Year 11 Methods Chapter 1

Chapter 1 • Linear functions 1

Chapter ContentS 1a Solving linear equations and inequations 1B Rearrangement and substitution 1C Gradient of a straight line 1d Sketching linear functions 1e Simultaneous equations 1F Finding the equation of a straight line 1G Distance between two points and midpoint of a segment 1h Linear modelling

Chapter 1

Linear functions

diGital doCdoc-969510 Quick Questions

1a Solving linear equations and inequationsA linear equation is one that involves a variable raised to the power of 1 only. Recall that x1 is the same as x, so a linear equation involving x would contain only x’s, and not x

1

2, x2, x3, x −1, x −2 and so on. For example, y = 7x − 3, ax + by = c and 8

3x + 1 = 9 are all linear equations.

Note that y = 1

x is not a linear equation, as the

1

x is really x −1.

To isolate a particular variable — known as ‘making x (or whatever the variable is) the subject’ — we focus on the variable by ‘undoing’ other terms and operations. Remember to do the same operation to both sides of an equation, in the reverse order to that originally used to make up the equation.

Though the focus of this chapter is linear equations, some other types of equations will be included for skills practice.

When there is only one variable involved in an equation, we may attempt to solve and fi nd a numerical value by rearranging to make the variable the subject.

Worked example 1

Solve the following linear equations.

a 7x − 4 = 17 b − + =x3 24

5 1 c −

=x

245

1 6

think Write

a 1 Write the equation. a − =x7 4 17

2 Add 4 to both sides. =x7 21

3 Divide both sides by 7. =

=

x

x

21

73

b 1 Write the equation. b− + =x3 2

45 1

2 Subtract 5 from both sides.− = −x3 2

44

ContentsLinear functions 1 1a Solving linear equations and inequations 1 exercise 1a Solving linear equations and inequations 3

1B rearrangement and substitution 4 exercise 1B rearrangement and substitution 6

1C Gradient of a straight line 8 exercise 1C Gradient of a straight line 10

1d Sketching linear functions 12 exercise 1d Sketching linear functions 14

1e Simultaneous equations 15 exercise 1e Simultaneous equations 17

1F Finding the equation of a straight line 18

2 Maths Quest 11 Mathematical Methods CAS

3 Multiply both sides by 4. − = −x3 2 16

4 Add 2 to both sides. = −x3 14

5 Divide both sides by 3. =−

x14

3

c 1 Write the equation. c −

=x

24

51 6

2 Divide both sides by 2. − =x4

51 3

3 Add 1 to both sides. =x4

54

4 Multiply both sides by 5. =x4 20


=

x

x

20

45

When an equation has variables on both sides, at some stage they must be gathered together on the same side of the equation.

Worked example 2

Solve:a 4x − 3 = 3(6 − x) b

+ = +x x9 32

13 73

c − =

++x x3

4

9( 7)

101.

think Write

a 1 Write the equation. a 4x − 3 = 3(6 − x)

2 Expand the right-hand side (RHS). 4x − 3 = 18 − 3x

3 Collect x’s on one side, for example, the side that results in a positive x term, in this case, the left-hand side (LHS). (That is, add 3x to both sides.)

7x − 3 = 18

4 Add 3 to both sides. 7x = 21


=

x

x

21

73

b 1 Write the equation. b + = +x x9 3

2

13 7

3

2 Find the lowest common denominator for both terms. LCD = 6

3 Write all terms with the common denominator, adjusting numerators accordingly (so that numerator and denominator have been multiplied by the same amount).

+ = +x x3(9 3)

6

2(13 7)

6

4 Now that all terms have the same denominator, the numerators must be equal. (Multiply each side by 6.)

+ = +x x3(9 3) 2(13 7)

5 Expand all brackets. + = +x x27 9 26 14

6 Collect x’s on the LHS and numbers on the RHS. − = −x x27 26 14 9

7 Simplify and solve. x = 5

c 1 Write the equation. c− = + +x x3

4

9( 7)

101

2 Find the lowest common denominator for all three terms. LCD = 20


3 Write all terms with the common denominator, adjusting numerators accordingly (so that numerator and denominator have been multiplied by the same amount).

