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C H A P T E R
16
MODULE 3
PL
E
Constructing and
interpreting linear graphs
How do we find the gradient of a line, given the coordinates of two points?
How do we sketch straight line graphs using gradient and y-axis intercept?
How do we obtain the equation for a straight line given:
r the gradient and y-axis intercept?
r the coordinates of two points?
How do we construct straight line graphs?
M
How do we solve linear simultaneous equations for two unknowns?
How do we apply linear simultaneous equations to break-even analysis, where cost
and revenue functions are linear?
The gradient of a straight line
Through any two points it is possible to draw only a single straight line. Therefore a straight
line is defined by any two points on the line.
From previous work, you should be familiar with the
rise
concept of the gradient or slope of a line. The symbol
used for gradient is m. This may be defined as:
run
rise
gradient =
run
Hence, given any two points on the line, A(x1 , y1 )
y
and B(x2 , y2 ), the gradient of the line can be found.
B(x2, y2)
y2 − y1
gradient m =
x2 − x1
rise = y2 – y1
A(x1, y1)
SA
16.1
run = x2 – x1
0
x
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442
Essential Further Mathematics – Module 3 Graphs and relations
Example 1
Finding the gradient of a line given two points
y
Find the gradient of this line.
2
1
–2 –1
Solution
E
Let (x 2 , y 2 ) = (0, 2) and let (x 1 , y 1 ) = (−2, 0)
y2 − y1
2−0
Gradient m =
=
x2 − x1
0 − (−2)
2
= =1
2
PL
1 From the graph, identify two points where
the coordinates can be easily determined.
In this case the axes intercepts are chosen
(2, 0) and (0, 2).
y2 − y1
2 Use m =
x2 − x1
x
0
Example 2
Finding the gradient of a line given two points
y
Find the gradient of this line.
3
2
1
Solution
0
2
x
Let (x 1 , y 1 ) = (0, 3) and (x 2 , y 2 ) = (2, 0)
y2 − y1
0−3
m=
=
x2 − x1
2−0
3
=−
2
M
1 From the graph identify two points where
the coordinates can be easily determined.
In this case the axes intercepts are chosen
(2, 0) and (0, 3).
3
y2 − y1
=−
2 Use m =
x2 − x1
2
SA
1
The gradient of a line that slopes upwards from left to right is positive, and the gradient
of a line that slopes downwards from left to right is negative.
The gradient of a horizontal line is zero, since y2 − y1 = 0.
The gradient of a vertical line is undefined, since x2 − x1 = 0.
Exercise 16A
1 Calculate the gradient of each of the following.
a
b
y
c
y
y
6
4
2
6
4
0
2
–4
–2
0
–2
x
2
(3, 0)
(0, –4)
x
–2
0 2
x
4
6
4
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Chapter 16 – Constructing and interpreting linear graphs
443
d
e
y
f
y
y
(6, 0)
(–3, 0)
x
0
x
–5
–5
(0, –4)
2 Sketch a graph of a line with gradient 1.
3 Sketch a graph of a line with gradient 0.
0
x
E
(–2, 0)
4 Find the gradients of the lines that pass through each of the following pairs of points:
d (10, 16) (12, 0)
h (−5, 25) (−8, 64)
The general equation of a straight line
The general equation of a straight line is y = mx + c, where m is the gradient of the line. This
form, expressing the relation in terms of y, is called the gradient form.
If we let x = 0, we get
M
y = m(0) + c
y=c
That is, the y-axis intercept is equal to c.
Example 3
Interpreting the equation of a line given the form y = mx + c
Find the gradient and y-axis intercept of the graph of y = 3x − 4.
SA
16.2
b (−6, 8) (2, −12) c (3, 3.5) (5.5, −1.5)
f (0, −3) (−3, 0) g (3, 9) (4, 16)
PL
a (12, 6) (4, 8)
e (3, 0) (−3, 0)
Solution
The equation of the line is in the form
y = mx + c where m is the gradient
and c the y-axis intercept.
The value of m is 3 and the value of c is −4.
Therefore the gradient of this line is 3 and the
y-axis intercept is −4.
If we are given the rule of a straight line, we can sketch the graph using the gradient and the
y-axis intercept.
Example 4
Sketching straight-line graphs
Sketch the graph of y = 3x − 1.
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Essential Further Mathematics – Module 3 Graphs and relations
Solution
1 The equation is written in the form y = mx + c. Hence the gradient and y-axis intercept can
be read directly.
3
rise
=
Gradient = 3, i.e.
run
1
y-axis intercept = −1
y
(1, 2)
2
E
2 Plot the point (0, −1), the y-axis intercept.
3 A gradient of 3 means that for every unit across in
the positive x-axis direction you go up 3 in the positive
y-axis direction.
