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n Monday night, 7 January 2002, the asteroid, known as
2001 YB5, passed within 600 000 kilometres of Earth.
In galactic terms, this is a very near miss. In 1908, a
60-metre-wide asteroid smashed into Siberia and released
the same amount of energy as 600 nuclear bombs. 2001 YB5
was 500 metres wide!
If 2001 YB5 had hit Earth near Melbourne, then nothing
would have survived within a 150-kilometre radius of the
impact point and the waves of destruction would have
Astronomers and physicists use complex algebraic
mathematical models to predict the likelihood and location
of impact of large asteroids. Thousands of small asteroids
are destroyed each year entering Earth’s atmosphere, but
large ones like 2001 YB5 have the potential for mass
destruction.
47
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Prepare for this chapter by attempting the following questions. If you have difficulty with
a question, click on the Replay Worksheet icon on your Exam Café CD or ask your teacher
for the Replay Worksheet.
e
Worksheet R2.1
e
Worksheet R2.2
2 In the equations below, the equals sign is incorrectly positioned. Reposition the equals sign so
that the left-hand side of the equation equals the right-hand side.
(a) 7 + 4 + 6 = 5
(b) 6y + 8 + 5 + 3 = 5y + y
(c) (10 ÷ 2) = (6 × 3) + (5 × 4) + (9 ÷ 3)
(d) 5x + 6 − 2x + 8x = 9 − 3 − 5x
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Worksheet R2.3
3 Expand each of the following expressions and simplify where possible.
(a) 2(a + 4)
(b) x(x + 5)
(c) −3(p − 4)
(d) x(y − x)
(e) (y + 3)(y − 2)
(f) (m + 3)(m − 3)
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Worksheet R2.4
4 Solve each of the following equations.
(a) x + 4 = 11
(b) 2x = 9
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Worksheet R2.5
5 Sketch each of the following on separate axes.
(a) y = 3x − 2
(b) y = − 2--3- x + 4
(c) f − 2g
(f) 6(e + 2f )
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1 Given e = 6, f = −2 and g = 3, find the values of:
(a) e + f
(b) g − f
(d) 6e + 2f
(e) 2f 2
m
(c) ---- = 3
5
m
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(c) 2x + 3y = 6
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Order of operations:
1. Brackets.
2. Addition and subtraction (work from left to right).
3. Multiplication and division (work from left to right).
Expanding: a(b + c) = a × b + a × c.
y
Sketching linear graphs:
c
Form: y = mx + c
rise
run
rise
y-intercept = c, gradient = m = --------- .
run
x
Form: ax + by = d
Substitute x = 0, to find y-intercept.
Substitute y = 0, to find x-intercept.
y
y-intercept
x
x-intercept
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2.1
Solving linear equations
Operation
+
−
×
Opposite operation
−
+
÷
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Algebra enables the efficient summary of patterns and relationships. Algebraic techniques are
the processes by which many mathematical and real-world problems are solved.
A linear equation contains one or more variables raised only to the power of one.
1
Hence, a linear equation will not contain x2, x3, --- , x etc.
x
For example, 2x − 1 = 7 is a linear equation. An equation is a statement that is only true for
certain values of the variable. The equation 2x − 1 = 7 is only true for x = 4. When we are solving
any equation we are finding the value or values of the pronumeral that make the equation true;
that is, the value on the left-hand side (frequently abbreviated to LHS) of the equation equals
the value on the right-hand side (RHS) of the equation. A linear equation will have only one
value for the pronumeral that makes the equation true. One method to solve equations is to
perform opposite operations to both sides of the equals sign to ‘undo’ the equation and get
back to the value of the pronumeral.
Whatever operation we want to remove, we simply perform the opposite operation to both
sides of the equals sign. Examples of the most common operations are given in the table below.
÷
×
square
±
square
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Opposite operations are more formally called inverse operations. If we have built up the
equation, the last operation performed is usually the first operation we undo in solving the
equation. This undoing procedure is continued to find the value of the pronumeral that makes
the equation true. To describe the steps involved in solving an equation we need to use the
words ‘coefficient’ and ‘constant’. In the expression 2x − 1, the number multiplying the
variable x is called the coefficient of x (in this case, 2) and the number added or subtracted
is the constant or the term independent of x (in this case, −1).
worked example 1
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Solve the equation 2x − 1 = 7.
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Steps
1. Add the value of the constant to both sides. In
this case, to undo subtract 1, we add 1 to both
sides.
2. Divide both sides by the coefficient, in this
case, 2.
4. Check the solution by substituting into the
original equation.
Solution
2x − 1 = 7
2x − 1 + 1 = 7 + 1
∴ 2x = 8
2x
------ = 8
--2
2
∴x=4
2×4−1=7
Often, the individual steps in solving an equation involve simple arithmetic; therefore, the
steps shown in red in the above example are frequently omitted.