− =× +

+x x5(3 )

20

2 9( 7)

20

20

20

4 Now that all terms have the same denominator, the numerators must be equal. (Multiply each term by 20.)

5(3 − x) = 18(x + 7) + 20

5 Expand all brackets. 15 − 5x = 18x + 126 + 20

6 Collect x’s on the RHS and numbers on the LHS. − − = +

=−

x x

x

15 126 20 18 5

131 23

7 Simplify and solve. =−

x131

23

=−

x131

23

There is only one solution to a linear equation. For example, the equation 3x + 2 = 8 has the solution x = 2. The solution to a linear equation can be represented as a single point on a number line. If the equals sign is replaced with an inequality sign, the solution is a portion of a number line. These expressions are referred to as linear inequations (sometimes called inequalities).

Worked example 3

Solve the following linear inequations.a 6x − 7 ≥ 3x + 5 b −3m + 5 < −7

think Write

a 1 Write the inequations. a 6x − 7 ≥ 3x + 5

2 Combine the variable terms by subtracting 3x from both sides.

3x − 7 ≥ 5

3 Add 7 to both sides. 3x ≥ 12

4 Divide both sides by 3. x ≥ 4

b 1 Write the inequations. b −3m + 5 < −7

2 Subtract 5 from both sides. −3m < −12

3 Divide both sides by −3. Reverse the inequality sign, as you are dividing by a negative number.

m > 4

exercise 1a Solving linear equations and inequations 1 We1 Solve the following linear equations.

a 3x − 19 = −13 b 4x + 25 = −7 c 9x + 19 = −2

d x3 1

45

− = e x12 3

35

− = f x4 6

37 3

− − =

g x7 4

38 9

+ − = − h −

=x

23

51 10 i −

=x

7 84

77

2 We2 Solve the following linear equations.a 2x − 9 = 3(2x − 11) b 7x − 1 = 17(3x − 13) c x + 11 = 2(x + 12) d 3x − 7 = 3(35 − 2x)

e x x2

6

5

3

+ = −f

x x11

3

2( 14)

9

+ = +g

x x4 66

3

13 3

4

+ = −h

x x10

9

2(7 3 )

5

+ = −

i x x6 7

5

5 1

41

+ = + + j x x2 29

3

44

82

+ = + + k x x7 9

9

21

318

− = − − l x x17

2

49 2

55

− = + +

diGital doCdoc-9696equation solvers


3 We3 Solve the following linear inequations.

a 15 − 6x ≤ 2 b x11 3

76

− ≥ c x19

43 1

− + >−−

d 16 − 4x < 7(1 − x) e 3x − 7 ≥ 2(35 − 2x) f x x4

3

2( 1)

6

− ≤ +

1B rearrangement and substitutionWhen there is more than one variable involved in an equation, we may rearrange the equation to make a particular variable the subject using the same rules of equation solving described in the previous section.

Worked example 4

Rearrange each of the following to make the variable in parentheses the subject.

a 6x + 8y − 48 = 0 (y) b = +s ut at k12

2 (u) c π=Tmkk2 (k)

think Write

a 1 Write the equation. a 6x + 8y − 48 = 0

2 Add 48 to both sides. 6x + 8y = 48

3 Subtract 6x from both sides. 8y = 48 − 6x

4 Divide both sides by 8. = −y

x48 6

8

5 Cancel if possible. Here, divide the numerator and denominator by 2.

= −y

x24 3

4

6 Other ways of representing the answer are shown opposite.

= −

= −

= +−

y x

x

x

or24

4

3

4

63

43

46

b 1 Write the equation. b s ut at12

2= +

2 Subtract at12

2 from both sides. s at ut12

2− =

3 Multiply both sides by 2. − =s at ut2 22

4 Divide both sides by 2t. − =s at

tu

2

2

2

5 Write the equation with the desired variable on the left. = −u

s at

t

2

2

2

c 1 Write the equation. c π=Tm

k2

2 Divide both sides by 2π.π

=T m

k2

3 Square both sides.π

=T m

k4

2

2

4 Form the reciprocal of both sides to make k the numerator.

π =T

k

m

4 2

2


5 Multiply both sides by m. π =m

Tk

4 2

2

6 Write the equation with the desired variable on the left.π=k m

T

4 2

2

Once a variable is isolated, we may substitute values of other variables to calculate various values of the isolated variable. The following worked example illustrates some practical applications.