From the y-axis intercept move across 1 (run) and
up 3 (rise) to plot the point (1, 2).
4 Draw a line through the points (0, −1) and (1, 2) to
sketch the graph.
1
3
0
–1
x
1
PL
–1
1
If the equation for the straight line is not written in gradient form, to use the above method for
sketching a graph the equation must first be transposed into gradient form.
Example 5
Sketching a straight-line graph
M
Sketch the graph of 3y + 6x = 9.
Solution
SA
1 First rearrange the equation into
gradient form: y = mx + c
3y + 6x = 9
∴ 3y = 9 − 6x (subtracting 6x from both sides)
9 − 6x
(dividing both sides by 3)
y =
3
y = 3 − 2x
∴ y = −2x + 3
2 The equation is now written in
the form y = mx + c. Therefore m = −2
and c = 3
3 A gradient of −2 means that for every unit
across in the positive x-axis direction you go
down 2 in the negative y-axis direction. From
the y intercept (0, 3) move across 1 (run) and
down 2 (rise) to plot the point (1, 1).
4 Draw a line through the points (0, 3) and
(1, 1) to sketch the graph.
y
1
3
–1
2
2
1
(1, 1)
0
–1
x
1
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Chapter 16 – Constructing and interpreting linear graphs
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Parallel lines
y
If the value of m is the same for two rules, then the
lines are parallel.
For example, consider the lines with rules
y = 2x + 3
4
Exercise 16B
1
2
–4
PL
1 Sketch the graph of each of the following:
x
–1 0
–2
E
–2
They are parallel because they both have the same
gradient, m = 2.
a y = x +1
d y =4−x
y = 2x – 4
2
y = 2x + 3
y = 2x − 4
b y = 2x + 1
e y = 4 − 2x
c y = 3x − 6
f y = −x + 6
2 For which of the following equations do the lines pass through the origin?
a y+x =1
b y+x =0
c 2y = 3x
d y + 3x = 3(x + 1)
e x−y=1
3 Give the gradient of each of the lines in Question 2.
4 Sketch the graph of each of the following by first expressing it in the form y = mx + c.
c 6y − 4x = 24
f 2y + 3x = 6
i 2x − 3y = 6
Finding the equation of a straight line
The equation of a straight line can be found if the gradient and y-axis intercept are known.
SA
16.3
b 6x + 3y = 12
e y − 2x = 6
h y − 2x = 1
M
a 4x + 2y = 12
d x+y=3
g x + 2y = 1
Example 6
Finding the equation of a straight line
Write down the equation of the straight line if m = −3 and c = 10.
Solution
The equation is y = −3x + 10.
If, however, we are not told the gradient and/or the y-axis intercept, they can be determined
from other information.
y
Case (i) We are given any two points A(x 1 , y1 )
B(x2, y2)
and B(x2 , y2 ). Using these two points, we can first
P(x, y)
calculate the gradient of the line AB:
A(x1, y1)
y2 − y1
x
m=
0
x2 − x1
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Essential Further Mathematics – Module 3 Graphs and relations
Using the general point P (x, y), also on the line, we have:
m=
y − y1
x − x1
Therefore the equation of the line is:
y − y1 = m(x − x1 ) where m =
Example 7
y2 − y1
x2 − x1
E
Finding the equation of a straight line
Find the equation of the straight line passing through the points
(1, −2) and (3, 2).
y
2
1
PL
0
–2
(3, 2)
2
3
x
4
(1, –2)
–4
Solution
1 First find the gradient using
m=
y2 − y1
x2 − x1
x 1 = 1, y 1 = −2, x 2 = 3 and y 2 = 2
2 − (−2)
4
= =2
3−1
2
M
2 Now use y − y1 = m(x − x1 ) to determine y − (−2) = 2(x − 1)
∴ y = 2x − 4
the equation.
3 Write down your answer.
The equation of the straight line is y = 2x − 4.
We are given the gradient m and one other point, A(x1 , y1 ).
Since we already know the gradient m we can find the rule using
Case (ii)
SA
y − y1 = m(x − x1 )
Example 8
Finding the equation of a straight line
Find the equation of the line that passes through the point (3, 2) and has a gradient of −2.
Solution
1 Since we already know the gradient m and
a point on the line, we can find the rule,
using y − y1 = m(x − x1 ). Write down the
values of m, x1 and y1 . Substitute and
reorganise the equation to make y the subject.
2 Write down your answer.
m = −2, x 1 = 3 and y 1 = 2
y − 2 = −2(x − 3)
y − 2 = −2x + 6
∴ y = −2x + 8
The equation of the line is
y = −2x + 8.
Note: The equation could also be expressed as y + 2x − 8 = 0.