Many of the more complicated equations require only one or two steps before they look like
Worked Example 1. It is a useful practice to look for the steps required to get back to an equation
of the form of Worked Example 1; that is, an equation that looks the same but has different
numbers.
The table on the next page gives a useful guide to the steps required to solve many common
equations to the stage of being in the same form as Worked Example 1. If you can solve
equations like this, look for the steps required to get to this stage. This table does not strictly
follow the rule of undoing operations in the opposite order. It takes an approach designed to
make the solution process as simple as possible.
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First steps in equation solving
Equation
First step
Result after first step
2x – 1
-------------- = 4
3
Multiply both sides by 3.
2x − 1 = 12
2
4(3p + 2) = 7
Expand brackets.
12p + 8 = 7
3
6m − 7 = 4m + 2
Subtract 4m from both sides.
2m − 7 = 2
4
5(2y − 5) = 3(y − 1)
Expand brackets.
10y − 25 = 3y − 3
Now, proceed as for equation 3.
5
3(2a – 1)
--------------------- = 4
7
Expand brackets.
6a – 3
-------------- = 4
7
Now, proceed as for equation 1.
6
2t – 1 5t – 7
------------- = ------------3
4
To avoid working in fractions
multiply by the LCD, which is 12.
4(2t − 1) = 3(5t − 7)
Now, proceed as for equation 4.
7
5(2x – 1) x + 2
--------------------- = ----------6
5
Expand bracket.
10x – 5 x + 2
----------------- = ----------6
5
Now, proceed as for equation 6.
8
2x + 3 x + 1
-------------- = ----------- + 2
4
3
To avoid working in fractions
multiply by the LCD, which is 12.
worked example 2
12(2x + 3) 12(x + 1)
------------------------ = --------------------- + 2 × 12
4
3
3(2x + 3) = 4(x + 1) + 24
Expand (equation 4) and collect like
terms before proceeding.
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Solve 3(2 − 3a) = 2(a − 8).
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Steps
1. Expand the brackets on both sides of the
equation.
2. ‘Remove’ the pronumeral from one side.
(Either add 9a to both sides, or subtract 2a
from both sides. In this case, adding 9a is
better because you then work in positives
rather than negatives.)
3. If necessary, ‘reverse’ the equation to get the
pronumeral on the LHS.
4. Eliminate the constant from the LHS.
5. Divide both sides by the coefficient of the
pronumeral.
6. Verify the solution by substituting the value
into both sides of the equation and checking
that LHS = RHS. Note: You don’t need to
actually write step 6, as you can usually check
50
Solution
3(2 − 3a) = 2(a − 8)
6 − 9a = 2a − 16
6 − 9a + 9a = 2a − 16 + 9a
6 = 11a − 16
11a − 16 = 6
11a = 22
11a 22
---------- = -----11
11
a=2
LHS = 3(2 − 3 × 2)
= 3(−4)
= −12
RHS = 2(2 − 8)
= 2(−6)
= −12
As LHS = RHS the answer is correct.
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When you reverse the equation, as in step 3 in Worked Example 2, be careful not to change the sign
of any term. If you reverse 6 = 16 − 11a, you get 16 − 11a = 6 not 11a − 16 = 6.
worked example 3
2e – 4 4e – 1
Solve the equation ---------------- = ---------------- .
3
7
Solution
2e – 4 4e – 1
---------------- = ---------------3
7
7(2e − 4) = 3(4e − 1)
14e − 28 = 12e − 3
14e − 28 − 12e = 12e − 3 − 12e
2e − 28 = −3
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Steps
1. To eliminate the fractions, cross-multiply.
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2t – 1 5t – 7
In equation 6 in the table on page 46, -------------- = -------------- our first objective is to eliminate the
3
4
fractions. To achieve this we can multiply both sides by the lowest common denominator,
(3 × 4 = 12), and then ‘cancel down’.
12(2t – 1) 12(5t – 7)
------------------------- = ------------------------ simplifies to 4(2t − 1) = 3(5t − 7)
3
4
Alternatively, we can convert both sides to a common denominator, and then multiply by
the denominator.
4(2t – 1)- --------------------3(5t – 7)--------------------=
also simplifies to 4(2t − 1) = 3(5t − 7).
12
12
These steps can be shortened by a process known as cross-multiplying.
Whenever an equation is of the form fraction = fraction, we can simply multiply the LHS
numerator by the RHS denominator and multiply the RHS numerator by the LHS
denominator.
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2. Expand the brackets.
3. Subtract the smaller value of the variable (in
this case, 12e) from both sides to eliminate the
variable from one side, and collect like terms.