Worked example 5

a The formula for converting temperature in degrees Fahrenheit (F) (which is the system used in

the USA) to degrees Celsius (C ) is C = −F5( 32)9

. i Make F the subject. ii What is the temperature in degrees Fahrenheit when the temperature measured in degrees

Celsius is 21° C?b The area (A) of a circle is given by A = π r 2, where r is the radius. Calculate the value of r correct

to 2 decimal places when A = 20 cm2.

think Write

a i 1 Write the equation. a i =−

CF5( 32)

9

2 Multiply both sides by 9. = −C F9 5( 32)

3 Divide both sides by 5. = −C

F9

532

4 Add 32 to both sides. + =C

F9

532

5 Write the equation with F first. Sometimes it may be appropriate to use a common denominator.

= +

=+

FC

FC

9

532

or9 160

5

ii 1 Replace C with 21. Note that 9C means 9 × C. ii =×

+F9 21

532

2 Evaluate F. F

F

F

189

532

37.8 32

69.8

= +

= +=

so 21°C = 69.8 °F

b 1 Write the equation. Solve for r (make r the subject) as follows.

b A = πr2

2 Divide both sides by π.π

=A

r2

3 Take the square root of both sides, and write r first. As r is the radius, we take the positive root only. π

=rA

4 Substitute A = 20 into the new formula. A

r

If 20,

20

π

=

=

5 Evaluate r. r = 2.52 cm, correct to 2 decimal places


Worked example 6

Note: In this example, m represents metres, m/s represents metres per second (velocity), and m/s2 represents metres per second per second (acceleration).

The fi nal velocity, v m/s, of an object that begins with velocity u m/s and accelerates ata m/s2 over a distance of s m is given by the equation = +v u as2 .2 2

a Find the value of v when u = 16, a = 2 and s = 36.b Rearrange the given equation to make s the subject.c Find the distance travelled by an object that begins with a velocity of 10 m/s and has a fi nal

velocity of 4 m/s while accelerating at −1 m/s2.

think Write

a 1 Write the given equation and then make v the subject by fi nding the square root of both sides.

a = +

= ± +

v u as

v u as

2

2

2 2

2

2 Substitute u = 15, a = 2 and s = 36. 16 2 2 36

256 144

400

20

2= ± + × ×

= ± +

= ±= ±

The fi nal velocity is 20 m/s.

3 Simplify and evaluate v.

b 1 Begin with the given equation. b = +v u as22 2

2 Subtract u2 from both sides. − =v u as22 2

3 Divide both sides by 2a. −=

v u

as

2

2 2

4 Reverse so that s is given on the left. =−

sv u

a2

2 2

c 1 Match the variables with the given information. c ==

= −

u

v

a

10

4

1

2 Write the formula that has s as the subject (see part b above).

=−

sv u

a2

2 2

3 Substitute the values given in step 1.=

−× −

4 10

2 1

2 2

4 Simplify and evaluate. =−

=

=

−

−

−

16 100

284

242

5 Explain the answer in words. The object travels 42 m in its initial direction.

exercise 1B rearrangement and substitution 1 We4 Each of the following is a real equation used in business, mathematics, physics or another

area of science. Make the variable shown in parentheses the subject in each case.a A = L + P (P) b A = lw (l)

c vdt= (t) d C = 2πr (r)


e E = αθ + βθ 2 (β) f FkQq

r2= (r)

g Fd mv mu1

2

1

22 2= − (v) h v rTγ= (γ)

i S = 2w(l + h) + 2lh (w) j S = 2πr2 + 2πrH (H)