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Chapter 16 – Constructing and interpreting linear graphs
447
The equation of a straight line can also be found from the graph by reading off two points and
using them to find the equation, as outlined above.
Example 9
Finding the equation of a line given its graph
y
Find the equation of the line shown in the graph.
A
E
4
3
2
1
0
Solution
x
The y-axis intercept is (0, 4), so c = 4.
The coordinates of A and B are (0, 4) and (2, 0):
PL
1 The equation of a straight line is
y = mx + c.
2 c is the y-axis intercept. Read
from the graph.
3 m is the gradient. Calculate using the
two points A and B.
4 Write down your answer.
B
1 2
y2 − y1
x2 − x1
0−4
4
=
= − = −2
2−0
2
∴ gradient m =
The equation of the line is y = −2x + 4.
M
Vertical and horizontal lines
If m = 0, then the line is horizontal and the equation is simply y = c, where c is the y-axis
intercept.
Example 10
Horizontal lines
SA
Sketch the graph of y = 2.
Solution
1 The equation of a straight line is y = mx + c.
2 For this equation the y-axis intercept is 2 and
the gradient is 0. The line is parallel to the x axis.
3 Draw a horizontal line passing through (0, 2).
y
3
(0, 2)
y=2
1
0
If the line is vertical, the gradient is undefined and its rule is given as x = a, where a is the
x-axis intercept.
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Essential Further Mathematics – Module 3 Graphs and relations
Example 11
Vertical lines
Sketch the graph of x = 3.
Solution
y
1 Note that the equation of a vertical line is not
in the form y = mx + c.
2 For this equation, the graph is the line which consists
of all points with coordinates (3, y) and y varies.
3 Draw a vertical line passing through (3, 0).
E
x=3
(3, 0)
0
PL
Exercise 16C
1
2 3
x
4
1 Find the equation of each of the following straight lines.
y
a
M
SA
2
y
g
–3
y
x
x
x
0
e
0
0 1
2
x
0
y
d
y
c
6
2
–1
y
b
y
f
2
x
0
–3
–2
0
x
y
h
3
01
2
x
0
x
8
2 Find the equation of each of the following straight lines defined by:
a gradient 12 , passing through (0, 1)
c gradient 0, passing through (0, 6)
e gradient −2, passing through (1, 3)
b gradient 2, passing through (1, 4)
d gradient 1, passing through (2, −4)
3 Write in the form y = mx + c the equation of the line that has the given gradient and passes
through the given point:
b 12 , (0, 6)
c −3, (1, 4)
a 32 , (0, 1)
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Chapter 16 – Constructing and interpreting linear graphs
449
4 Find the equation of the line that passes through each of the following pairs of points:
b (−3, 0)(0, −6)
e (1, 4) (5, 6)
a (0, 3) (3, 0)
d (0, 1)(−1, 0)
Equation of a straight line in intercept form
ax + by = c
E
Often we encounter a linear relation that is not expressed in the form y = mx + c. An
alternative standard notation is
PL
This is sometimes referred to as intercept form.
While it is necessary to transpose into gradient form if you wish to find the gradient, it is
often convenient to work with linear relations in the intercept form.
Sketching graphs in intercept form
A convenient way to sketch graphs of straight lines is to plot the two axes intercepts.
Example 12
Sketching the graph of a line given in intercept form
Sketch the graph of 2x + 4y = 10.
Solution
x-axis intercept (y = 0):
M
1 Find the x-axis intercept.
∴ 2x + 4(0) = 10 or x = 5
y-axis intercept (x = 0):
2 Find the y-axis intercept.
∴ 2(0) + 4y = 10 or y = 2.5
y
3 Use the two points to sketch the line.
SA
16.4
c (0, 4) (4, 2)
2.5
0
x
1
2
3 4 5
When finding the equation of a straight line, it is also sometimes more convenient to express it
in intercept form.
Example 13
Finding the equation of a line given two points
Find the equation of the line passing through the points A (2, 5) and B (6, 8).
Solution
1 Find the gradient of the line.
m =
3
8−5
=
6−2
4
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Essential Further Mathematics – Module 3 Graphs and relations
2 Use y − y1 = m(x − x1 ) to find the equation.
Here y1 = 5 and x1 = 2. Substitute for m, x1
and y1 , and rearrange into the intercept form.