4. Eliminate the constant from the side involving
the pronumeral. (Add 28 to both sides.)
5. Divide both sides by the constant (2).
6. Check the answer by substituting into the
original equation.
2e = 25
e = 12 1--22(12 1--2 ) – 4 25 – 4 21
- = ---------------- = ------ = 7
LHS = ------------------------3
3
3
1
4(12 --2 ) – 1 50 – 1 49
- = ---------------- = ------ = 7
RHS = ------------------------7
7
7
∴ LHS = RHS
2x + 5- = 3x
– 3- + 2
------------You cannot use cross-multiplying when the equation is of the form ------------2
6
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worked example 4
A rectangle has a perimeter of 72 cm. If the length is 2 cm more than the width, find the dimensions of
the rectangle.
Steps
1. Introduce a pronumeral to represent an unknown
quantity.
2. Express other quantities in terms of this unknown.
3. If possible, draw a diagram. You can sometimes
define the unknown quantities on the diagram and
omit steps 1 and 2.
Solution
Let x represent the width of the rectangle.
Length will be x + 2
x
x+2
6. Solve the equation.
7. Interpret and verify the solution.
8. State the final answer in words.
2x + 2(x + 2) = 72
2x + 2x + 4 = 72
4x + 4 = 72
4x + 4 − 4 = 72 − 4
4x = 68
x = 17
If x = 17, then x + 2 = 19
2(17) + 2(19) = 72
The dimensions of the rectangle are width
17 cm and length 19 cm.
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4. Express the information as an equation. Twice the
width plus twice the length equals the perimeter.
5. If possible, simplify the equation.
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Solving equations using CAS
Your CAS has a Solve function that enables you to check the solutions you have found.
2e – 4 4e – 1
The equation from Worked Example 3, -------------- = --------------, will be used to illustrate the process.
3
7
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Using the TI-Nspire CAS
Solve is found by pressing b > Algebra >
Solve and we would be wise to use the fraction
template, / p , to make correct entry
easier. Notice that you need to put the variable
for which you are solving after a comma.
52
Solve is found by tapping Action > Equation/
Inequality > solve and we would be wise to use
the fraction template, found in the )
key pad, to make entry easier. You should use
the V key pad to enter the variables. The
variable of interest is included at the end after a
comma.
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Solving linear equations
1 Solve the following equations.
(a) 3m − 2 = 7
(b)
(k) 4(3p + 2) = 7
(l) 5 − 2--7- d = 3
(b) 5 − 2a = 5a + 2
(e) 4 − 3x = 6 − 2x
(h) 8 − 2(3 − 4e) = 5(e − 4)
(k) −2(3 − 5d) + 1 = 2d − 1
(c) 5e + 7 = 2e − 5
(f) 3(3 − 2f ) = f + 5
(i) −4(3m + 1) = 5(2 − 3m)
(l) 4 − 3(2b − 1) = 3(2b − 1)
(b)
(e)
(h)
(k)
e
10x – 5 x + 2
----------------- = ----------6
5
6 – 2y y + 1
--------------- = -----------3
5
3(5 – f ) 3f + 2
------------------- = -------------2
6
7(4 – 2h) h
----------------------- = --3
4
Hint
Worked Example 1
Worked Example 2
Worked Example 3
3m – 3 3m + 2
(c) ---------------- = ----------------5
3
−
2(2x + 4)
5(3x – 5)
(f) ----------------------- = ------------------------3
7
− 4(g + 3)
g
(i) ---------------------- = --5
2
3(2a – 1) a
(l) ----------------------- = --2
5
Worked Example 4
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5 – 2y
(g) --------------- = 2
3
3(2a – 1)
(j) ----------------------- = 4
7
2 Solve the following equations.
(a) 6m − 7 = 4m + 2
(d) 5(2y − 5) = 3(y − 1)
(g) 5(2y + 1) = 3y + 8
(j) −(7 − 2a) = 7 − 2a
3 Solve the following equations.
2x + 1 3x – 2
(a) --------------- = -------------3
4
5(2x – 1) x + 2
(d) ----------------------- = ----------6
5
2(3 – 4x) x + 1
(g) ----------------------- = ----------5
2
− 7(2 – y)
y
(j) ---------------------- = --3
4
4 The perimeter of a rectangular
number plate is 92 cm. If the
width is 6 cm less than the
length, find the dimensions of
the number plate.
