same in both equations.
- then eliminate that variable (𝑥 or 𝑦) to produce a single equation in one unknown.
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Example: solve the above system of linear equation using elimination method
We will eliminate 𝑥, so we multiply the first equation by 2 and the second equation by 3
6𝑥 + 4𝑦 = 42
(3)
6𝑥 + 15𝑦 = 9
(4)
(4) − (3) we gain: (6𝑥 − 6𝑥) + (4𝑦 − 15𝑦) = 42 − 9
11𝑦 = −33
𝑦=−
33
= −3
11
Substitute 𝑦 = −3 to equation (1) to solve for 𝑥
3𝑥 + 2 × (−3) = 21
3𝑥 − 6 = 21
3𝑥 = 21 + 6 = 27
𝑥=
27
=9
3
Therefore, 𝑥 = 9, 𝑦 = −3 is the solution. We can double check by substitute this solution to
the second equation to see if it is correct.
Method 2: Substitution
- from one equation, express one variable in term of another variable
- then substitute in to the remaining equation to produce a single equation in one
unknown.
Example: solve the above system of equation again using substitution method.
3𝑥 + 2𝑦 = 21 (1)
2𝑥 + 5𝑦 = 3 (2)
From (1), we express 𝑦 in terms of 𝑥:
2𝑦 = 21 − 3𝑥
𝑦=
Then substitute 𝑦 =
21−3𝑥
2
21 − 3𝑥
2
to the second equation, we have
2𝑥 + 5 ×
21 − 3𝑥
=3
2
Multiply both sides with 2 to get rid of the denominator then solve for 𝑥
2 × 2𝑥 + 5(21 − 3𝑥) = 3 × 2
4𝑥 + 105 − 15𝑥 = 6
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−11𝑥 = 6 − 105 = −99
𝑥=
−99
=9
−11
Substitute 𝑥 = 9 back to 𝑦 expression to find 𝑦
𝑦=
21 − 3𝑥 21 − 3 × 9 21 − 27
6
=
=
= − = −3
2
2
2
2
Therefore, 𝑥 = 9, 𝑦 = −3 is the solution. Double check the answer if you have time!
More examples: solve the following system of linear equations
a. 7𝑥 – 6𝑦 = 20
3𝑥 + 4𝑦 = 2
b. 𝑐 + 4𝑑 = – 16
3𝑐 + 2𝑑 = 20
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c. A total of 78 seats for a concert are sold, producing a total revenue of $483. If seats
cost either $2.50 or $10.50, how many $2.50 seats and how many $10.50 seats were
sold?
5.3 Quadratic equation
General form of a quadratic equation:
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
There are different ways to solve this equation. We will learn 2 typical ways.
Method 1: Factorisation
- Factorise 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = (𝑎1 𝑥 + 𝑚)(𝑎2 𝑥 + 𝑛) = 0
- Let each factor be 0 and solve for 𝑥
𝑚
𝑎1 𝑥 + 𝑚 = 0, so 𝑥 = − 𝑎
1
𝑛
𝑎2 𝑥 + 𝑛 = 0, s0 𝑥 = − 𝑎
2
Examples: Solve the following equations using factorisation method
a.
𝑥 2 + 12𝑥 + 35 = 0
Factorise: 𝑥 2 + 12𝑥 + 35 = (𝑥 + 5)(𝑥 + 7) = 0
𝑥 + 5 = 0 or 𝑥 + 7 = 0
𝑥 = −5, −7 are the solutions.
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b. 3𝑦 2 – 7𝑦 + 2 = 0
c. 4𝑚2 – 4𝑚 – 3 = 0
d. 𝑥 2 − 6𝑥 + 8 = 0
Method 2: Quadratic formula
Consider the general quadratic equation
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
Then the solutions are:
𝑥=
•
•
•
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
If 𝑏 2 − 4𝑎𝑐 > 0, we have 2 roots/solutions
If 𝑏 2 − 4𝑎𝑐 = 0, we have 1 root
If 𝑏 2 − 4𝑎𝑐 < 0, we have no solution
Examples: solve the following equations:
a. 3𝑦 2 – 7𝑦 + 2 = 0
𝑎 = 3, 𝑏 = −7, 𝑐 = 2
𝑦=
−𝑏 ± √𝑏 2 − 4𝑎𝑐 −(−7) ± √(−7)2 − 4 × 3 × 2
=
2𝑎
2×3
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𝑦=
7 ± √25 7 ± 5
=
6
6
𝑦 = 2,
1
3
b. 2𝑥 2 + 3𝑥 + 7 = 0
c. 2𝑥 2 + 3𝑥 – 1 = 0
d. The height ℎ (above the ground level) of a ball thrown vertically upwards is given
by
ℎ = −4.9𝑡 2 + 55𝑡 + 12
where 𝑡 is time in second. Find the time taken to reach the ground.
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6. Homework
1. Express as a sing fraction in its simplest form
a.
b.
𝑥+5
𝑥+3
+ 2𝑥+1
2−𝑥
2𝑥
𝑥 2 +𝑥−6
1
+ 𝑥−2
2. When two electrical resistances of magnitudes 𝑟1 and 𝑟2 respectively are connected in
parallel the total effective resistance 𝑅 is given by the formula
1 1 1
= +
𝑅 𝑟1 𝑟2
Find the effective resistance 𝑅 when 𝑟1 = 8 ohms and 𝑟2 = 3 ohms
3. Make 𝑛 the subject of the formula
𝐼=
𝑛𝐸
𝑅 + 𝑛𝑟
4. Express the following in simplest surd form
a. 2√50 − 3√18
b. √5 + √2 − √45 + √8
c.
√15
√5
d. √2(√2 + √8)
e.
1
√5−2
−
√2
√3−√2
5. Divide 2𝑥 3 + 3𝑥 2 − 5𝑥 + 15 by 𝑥 + 3
6. Solve:
𝑥−3
2𝑥+1
a. 5 = 4
b.
6
𝑥+2
2
3
− 𝑥−2 = 0
c. 3𝑥 − 14𝑥 + 8 = 0
d. 𝑥 2 − 𝑥 − 2 = 0
e. 2𝑥 2 + 9𝑥 − 5 = 0
7. The currents 𝐼1 and 𝐼2 satisfy the following system of equations:
16 𝐼1 − 9 𝐼2 = 110
20 𝐼2 − 9 𝐼1 + 110 = 0
Find 𝐼1 and 𝐼2
8. An analysis of a circuit shown yields the quadratic equation for
the current 𝐼 as 3𝐼 2 + 6𝐼 = 45, where 𝐼 is in amps. Find 𝐼
More questions can be found on vUWS and text book.
Page 12 of 13
Answers:
𝑥 2 +10𝑥+11
3(𝑥+1)
1. a. (2−𝑥)(2𝑥+1) b. (𝑥+3)(𝑥−2)
𝑟 𝑟
24
2. 𝑅 = 𝑟 1+𝑟2 ; 𝑅 = 11 ohm.
1
2
𝐼𝑅
3. 𝑛 = 𝐸−𝐼𝑟
4. a. √2; b. 3√2 − 2√5; c. √3; d. 6; e. √5 − √6
5. 2𝑥 3 + 3𝑥 2 − 5𝑥 + 15 = (2𝑥 3 − 3𝑥 + 4)(𝑥 + 3) + 3
17
2
1
6. a. 𝑥 = − 6 ; b. 𝑥 = 6; c. 𝑥 = 3 , 4; d. 𝑥 = −1, 2; e. 𝑥 = −5, 2
7. 𝐼1 =
1210
239
, 𝐼2 = −
770
239
8. 𝐼 = −5, 3
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