3
y − 5 = (x − 2)
4
∴ 4(y − 5) = 3(x − 2)
∴ 4y − 20 = 3x − 6
4y − 3x = 14
3 Write down your answer.
The equation of the line is
4y − 3x = 14 or −3x + 4y = 14
Exercise 16D
E
450
1 Sketch the graph of each of the following linear relations:
b 4x − y = 8
e −3x + 4y = 15
c 3x − 4y = 24
f 7x − 2y = 14
PL
a 2x − 3y = 12
d 2x − 5y = 20
2 Transpose each of the following linear relations from intercept form to gradient form and
hence state the gradient:
a 2x − y = 6
d 5x − 2y = 10
16.5
b x + 4y = 12
e x − 5y = 10
c −x − 2y = 6
f −x + 2y = 8
Linear models
M
In many practical situations a linear rule can be used.
Example 14
Application of linear rules and graphs
SA
Austcom’s rates for local calls from private telephones consist of a quarterly rental fee of $40
plus 25c for every call. Construct a linear rule that describes the quarterly telephone bill and
sketch the graph.
Solution
Strategy: Construct a linear rule of the form C = mn + c by finding the gradient m and the
C-axis intercept.
Let
C
n
C -axis intercept
Gradient m
∴C
= cost ($) of quarterly telephone bill
= number of calls
= 40
= 0.25
= 0.25n + 40
C
100
80
(200, 90)
60
40
20
50 100 150 200
n
Note: The graph should be a series of discrete points rather than a continuous line. With the scale used it is not
practical to show it correctly.
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Chapter 16 – Constructing and interpreting linear graphs
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Example 15
Application of linear rules and graphs
The tyres on a racing car had lost 3 mm of tread after completing 250 km of a race and 4 mm
of tread after completing 1000 km. Assuming that the loss of tread was proportional to the
distance covered, find the total loss of tread, d mm, after s km from the start of the race. What
would be the tread loss by the end of a 2000 km race?
Solution
d (mm)
PL
Rule : d = m s + c
rise
Gradient, m =
run
1
4−3
=
=
1000 − 250
750
1
s+c
∴ d=
750
When
s = 250, d = 3
1
2
∴ 3 = + c or c = 2 3
3
E
Strategy: Construct a rule of the form d = ms + c by finding the gradient m and the d-axis
intercept.
∴ Total loss of tread after s km, d =
0 250
When
rise
run
1000
s (km)
1
2
s + 23
750
M
s = 2000
1
2
d=
× 2000 + 2 3
750
1
=5
3
4
3
1
SA
The loss of tread at the end of a 2000 km race is 5 3 mm.
Exercise 16E
1 The weekly wage, $w, of a vacuum cleaner salesperson consists of a fixed sum of $350 plus
$20 for each cleaner sold. If n cleaners are sold per week construct a rule that describes the
weekly wage of the salesperson.
2 The reservoir feeding an intravenous drip contains 500 mL of a saline solution. If the drip
releases the solution into a patient at the rate of 2.5 mL/min:
a construct a rule that relates v, the amount of solution left in the reservoir after time,
t minutes
b state the possible values of t and v
c sketch the graph of the relation
3 The cost ($C) of hiring a taxi consists of two elements, a fixed flagfall and a figure that
varies with the number (n) of kilometres travelled. If the flagfall is $2.60 and the cost per
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Essential Further Mathematics – Module 3 Graphs and relations
4 A car rental company charges $85, plus an additional amount of 24 cents per kilometre.
a Write a rule to determine the total charge $C for hiring a car and travelling x kilometres.
b What would be the cost to travel 250 kilometres?
5 The table shows the extension
(in cm) of a spring when weights
(in g) are attached to it.
extension (x) 0
1
2
3
4
5
6
weight (w)
50 50.2 50.4 50.6 50.8 51.0 51.2
E
a Sketch a graph to show the relationship between x and w.
b Write a rule that describes the graph. c What will the extension be if w = 52.5 g?
6 A printing firm charged $35 for printing 600 sheets of headed notepaper and $47 for printing
800 sheets.
PL
a Find a formula, assuming the relationship is linear, for the charge, $C, in terms of number
of sheets printed, n.
b How much would they charge for printing 1000 sheets?
7 An electronic bankteller registered $775 after it had counted 120 notes and $975 after it had
counted 160 notes.
a Find a formula for the sum registered ($C) in terms of the number of notes (n) counted.
(Assume the formula connecting C and n is of the form C = an + b).
b Was there a sum already on the register when counting began? If so, how much?
M
8 An electrician charges $50 to call and $80 per hour. Assuming that the relationship between
charge and time is linear:
a find a formula for $C, the charge in terms of the time spent (t hours)
b draw a graph of C against t
c calculate the cost of a job that takes 2 hours 45 minutes to complete
SA
9 The table shows the conversion from
British pounds to Australian dollars
for various amounts of money.
a
b
c
d
16.6
Australian dollars ($) 46
British pounds (£)
20
92 184
40 80
Plot these points and draw a straight line through them.
What does the gradient of the straight line represent?
State the gradient.
Find a formula for $A in terms of £P, British pounds.