(c) 3 − 2x = 5
2(g – 3)
(f) ------------------- = 6
5
2d
(i) 3 − ------ = 7
5
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(d) 6d = 2d + 4
− 5 = −8
2x – 1
(e) -------------- = 4
3
− 4(1 – x)
(h) ---------------------- = 8
3
−2e
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exercise 2.1
5 Solve the following. Hint: Why can’t you cross-multiply these equations?
2x + 3 x + 1
2(4y + 3) 3y + 2
2e + 3
–4
(a) --------------- = ----------- + 2
(b) ----------------------- = --------------- − 1
(c) 3e
-------------- + 2--- = --------------4
3
5
3
3
7
3
n
c
3m
–
5
3m
3
–
n
2
2
–
3c
1
(d) ---------------- – ------- = 2m + 1
(e) ------------ + --- = --(f) -------------- + --- = --6
4
2
5 2
4
5 2
Multiple choice
− 5(x – 2)
6 The solution to the equation ---------------------- = 3 is:
4
14
−
−
-----B ----C 22
A 2
5
5
3x – 4 2x + 1
7 The solution to the equation -------------- = --------------- − 1 is:
3
5
20168-----A ----B
C
3
3
9
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Hint
− 22
-----5
D
− 2--5
E
D
22----9
E 4
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e
Hint
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8 The repair charges to Jo’s hot water
heater were a fixed service call fee of \$55
covering the first 30 minutes of service,
plus an additional charge for each
15 minutes of service beyond the initial
30 minutes.
(a) If the additional charge was \$20 per
each 15 minutes beyond the initial
30 minutes, what would be the
charge for a repair that took 1 hour?
(b) If the service call took 1 hour
15 minutes and the bill was
\$122.50, what was the charge for
each 15 minutes beyond the initial
30 minutes?
(c) Another plumber charges a fixed
service call of \$35, covering the first
20 minutes of the service, plus an
additional charge of \$20 for each
additional 15 minutes, or part of
15 minutes. How much will this
plumber charge for a service call
that takes 1 hour 15 minutes?
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2.2 Formulae and substitution
Formulae
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A formula is an equation or rule that mathematically expresses the relationship between two
or more quantities. A formula is therefore a shorthand form of expressing a relationship, thus
enabling us to more easily calculate solutions and analyse and solve practical problems. Some
common formulae that you might recognise include:
A = --12- bh
area (A) of triangle of base b and height h
A = πr2
area (A) of circle of radius r
2
F = 9--5- C + 32 converts degrees Celsius to
V = πr h volume (V ) of a cylinder of radius r and height h
degrees Fahrenheit
PrT
I = --------calculates the simple interest earned (I ) when an amount (\$P) is invested at a rate of r% for T years
100
T = 2π -l- period (T) in seconds of a swinging pendulum of length l metres
g
d
v = --average speed (v) of a moving object that travels d metres in t seconds
t
The letters used in equations are called pronumerals, as they represent numbers. If the base
of a triangle is 6 cm and the height is 8 cm the area is given by:
A = 1--2- bh, then A = 1--2- × 6 × 8 = 24 cm2
Note: A pronumeral can represent either an unknown constant, which is a fixed value, or
a variable, which can take different values. In the equation y = 2x + 3, 2 and 3 are constants
and x and y are variables.
l
The formula for calculating the period of a pendulum is: T = 2π -g
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The pronumeral, g, represents the gravitational constant, which is assumed at the Earth’s
surface to have a value of 9.8, and π = 3.141 592 6…; hence, π and g are constants and the
period, T, and pendulum length, l, are variables. If we were conducting an experiment to
examine the period of a pendulum, we would vary the length, l, of the pendulum to determine
its effect on T; hence, the values of l and T would vary.
Conventions
Conventions exist for choosing which letters to use for pronumerals and for the formatting of
formal typed or word-processed formulae.
• Letters at the beginning of the alphabet (a, b, c, …) are usually used to denote constants.
• Letters at the end of the alphabet (… x, y, z) are usually used to denote variables.
• Where possible, letters are chosen that help to relate to the quantity they represent, such as
A for area and V for volume.
• This textbook follows the convention of presenting all pronumerals in italic.
Substitution
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Substitution is the process of replacing one quantity or number by another. The most common
mathematical use of substitution is to replace a pronumeral by the number that it represents.
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Before substituting values into a formula, check to make sure the units are compatible. For example,
if one given side of a rectangle was in centimetres and the other in metres, you would have to
convert them both to centimetres or both into metres before proceeding to calculate perimeter, area,
volume etc. Common sense and practicality usually dictate the sensible choice of units. We
wouldn’t, for example, measure the distance from Melbourne to Perth in millimetres!
worked example 5
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This formula is used to convert temperature in degrees Fahrenheit (°F) to temperature in degrees
Celsius (°C):
C = 5-9 (F − 32)
Use the formula to convert 100°F to degrees Celsius.