2 We5 Calculate the value of the subject (the first mentioned variable), given the values of the other variables.

a Ik

d 2= k = 60, d = 15

b E = K + mgh K = 250, m = 2, g = 10, h = 5

c D n( )12

λ= − n = 3, λ = 2.8

d E = hf0 − W h = 6.62, f0 = 5000, W = 20 000

e v r y2 2ω= − ω = 2, r = 1.6, y = 1

3 Make the variable in parentheses the subject and find its value using the given information.

a A = l 2 (l) A = 60

b V r43

3π= (r) V = 1000

c v = u + at (a) v = 25, u = 0, t = 6

d Tl

g2π= (l) T = 4, g = 9.8

e Kc

1

2αα

=−

(c) K = 6.9, α = 0.05

4 We6 The perimeter, P, of a rectangle of length l and width w may be found using the equation P = 2(l + w).a Find the perimeter of a rectangle of length 16 cm and width 5 cm.b Rearrange the equation to make w the subject.c Find the width of a rectangle that has perimeter 560 mm and length 240 mm.

5 The area of a trapezium (Figure A) is given by Aa b

h2

=+

, where a and b are the lengths

of the parallel sides and h is the height.a Find the area of the trapezium shown in Figure B.b Using Figure A, fi nd an equation for determining side a in terms of the area A and side b.c Find a in Figure C.

Figure A

a

h

b

AreaA

Figure C

62 cm50 cm

a

Area = 2000 cm2

9 m

16 m

21 m

Figure B

6 The size of a 2-year investment account with a particular bank is given by A Dr

1100

2

= +

,

where A is the amount ($) in the account after two years, D is the initial deposit ($) and r is the interest rate (%).a Find the amount in such an account after two years if the initial deposit was $1000 and the

interest rate was 6%.b Make r the subject of the equation.c Find the rate required for an initial deposit of $1000 to grow to $2000 after 2 years.


7 The object and image positions for a lens of focal length f are related by

the formula u v f

1 1 1,+ = where u is the distance of the object from the lens

and v is the distance of the image from the lens.a Make f the subject of the equation.b Make u the subject of the equation.c How far from the lens is the image when an object is

30 cm in front of a lens of focal length 25 cm?

8 The length of a side of a right-angled triangle can be found using Pythagoras’ theorem: c2 = a2 + b2, where c is the length of the longest side, and a and b are the lengths of the two shorter sides. Find the value of b in the triangle above.

9 The volume of a cone is given by the rule V = 13 π r2h, where r is the radius of the widest

part of the cone and h is the vertical height of the cone. Given that the volume of a cone is 100 cm3 and its radius at the widest point is 12 cm, find the height of the cone, expressing your answer in terms of π.

1C Gradient of a straight lineThe gradient of a line describes its slope or steepness.

Positivegradient

y

x

Negativegradient

y

x

Zerogradient

y

x

In�nitegradient

y

x

The gradient may be calculated using the formula:

m = riserun or m

y y

x x2 1

2 1=

−−

These terms are illustrated at left. Here are two examples of where gradient can affect our everyday lives. Can you think of others?

A cliff face with a steeper gradient provides a greater challenge for climbers.

Scientists calculate the required gradient of solar panels so that the maximum amount of energy is absorbed.

32

b

4

diGital doCdoc-9697

Career profile rick morris – vigneron

r

h

cone

y

x

(x1, y1)

(x2, y2)

Run

Rise


Worked example 7

Calculate the gradient of this linear graph using the intercepts shown.

think Write

1 Identify the rise and run. Rise = 14, run = 2.

2 Calculate m = riserun

. =

=

m14

27

Worked example 8

Calculate the gradient of the line passing through the points (3, −6) and (−1, 8).

think Write

1 Use the formula my y

x x2 1

2 1= −

−. = −

−m

y y

x x2 1

2 1

2 Match up the terms in the formula with the values given.(x1, y1) (x2, y2)(3, −6) (−1, 8)

3 Substitute the given values. = −−

=

=

−

−

−

−

m8 6

1 314

47

2

4 Simplify. = −14

4

5 Cancel if possible. =−7

2

If the angle a line makes with the positive direction of the x-axis is known, the gradient may be found using trigonometry applied to the triangle shown below.