Simultaneous equations
A linear equation that contains two unknowns (e.g. 2y + 3x = 10) does not have a single
solution. Such an equation actually expresses a relationship between pairs of numbers, x and y,
that satisfy the equation. If we represent graphically all the possible pairs of numbers (x, y) that
will satisfy the equation, the result is a straight line. Hence the name linear relation.
If the graphs of two such equations are drawn on the same set of axes, and they are
non-parallel, the lines will intersect at one point only. Hence there is one pair of numbers that
will satisfy both equations simultaneously.
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Chapter 16 – Constructing and interpreting linear graphs
453
–y
4
3
2
–3 –2 –1
1
1
2
3
x
0
–1
–2
x+
E
–3
–4
(1, –2)
2y
= –3
Solving a pair of simultaneous equations
PL
Example 16
=4
y
2x
Finding the intersection of two straight lines
can be done graphically; however, the accuracy
of the solution will depend on the accuracy of the
graphs. Alternatively this point of intersection can
be found algebraically by solving the pair of
simultaneous equations.
We will look first at two techniques for solving
simultaneous equations.
Solve the equations 2x − y = 4 and x + 2y = −3.
Solution
By substitution
1 Write down the equations and label them 1 and 2.
M
2 Express one unknown from either equation
in terms of the other unknown.
3 Then substitute this expression into the
other equation, reducing it to one equation
in one unknown.
4 Solve this equation
SA
5 Substitute the value of y into (2):
6 Check in Equation 1
2x − y = 4
x + 2y = −3
(1)
(2)
From equation (2) we get:
x = −3 − 2y
Equation (1) then becomes:
2(−3 − 2y ) − y = 4
∴ −6 − 4y − y = 4
∴
−5y = 10
∴
y = −2
x + 2(−2) = −3
∴
x =1
Check in (1) LHS = 2(1) − (−2) = 4
RHS = 4
Note: This means that the point (1, −2) is the point
of intersection of the graphs of the two linear relations.
By elimination
Strategy: If the coefficient of one of the unknowns is the same in both equations, we can
eliminate that unknown by subtracting one equation from the other. It may be necessary to
multiply one of the equations by a constant to make the coefficients of x or y the same for the
two equations.
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Essential Further Mathematics – Module 3 Graphs and relations
2x − y = 4
(1)
x + 2y = −3
(2)
2x + 4y = −6 (2 )
1 Write down the equations and label them 1 and 2.
2 To eliminate x, multiply equation (2) by 2 and
subtract the result from equation (1).
3 Subtracting (1) − (2 )
∴
−5y = 10 (1) − (2 )
or y = −2
2x − (−2) = 4
or x = 1
E
4 Now substitute for y in (1) to find x
5 Check as in the substitution method.
∴
PL
How to to solve simultaneous equations using the TI-Nspire CAS
Solve the following pair of simultaneous equations:
24x + 12y = 36
45x + 30y = 90
Steps
1 Start a new document and select
1:Add Calculator.
M
2 Press b /3:Algebra/1:Solve
to paste the solve( command.
SA
and select the template
3 Press / +
shown opposite.
4 Enter the values as shown. Use the
key to move between entry boxes.
5 Press
to move to the end of
.
the equations and type
6 Press
enter
to display the solution.
7 The solution x = 0 and y = 3 can
be checked by substitution.
24 × 0 + 12 × 3 = 36
45 × 0 + 30 × 3 = 90
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Chapter 16 – Constructing and interpreting linear graphs
455
How to solve simultaneous linear equations algebraically using the ClassPad
Steps
1 From the application menu screen, open
the Main application,
.
PL
2 Press k on the front of the
calculator to display the built-in
keyboard.
Tap the ) tab and locate and tap the
.
simultaneous equations icon
Enter the information
24x + 12y = 36 45x + 30y = 90 E
Solve the following pair of simultaneous equations:
24x + 12y = 36
45x + 30y = 90
x,y
M
Press E to display the solution x = 0
and y = 3.
3 The solution x = 0 and y = 3 can be checked by substitution.
SA
24 × 0 + 12 × 3 = 36
45 × 0 + 30 × 3 = 90
Exercise 16F
1 Solve each of the following pairs of simultaneous linear equations using either the
substitution or elimination method. Check your answers with your calculator.