Steps
1. Write the formula.
Solution
C = 5-9 (F − 32)
2. Substitute known values.
3. Evaluate, using a calculator if necessary.
(Calculator shows 37.777 777 78, but we
would never write down all these digits.)
4. State the final answer, including units, to an
appropriate degree of accuracy.
C = 5-9 (100 − 32)
Display shows 37.777…
C = 37.8°C
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Defining a function to aid substitution
Your CAS can be very helpful with substitutions, especially if you need to substitute a lot of
numbers into the same expression. One of the ways your CAS can help you is by defining the
expression. The degree conversion from Worked Example 5 will be used to illustrate this.
Once you have defined the rule you can just
enter whatever value you like for f. If you don’t
want the exact answer you can get an
approximation by tapping D . .
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Once you have defined the rule you can just
enter whatever value you like for f. If you don’t
want the exact answer you can get an
approximation by pressing / · .
You find Define by tapping Action > Command
> Define. Notice the need to include the
variable, f, on the LHS of the definition. Of
course, you could use whatever variable you
liked as long as you were consistent. You do
need to be careful to use c from the 0
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Using the TI-Nspire CAS
You find Define by pressing b > Actions >
Define. Notice the need to include the variable, f,
on the LHS of the definition. Of course, you could
use whatever variable you liked as long as you
were consistent.
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Don’t forget to include a multiplication between
the fraction and the bracket.
Degree of accuracy
Depending on the situation, an appropriate degree of accuracy is determined either by
common sense in the particular situation or, if we are referring to anything measured with an
instrument, by the accuracy limitation of the particular instrument. You would not normally
need to know your height in millimetres—to the nearest centimetre would usually be accurate
enough. For a scientific application, an appropriate degree of accuracy would usually be
determined by the accuracy limitations of the measuring instruments. With a ruler, for example,
we could probably measure to an accuracy of 0.5 mm. If we needed greater accuracy we would
require a more accurate measuring instrument.
If you are going to perform a calculation on measured figures, your answer cannot be more
accurate than the figures you used to calculate the answer. Therefore, if the given figures in the
same units were to two decimal places, an answer you calculated from these figures could not
be to more than to two decimal places.
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Sometimes you will be instructed to state your final answer to a given number of decimal
places. When such a situation occurs, assume this instruction overrides any other considerations.
worked example 6
The period of a pendulum, T seconds, which is the time taken for a pendulum of length, l metres, to swing
one to-and-fro motion, is given by T = 2π --l- , where g = 9.8. Find the period of a pendulum of length 6 cm.
g
Solution
1. Write the formula.
2. Convert units if necessary. The rule states that
l is measured in metres, so we must convert
the length into metres.
3. Substitute known values.
4. Evaluate and state the answer correct to
appropriate accuracy (one decimal place here,
as this is the accuracy of the input values).
exercise 2.2
l
T = 2π --g
l = 6 cm
6
= ---------- m
100
= 0.06 m
0.06
T = 2π ----------9.8
T = 0.5 seconds
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Steps
Formulae and substitution
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1 The formula C = --59- (F − 32) is used to convert temperature in Fahrenheit to temperature in
Celsius. Use the formula to convert 68.0°F to degrees Celsius.
l
2 The period of a pendulum, T seconds, of length, l metres, is given by T = 2π -g
where g = 9.8. Find the period of a pendulum of length 15.4 cm.
Worked Example 5
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Hint
Worked Example 6
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3 Write each of the following statements as a mathematical
formula.
(a) The area (A) of a rectangle is the product of its
length (l) and width (w).
(b) The circumference (C) of a circle is the product of its
(c) The average speed (v) of a moving body is the
distance (d) travelled divided by the time (t) taken.
(d) The charges (C) to repair a video camera are \$35 per
service, plus \$12 per 15 minutes of repair time (n).
(e) In a right-angled triangle, the square of the
hypotenuse (c) is equal to the sum of the squares
of the other two sides. Let the other two sides be
a and b.
(f) The potential difference (V) across the terminals of
an ohmic conductor is given by the product of the
current (I) flowing through the device and the
resistance (R) of the conductor.
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(g) The daily wage (w) Laura earns selling
ﬂowers at a ﬂorist shop is \$16 per day plus
\$1.20 per bunch (n) of ﬂowers she sells.
(h) The cost (C) per day of hiring a car is a
ﬁxed charge of \$22 per day, plus 40 cents
per kilometre (d ) travelled.
(i) The volume (V ) of a prism is the product
of the cross-sectional area (A) and the
height (h).
(j) The volume (V ) of a sphere is the product
of 4--3- π and the cube of the radius (r).
e
Hint
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Hint
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4 For each of the following, ﬁnd the value of the ﬁrst named pronumeral by substituting the
stated values into the given formula.