y

x

rise

run

tan = opposite = = madjacentriserun

( )

θ

θ

Worked example 9

a Calculate the gradient (accurate to 3 decimal places) of a line making an angle of 40° to the positive x-axis.

b Calculate the gradient of the line shown at right. Express your answerto 2 decimal places.

think Write

a Since the angle the line makes with the positive x-axis is given, the formula m = tan (θ) can be used.

a m = tan (θ)= tan (40°)= 0.839, correct to

3 decimal places

y

x−2

14

y

x60°

y

x40°


b 1 The angle given is not the one between the graph and the positive direction of the x-axis. Calculate the required angle θ.

b θ = 180° − 60°= 120°

2 Use m = tan (θ) to calculate m to 2 decimal places.

m = tan (θ)= tan (120°)= −1.73

exercise 1C Gradient of a straight line 1 We7 Calculate the gradient of each of the following linear graphs using the intercepts shown.

a

6

−3

y

x

b

2

6

y

x

2 We8 Without drawing a graph, calculate the gradient of the line passing through:

a (2, 4) and (10, 20) b (4, 4) and (6, 14)c (10, 4) and (3, 32) d (5, 31) and (− 7, 25).

3 We9a Calculate the gradient (accurate to 3 decimal places) of a line making the angle given with the positive x-axis.a 50° b 72°c 10° d − 30°e 150° f 0°g 45° h 89°

4 We9b Calculate the gradient of each line below. Give answers to 2 decimal places.

a y

x43°

b y

x69°

c y

x28°

d y

x

15°

5 Which of these lines has:a a non-zero positive gradient?b a negative gradient?c a zero gradient?d an undefi ned gradient?

y

x60° θ

diGital doCdoc-9698

Gradient of a straight line

54321

−1−2−3−4−5

−5−4−3−2−1 1 2 3 4 50 x

y

A

C

B D


6 mC

a Which of the following lines has a gradient of −2?

54321

−1−2−3−4−5

−5−4−3−2−1 1 2 3 4 50 x

yA C

E

B D

b Which of the following lines has a gradient of 3?

54321

−1−2−3−4−5

−5−4−3−2−1 1 2 3 4 50 x

yAC

E

B

D

7 Burghar plots the coordinates of a proposed driveway on a plan that is shown below. What is the gradient of the proposed driveway?

Driveway

Garage

2 m

17 m

8 An assembly line is pictured below. What is the gradient of the sloping section? (Give your answer as a fraction.)

0.85 m

15 m

BOFFOMade inAustraliaBOFFOMade inAustralia

BOFFOMade inAustralia

9 Determine the value of a in each case so the gradient joining the points is equal to the value given.a (3, 0) and (5, a), gradient 2b (2, 1) and (8, a), gradient 5c (0, 4) and (a, −11), gradient 3d (a, 5) and (5, 1), gradient −2

10 For safety considerations, wheelchair ramps are constructed under regulated specifications. One regulation requires that the maximum gradient of a ramp exceeding 1200 mm in length is 1

14.

a Does a ramp 25 cm high with a horizontal length of 210 cm meet the requirements?

b Does a ramp with gradient 118

meet the specifi cations?

c A 16 cm high ramp needs to be built. Find the horizontal length of the ramp required to meet the specifi cations.

diGital doCdoc-9699SkillSHEET 1.1Using a gradient to find the value of a parameter


1d Sketching linear functionsThe general form for linear equations is y = mx + c, where m is the gradient of the line and c is the y-intercept.

y = mx + c

Gradient y-intercept

y

x

Gradien

t m

y-intercept

x-intercept

y

x

These lines have identicalgradients (equal m values).

To sketch a graph from a linear equation expressed in general form, follow these steps.Step 1 Plot the y-intercept on a set of axes.Step 2 Find and plot a second point on the line. Do this by substituting any value of x into the equation

and fi nding the corresponding y-value.Step 3 Join the two points.

Alternatively, you can use a CAS calculator or other graphing technology.To fi nd the equation of a line given the gradient and y-intercept, simply substitute the values of m and

c into y = mx + c.