a x + 2y = 6
x + 3y = 4
b 7x + 6y = 0
5x − 6y = 48
c 3x + 2y = 12
x + 2y = 8
d x − 2y = 6
4x + 2y = 14
e x + 3y = 12
x+y=8
f 9x + 2y = 48
x − 2y = 2
g 2x + 3y = 13
2x + 5y = 21
h 3x − y = 10
x + y = −2
i 3 p + 5q = 17
4 p + 5q = 16
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456
Essential Further Mathematics – Module 3 Graphs and relations
j 2x + 3y = 12
5x + 4y = 23
l 9x + 8y = 17
2x − 6y = −4
m y =6−x
2x + y = 8
n 9+x = y
x + 2y = 12
o 2y = 4 + x
y = x +8
p x +4= y
y = 10 − 2x
q y =4−x
y = x +6
r y = 4 + 2x
y − 2x = 6
Problems involving simultaneous
linear equations
E
16.7
k 5x + 4y = 21
3x + 6y = 27
Example 17
Application of simultaneous linear equations
Solution
PL
There are two possible methods for paying gas bills:
Method A: A fixed charge of $25 per quarter + 50c per unit of gas used
Method B: A fixed charge of $50 per quarter + 25c per unit of gas used
Determine the number of units that must be used before method B becomes cheaper than
method A.
M
1 Set up the equations.
C 1 = charge in $ using method A
C 2 = charge in $ using method B
x = number of units of gas used
Now C 1 = 25 + 0.5x
C 2 = 50 + 0.25x
Let
C
100
Dollars
SA
2 Graph the two straight lines and
find the point of intersection.
75
C1 = 0.5x + 25
(100, 75)
C2 = 0.25x + 50
50
25
0
x
25 50 75 100 125 150
Units
3 Write down your answer.
It can be seen from the graph that if
the number of units exceeds 100
then method B is cheaper.
Note: The solution could also be obtained by
solving simultaneous linear equations:
C1
25 + 0.5x
0.25x
∴x
= C2
= 50 + 0.25x
= 25
= 100
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Chapter 16 – Constructing and interpreting linear graphs
457
Example 18
Application of simultaneous linear equations
If 3 kg of jam and 2 kg of butter cost $29, and 6 kg of jam and 3 kg of butter cost $54, find the
cost of 1 kg of jam and 1 kg of butter.
Solution
Then
and
3x + 2y = 29 (1)
6x + 3y = 54 (2)
Multiply (1) by 2 :
6x + 4y = 58 (1 )
Subtract (1 ) from (2) :
PL
2 Solve the equations, or use your
calculator.
Let the cost of 1 kg of jam = x dollars
and the cost of 1 kg of butter = y dollars.
E
1 Set up the equations.
3 Write down your answer.
−y = −4
y =4
Substituting in (2) gives : 6x + 3(4) = 54
6x = 42
∴x =7
∴ The jam costs $7 per kilogram and
the butter $4 per kilogram.
M
Exercise 16G
1 Find two numbers whose difference is 9 and whose sum is 11.
2 The sum of two numbers is 73 and their difference is 37. Find the numbers.
SA
3 The sum of two numbers is 20 and their difference is 4. Find the numbers.
4 Find two numbers where twice the first added to the second is 26 and the first added to three
times the second is 28.
5 A cup and a saucer cost $5.25 together. A cup and two saucers cost $7.50. Find the cost of a
cup.
6 A shop sells bread rolls. If five brown rolls and six white rolls cost $2.94 and three brown
rolls and four white rolls cost $1.86 find the cost of each type of roll.
7 In a test, the sum of Anne’s mark and David’s mark is 42. Sheila has twice as many marks as
David and the sum of Anne’s and Sheila’s marks is 52. What are the marks of the three
students?
8 A herbalist wishes to prepare a mixture from two herbs, A and B. The mixture costs
$6 per kilogram. If herb A costs $5 per kilogram and herb B costs $8 per kilogram and
60 kilograms of the mixture is to be prepared, how many kilograms of each herb should be
used?
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458
Essential Further Mathematics – Module 3 Graphs and relations
9 The cost ($C) of printing a pamphlet is given by the expression C = a + bn, where n is the
number of pamphlets, $a is the fixed cost of printing and $b is the cost per pamphlet.
a If it costs $460 to print 80 pamphlets and $1060 to print 200 pamphlets find:
i a
ii b
b Find the cost of printing 500 pamphlets.
c Sketch the graph of C against n.
16.8
Break-even analysis
0
E
demand curve
Quantity
0
For a price of $b consumers purchase a units.
Example 19
supply curve
$ Price/unit
(a, b)
(c, d)
PL
$ Price/unit
For each price of a product a corresponding quantity of the product will be demanded
(purchased) during some time interval. The quantities of a product that can be demanded or
supplied can be represented graphically.
Quantity
If c units are produced the price is $d.
Break-even analysis
M
A firm sells its product at $20 per unit. The cost of production ($C) is given by the rule
C = 4x + 48, where x is the number of units produced. Find the value of x for which the cost
of production of x units is equal to the revenue received by the firm for selling x units.