(a) v = u + at, u = 4, a = 3, t = 7
(b) A = lw, l = 4.2, w = 6.7
(c) V = 4--3- πr 3 , r = 3
(d) S = ut + 1--2- at 2 , u = 3.2, t = 3, a = 2
(e) E = mc2, m = 5, c = 3 × 108
(f) V = IR, I = 2.4, R = 400
(g) S = 2πr (r + h), r = 4.6, h = 8.9
(h) F = ma, m = 8.7, a = 4.7
n
(i) t = ar , a = 3, r = 2, n = 4
(j) F = --95- C + 32, C = 30
2
πd
(k) A = --------, d = 6
(l) T = 2π -l-, l = 16.5, g = 9.8
4
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Multiple choice
l
5 The period, in seconds, of a swinging pendulum of length, l metres, is given by T = 2π -g
where g = 9.80. The period, in seconds, of a pendulum of length 8 cm is closest to:
A 0.05
B 0.18
C 0.57
D 5.13
E 5.68
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S – 2πr 2
6 Substituting S = 64 and r = 3 into the formula h = -------------------- produces a value of h closest to:
2πr
A 0.395
B 2.395
C 3.902
D 23.641
E 34.391
7 Bernie, the organic dairy farmer, wants to
fence off a new rectangular paddock to
quarantine his new bull, named Dunny
Doo, from the cows. He only needs to fence
three sides, as the new paddock will be
adjoining an existing fence. Bernie has a
200-metre roll of fencing wire. The wire is
multi-strand hinge-joint, so Bernie doesn’t
need to consider the strands of wire in
determining his fencing requirements.
Let x = the length of the longer side of the
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The diagram opposite shows a quadrilateral drawn on a grid of square dots. For the given
shape, the number of enclosed or inside dots (I) is 4, and the number of perimeter dots (P)
is 9. As the shape is a trapezium the enclosed area (A) is 1--2- (3 + 2)3 = 7 1--2- squares.
(i) Using 1 cm graph paper, taking the points of intersection of the 1 cm lines as
representing ‘dots’, construct squares of increasing size, for example, (1 × 1),
(2 × 2), up to (5 × 5), and record the results in a table as shown below.
Square size
Area (A)
Perimeter dots (P)
Inside dots (I)
1×1
1
4
0
2×2
4
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(a)
1
3×3
4×4
5×5
8
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(ii) Find a relationship connecting all three variables, the perimeter dots (P), inside dots (I) and area (A).
(b) Systematically construct rectangles of varying sizes and find a relationship between P, I and A for
rectangles.
(c) Find the relationship between P, I and A if the shape is a trapezium.
(d) Generalise the relationship between P, I and A for any quadrilateral drawn on a square grid of dots.
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Westland
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Part A
At the Westland Shopping Centre, there are two florist shops. Both have offered Laura a part-time Saturday
morning job, standing outside the florist shop selling cut flowers. The main difference between the two jobs is
the way in which Laura will be paid.
At the ‘Poise and Ivy’ florist shop she will be paid \$16 every Saturday morning, plus a bonus of \$1.20 for each
bunch of cut flowers she sells.
At the ‘Friendly Triffid’ florist shop she will be paid \$32 every Saturday morning, plus a bonus of 40 cents for
each bunch of cut flowers she sells.
1 Let w represent Laura’s wage, and n represent the number of bunches of flowers sold. Express the given
information as two simultaneous equations.
2 How many bunches of flowers would Laura need to sell so that she makes the same amount of money
whichever job she accepts?
3 How much money would Laura make if she did sell this many bunches of flowers?
4 What information should Laura consider when deciding which job to accept?
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Part B
Next door to the ‘Friendly Triffid’ at
Westland Shopping Centre is ‘Mario’s
Wholesale Green-grocers’, which is
having a sale on boxes of fruit and
vegetables.You can buy two different boxes
for the following combinations and prices.
• A box of apples and a box of bananas
for \$21.
• A box of bananas and a box of carrots
for \$24.
• A box of carrots and a box of dill for \$32.
• A box of dill and a box of eggplant
for \$37.
• A box of eggplant and a box of figs
for \$31.
• A box of figs and a box of garlic for \$25.
• A box of garlic and a box of apples
for \$26.
1 If Mario allows you to buy the boxes individually for the same comparative price as
the sale price, find the cost of each individual box.
Part C
Mario sells apples in a number of different-sized boxes. Five different-sized boxes of apples are now weighed
two at a time, in all possible combinations. The weights obtained are 16, 18, 19, 20, 21, 22, 23, 24, 26 and
27 kilograms.
1 Find the weight of each individual box of apples.
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Summary
Solving simultaneous equations
• Simultaneous equations can be solved in three ways:
1. Graphically: Sketch the two lines on the same set of
axes and then read off the point of intersection. This
method is good for analysis, but is tedious and
possibly inaccurate.
2. Substitution: Substitute one equation into the other
equation and then solve. This is a versatile
technique that can be applied to non-linear
3. Elimination: The coefficients of the variable being
eliminated must be the same in both equations. This
variable is then eliminated by either adding or
subtracting the equations. If the coefficients of the
equations, then subtract the equations; if they are
opposite signs, then add the equations.
• Simultaneous equations can be checked on the CAS.
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Transposition and substitution
• Transposing an equation is the same process as solving
an equation. We perform the same steps on
pronumerals as we do with numbers.
• It is more efficient and usually easier to transpose an
equation to make the desired quantity the subject,
before substituting known values.
• Substitution is the process of replacing one number or
pronumeral by another. Before substituting, check to
ensure that all quantities are measured in the same or
appropriate units.
Checking algebraic processes
• Algebraic simplifications and factorisations can be
checked by making up values for the pronumerals and
substituting to determine if LHS = RHS.
• Solutions to equations can be checked by:
1. Substituting the answer and determining if
LHS = RHS.
2. Using solve on a CAS.
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Solving linear equations
• When solving linear equations we often expand
brackets first.
• If fractions are present, multiply the equation by the
lowest common denominator.
• If the equation is of the form fraction = fraction, then
to eliminate the fractions you can cross-multiply.
• Collect all the unknowns on one side and then undo
or reverse the operations.
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Developing formulae from
description
• To develop a formula, first define the variables.
• If the situation allows, draw a large, clear diagram and,
if possible, connect your variables on the diagram so
that the number of variables is reduced.
• An even number can be represented by 2n and an odd
number by 2n + 1.
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Use the following to check your progress. If you need more help with any questions, turn back to the
section given in the side column, look carefully at the explanation of the skill and the worked examples,
and try a few similar questions from the Exercise provided.
1 Solve each of the following equations.
2y – 1
(a) --------------- = 6
(b) 2(3m + 4) = 8
4
2.1
(c) 2(3e − 4) = 2e + 1
2(3x – 1)
(d) ----------------------- = 2x + 1
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PrT
2 The formula to calculate the simple interest earned is given by I = --------- where I is the interest on a
100
sum of money, P, invested for T years at an interest rate of r% per annum.
(a) Find the interest earned if \$850 is invested for 3 years at a simple interest rate of 3% per annum.
(b) Find the interest rate if the simple interest earned on \$620 invested for 4 years is \$62.
(c) How long will it take \$500 to earn \$50 interest if the interest rate is 3.5%? Round your answer
up to the next month.
3 The volume of a cylinder is given by V = πr2h. Find the radius, correct to two decimal places, of a
cylinder with a height of 2.5 metres and a volume of 25 cubic metres.
4 Transpose each of the following to make the letter in brackets the subject.
m(v – u)
GMm
(a) F = --------------------(u)
(b) F = -------------(m)
t
r2
5 An isosceles right-angled triangle has one side length of (x + 2) cm and the length of the hypotenuse
is 12 cm. Find the perimeter of the triangle.
6 Solve each of the following equations using your CAS, giving your answers to three decimal places.
(b) 5x2 − 3x = 8
(a) 5x3 + 2x − 3 = 0
h
7 The distance to the visible horizon, in kilometres, is given by d = 8 --- where h metres is the height
5
above sea level.
2.2
2.3
2.3
2.4
2.5
2.6
(a)
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Using a graphics calculator, generate a table of values and, hence, complete the following tables,
giving your values correct to one decimal place.
h metres
4
7
10
13
16
0.6
1.5
8
22
40
d kilometres
(b)