Solution
SA
1 Set up the equations.
Let the revenue for producing x units
be $R.
R = 20x
C = 4x + 48
2 Draw the graphs.
3 Find the point of intersection.
R, C
R = 20x
break-even point
Quantity
When
C = 4x + 48
x
C =R
20x = 4x + 48
∴ x =3
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Chapter 16 – Constructing and interpreting linear graphs
459
From the graph:
If less than three items are produced C > R.
If more than three items are produced C < R.
The point where R = C is called the break-even point.
Exercise 16H
E
1 Find the break-even point if sales revenue ($S) is given by the rule S = 0.75x, where x is the
number of items sold and the cost ($C) is given by the rule C = 0.25x + 100.
2 Find the break-even point for each of the following, where $C is the cost in producing
x units, and the sales price per unit is $n:
b C = 12x + 84; n = 40
e C = 30x + 350; n = 100
c C = 90x + 60; n = 100
f C = 400x + 800; n = 500
PL
a C = 2x + 16; n = 10
d C = 3x + 39; n = 16
3 A manufacturer sells his product at $72 per unit, selling all that he produces. His fixed cost
is $45 000 and the cost per unit is $22. Find:
a
b
c
d
the break-even point
the profit when 1800 units are produced
the loss when 450 units are produced
the sales volume required in order to obtain a profit of $90 000
M
4 A manufacturer of a product sells all that he produces. If his total revenue is given by
R = 7x (dollars) and his total cost is given by C = 6x + 800 (dollars), where x represents
the number of units produced and sold, then:
SA
a determine the level of production at the break-even point
b sketch the graph of C and R against x
c determine the level of production at the break-even point if the total cost increases by 5%
5 For each of the following $R is the total revenue and $C is the total cost. If x represents both
the number of units produced and the number of units sold, determine the break-even
quantity and sketch a graph of R and C against x for each.
a R = 3x
C = 2x + 4500
b R = 14x
c R = 1.05x
d R = 0.25x
22
x + 1200
C=
C = 0.85x + 600
C = 0.16x + 360
3
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Essential Further Mathematics – Module 3 Graphs and relations
Key ideas and chapter summary
rise
Gradient = run
rise
Gradient =
run
Gradient
rise
run
E
Given any two points on the line, A(x1 , y1 ) and B(x2 , y2 ), the
gradient of the line can be found.
Gradient, coordinate
definition
gradient m =
y
y2 − y1
x2 − x1
PL
B(x2, y2)
A(x1, y1)
rise = y2 – y1
run = x2 – x1
x
0
Gradient, sign of
The gradient of a line that slopes upwards from left to right is
positive, and the gradient of a line that slopes downwards from
left to right is negative.
M
Gradient–axis intercept form The general equation of a straight line is y = mx + c, where m
is the gradient of the line. This form, expressing the relation in
terms of y, is called the gradient form. The y-axis intercept is c.
Finding equation of straight
line given two points
SA
Review
460
Finding equation of straight
line given gradient and a
point
Given any two points A(x1 , y1 ) and B(x2 , y2 ), the equation of
the line is:
y2 − y1
y − y1 = m(x − x1 ) where m =
x2 − x1
y
B(x2, y2)
P(x, y)
A(x1, y1)
0
x
If we are given the gradient m and one other point, A(x1 , y1 ),
we can find the rule using:
y − y1 = m(x − x1 )
Horizontal lines
If m = 0, then the line is horizontal and the equation is simply
y = c, where c is the y-axis intercept.
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Chapter 16 – Constructing and interpreting linear graphs
461
Intercept form of a straight
line
ax + by = c
This is sometimes referred to as intercept form.
Simultaneous equations
Equations of two or more lines or curves in a cartesian plane,
the solution of which is the point of intersection of the pairs of
lines or curves.
Break-even point
The break-even point of a production
process is where the cost of
production is equal to the income
received.
E
If the line is vertical, the gradient is undefined and its rule is
given as x = a, where a is the x-axis intercept.
C = 4x + 48
break-even point
PL
R, C
R = 20x
Quantity
Skills check
M
Having completed this chapter you should be able to:
sketch a straight line given the coordinates of two points
sketch a straight line given gradient and a point
interpret linear graphs which model real-life situations
solve linear simultaneous equations by both algebraic and graphical techniques
use simultaneous equations to solve problems
use break even analysis to solve problems
SA
Multiple-choice questions
Questions 1 to 4 refer to the graph below
y
line A
5
4
3
2
1
–2 –1 –10
–2
–3
line B
–4
–5
1
2
1 The slope of line A is:
B −2
A − 12
3
4
5
line C
x
C 2
D − 32
E −5
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Vertical lines
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Essential Further Mathematics – Module 3 Graphs and relations
2 The equation of line B is:
A y − 2x = 3
D −2y − x = 6
B x−y=3
E y+x =3
3 The equation of line C is:
A y = −x
B y = −1
C y−x =3
C x =1
D x = −y
E y=1
D (−1, −4)
E (0, 1)
E
4 Lines B and C intersect at the point:
A (1, 4)
B (4, 1)
C (−4, −1)
5 The solution of the simultaneous equations 2x − y = 10 and x + 2y = 0 is:
A x = −2 and y = 3
B x = 2 and y = −3
C x = 4 and y = −2
D x = 6 and y = 2
E x = 1 and y = −8
PL
6 A straight line has gradient 3 and passes through the point with coordinate (1, 9). The
equation of the line is:
A y = x +9
B y = 3x + 9
C y = 3x + 6
1
1
E y = −3x + 6
D y = −3x + 1
7 If two lines 5x − y + 7 = 0 and ax + 2y − 11 = 0 are parallel then a equals:
A −5
B 5
C −10
D 10
E −1
2
M
8 The cost ($C) of hiring a car is given by the formula C = 2.5x + 65 where x is the number
of kilometres travelled. A person is charged $750 for the hire of the car. The number of
kilometres travelled was:
A 65
B 145
C 160
D 200
E 274
9 If Sue buys 3 packets of chips and 2 bottles of Coke it will cost her $13.20. However, if she
buys 2 packets of chips and 3 bottles of Coke it will cost her only $11.80. The price of each
item is:
A coke = $3.20, chips = $1.80
B coke = $2.10, chips = $3
C coke = $1.80, chips = $3.20
D coke = $2, chips = $2.90
E coke = $1.20, chips = $2
SA
Review
462
10 The equation of the line passing through (5, 9) and parallel to the line y = 3x + 7 is:
B y = − 13 x + 7
C y = 3x − 6
A y = − 13 x + 7 13
D y = 3x + 7
E y = 3x + 9
11 The cost ($C) of renting a truck is given by the formula C = 1.5x + 20 where x is the
number of kilometres travelled. A person is charged $320 for the rental of a truck. The
number of kilometres travelled was:
A 100
B 120
C 200
D 400
E 500
12 The graph of 3x − 2y = −6 is:
y
y
B
A
3
–2 0
y
C
y
D
3
3
x
0
y
E
x
2
0
–3
2
x
–3
0
–2
x
–2 0
–3
x
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Chapter 16 – Constructing and interpreting linear graphs
463
Review
13 Which of the following points lie on both the line with equation y = 3x − 1 and the line
with equation 4x + 2y = 8?
A (0, 0)
B (0, 4)
C (2, 0)
D (1, 2)
E (2, 1)
14 The gradient of the line with equation 2x + 4y − 6 = 0 is:
B 2
C 4
D −2
A − 12
E 1
x + y = 219
x + 2y = 361
E
15 The cost to enter a fun run was $2.00 per senior runner and $1.00 per junior runner. The
total number of runners was 219 and the total of the entry fees paid was $361. If the
number of senior entries was x and the number of junior entries was y, which one of the
following pairs of simultaneous equations is satisfied by x and y?
A x + y = 361
B x + y = 219
C x + y = 361
2x + y = 219
2x + y = 361
x + 2y = 219
PL
E x + y = 219
y
x + = 361
2
16 The equation of the straight line that passes through the points with coordinates (−2, 5) and
(2, −5) is:
A 5x + 2y = 0
B 7x − 5y + 45 = 0
C 5x − 7y = 45
D 7x − 5y = 45
E 5x + 7y + 45 = 0
D
SA
M
17 The equation of the line shown in the diagram is:
x
y
y
x
B
+ =1
+ =1
A
4
3
3
4
x
y
− = 1
D 4y − 3x = 12
C
4
3
E 3x + 4y + 25 = 0
y
(0, 4)
0
x
(3, 0)
y
18 The equation of the line passing through the point (2, 3) and
parallel to the x axis is given by:
A x+y=5
B y=3
C x =2
D x − y = −1
E 3x = 2y
(2, 3)
0
x
19 The solution of the system of equations 2x + y = 1 and y + 14 = 3x is:
A x =5 y=1
B x = 3 y = −5
C x =5 y=3
D x = −5 y = 3
E x = 1 y = −11
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Essential Further Mathematics – Module 3 Graphs and relations
M
PL
E
20 Given the straight line 2x + 3y − 6 = 0, which of the following statements is not true?
A If x increases then y decreases.
B The line has a negative gradient.
C The gradient of the line is −2.
D The line intercepts the x axis at 3 and the y axis at 2.
E The line is parallel to the line with the equation 2x + 3y − 4 = 0.
SA
Review
464
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