h metres
d kilometres
8 Solve each of the following sets of simultaneous equations.
(a) 2x + y = 4
(b) 2x + y = 7
(c) 3x − 2y = 10
y=x−1
3x − y = 3
5x − 2y = 18
(d)
+ 3y = 6
3x − 4y = 8
2.7
−2x
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9 The cost of two adult and two children’s tickets to attend the Australian Open Tennis Tournament is
\$124, while the cost of purchasing one adult and three children’s tickets is \$112. Find the cost of
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Multiple choice
10 The solution to 2--3- x − 1 = 7 is:
A 9
B 12
2y
–
4
11 The solution to --------------- = 2y is:
5
A 4--5B − 1--3-
2.1
C 4
D 8
E 11
2.1
C
−2 1--2
= −2, then
12 Given s = ut + 1--2- at 2 , when u = 4, t = 3 and a
A 21
B 9
C 3
D
− 1--2
s is equal to:
D 15
E
1
--3
E
−2
2.2
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13 In the equation S = 2(x + 16) 2x 2 + 2x + 1 + 2x(2x + 17), when x = 7, the value of S, to the nearest
whole number, is:
A 923
B 5632
C 15 593
D 489
E 212 220
14 When v = u + at is transposed to make a the subject, a is equal to:
v
v
u
v–u
A -- − u
B v−t−u
C --- − t
D v − --E -----------t
u
t
t
15 When S = 2πr(r + h) is transposed to make h the subject, h is equal to:
S
S – 2πr
S
S
A s−r
B --------- − r
C -----------------D --E ----------2- − 2πr
2πr
r
r
2πr
16 Bronwyn is three times as old as her sister, Jessica. In 4 years time, Bronwyn will only be twice as old
as Jessica. If j represents Jessica’s age now, this information can be represented algebraically as:
A 2j + 4 = 3j + 4
B 3j + 4 = 2j
C 6j + 4 = j + 4
D 2( j + 4) = 3j + 4
E j + 4 = 2(3j + 4)
x 2 – 4x + 4
17 The expression -------------------------- is equivalent to:
2–x
D x2 − 2x + 4
E 1
A 2−x
B x−2
C x2 − 2x + 2
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18 Given W = πd(r2h + --23- r 3 ), if d = 3 and h = 2, then the value of W when r = 5.7 is closest to:
A 312
B 736
C 817
D 1388
E 1776
19 The solution to the pair of simultaneous equations y = 2x − 7 and 3x + y = 8 is:
A (−3, −13)
B (−7, 8)
C (−15, −37)
D (3, 1)
E (3, −1)
−
20 The solution to the pair of simultaneous equations 2x − y = 4 and x − 2y = 1 is:
A (9, 5)
B (1, −1)
C (4, −1)
D (3, 2)
E (−1, −6)
21 The points (2, −3) and (4, 3) lie on the straight line with equation ax + by = 6. The values of a and b,
respectively, are:
A 1, − 2--3B 2, − 2--3C 0, 2
D 3, 0
E 1 1--2- , 0
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2.2
2.3
2.3
2.4
2.5
2.6
2.7
2.7
2.7
TECHNIQUES
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22
w
x
x
x
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x
21 cm
x
x
x
x
30 cm
23 The range, R, of a projectile, which is the horizontal
distance travelled, is given by the formula:
V 2 sin 2θ
R = -----------------------,
9.8
where V = the speed of projection, and θ = the angle
projected to the horizontal.
(a) Find the range of a projectile with projection
speed 10 m/s and angle to the horizontal of:
(i) 30°
(ii) 45°
(iii) 60°
(iv) 70°
(b) Explain why the maximum range occurs when
the object is projected at an angle of 45° to the
horizontal.
(c) Find the maximum range of an object projected
with an initial speed of 25 m/s.
(d) Find the speed of projection of a projectile with
a maximum range of 50 metres.
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The corners are cut from a 21 cm by 30 cm sheet of
paper as shown in the diagram above. The sides are
folded up and the shaded rectangle is folded over to
form the lid of a box.
The volume of a box is given by:
V = length × width × height
(a) Express the length, l, and width, w, of the box
in terms of x.
(b) Show that the volume of the box is given by
V = (21 − 2x)(15 − x)x.
(c) Find the volume of the box in cm3, to
two decimal places, when x = 4 cm.
(d) State, with reasons, the largest possible value
for x.
Use your CAS to sketch the graph of
v = (21 − 2x)(15 − x)x and use the graph to find
answers to the next two questions.
(e) Find the value for x that produces the box of
maximum volume.
(f) Find the maximum volume of the box.
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exam focus 3
VCAA 2004 Further Mathematics Units 3 & 4, Exam 1, Module 3, Question 6
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The cost, \$C, of hiring a boat for x hours is given by the equation C = ax + b where a is the hourly rate and
b is a fixed booking fee.
When the boat is hired for 4 hours the cost is \$320.
When the boat is hired for 6 hours the cost is \$450.
When the boat is hired for one hour the cost is
A. \$65
B. \$75
C. \$77
D. \$80
E. \$125
exam focus 4
VCAA 2004 Further Mathematics Units 3 & 4, Exam 2, Module 1, Question 2
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The purchase and installation of a basic heating system with five outlets costs \$3500. Each additional
outlet costs an extra \$80.
a. Determine the cost of installing a heating system with eight outlets.
1 mark
2 marks
c. Australian Heating recommends that a house with 20 squares of living area should have
12 heating outlets.
Using this recommended ratio, determine the cost of installing a heating system for
a house having 35 squares of living area.
2 marks
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b. A customer has \$4400 to spent on a heating system and outlets. Determine the
greatest number of outlets that can be bought with this heating system.
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TECHNIQUES
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