Sketching linear graphs using interceptsTo draw a graph, only two points are needed. A line may then be drawn through the two points, and will include all other points that follow the given rule. The two points can be chosen at random; however, it is often easier to sketch a graph using the points where the graph crosses the axes. These points are called x- and y-intercepts. The x-intercept occurs when y = 0, and the y-intercept occurs when x = 0.

Worked example 10

Sketch the graph of y = −x + 6, showing x- and y-intercepts.

think Write/draW

1 Find the y-intercept (when x = 0). Substitute x = 0 into the equation.

If x = 0: y = −1 × 0 + 6 y = 6 (0, 6)

2 Find the x-intercept (when y = 0).Substitute y = 0 into the equation.

If y = 0: x

x

0 6

6 (6, 0)

= +=

−

3 Mark the intercepts on a set of axes. y

x

(0, 6)

(6, 0)

4 Join the intercepts with a straight line.

diGital doCdoc-9700

WorkSHEET 1.1

y

x


Worked example 11

Sketch the graph of 3x − 2y = 12.

think Write/draW

1 Find the y-intercept (when x = 0). Substitute x = 0 into the equation.

If x = 0: 3 × 0 − 2y = 12−2y = 12

y12

2= −

y = −6

2 Find the x-intercept (when y = 0). Substitute y = 0 into the equation.

If y = 0: 3x − 2 × 0 = 12 3x = 12

x = 4

3 Mark the intercepts on a set of axes. y

x

−6

4

4 Join the intercepts with a straight line.

The graphs of some equations do not have two intercepts, as they pass through the origin (0, 0). Such equations are of the form y = k x or a x + by = 0.

To sketch graphs of such equations, we use (0, 0) and any other point, for example the point where x = 1. (We could choose any other non-zero value.)

Worked example 12

Sketch the graph for the equation 4 x − 3y = 0.

think Write/draW

1 Try substituting x = 0 to fi nd the y-intercept. If x = 0: 4 × 0 − 3y = 0−3y = 0

y = 0

2 Note that the graph passes through (0, 0). There is no point substituting y = 0, as we know we’ll get x = 0.

(0, 0)

3 Substitute another x-value. In this example we use x = 1.

If x = 1: 4 × 1 − 3y = 0 4 − 3y = 0 4 = 3y

y4

3=

(1, )43

4 Plot the points (0, 0) and (1, 43) on a set of axes. Note that 4

3 is 11

3, which is a little less

than 112.

y

x

4–3

(1, )

(0, 0)

1

1

2

tUtorialeles-1404Worked example 12


exercise 1d Sketching linear functions 1 Use a CAS calculator or other method to sketch graphs of the following equations on the same set of axes.

a y = x b y = 2x c y = 3x d y = −x e y = −2x

2 In question 1, what is the effect on the graph of the number in front of the x (the ‘x-coefficient’ or ‘gradient’)?

3 Use a CAS calculator or other method to sketch graphs of the following equations on the same set of axes.a y = x + 1 b y = x + 2 c y = x + 3 d y = x − 4

4 Use a CAS calculator or other method to sketch graphs of the following equations on the same set of axes.a y = 2x + 1 b y = 2x − 7 c y = −3x + 6 d y = 3x − 5

5 In questions 3 and 4, what is the effect on the graph of the number at the end of the equation (the ‘y-intercept’)?

6 Write the equation of a line having the following properties (where m = gradient and c = y-intercept).a m = 2, c = 7 b m = −3, c = 1 c m = 5, c = −2d m = 2

3, c = 1

3e m = 3

4

−, c =

12

−f y-intercept 12, gradient −2

7 Rearrange the following equations and state the gradient and y-intercept for each.

a 2y = 8x + 10 b 3y = 12x − 24 c −y = 3x − 1 d 16 − 4y = 8xe 21x + 3y = −27 f −10x + 5y = 25 g −11y − 2x = 66 h 8x + 3y − 2 = 0i 15 − 6y + x = 0 j 2y + 7 + 5x = 0

8 State the equation for each of the following graphs.

a 54321

−1−2

−1

6

1 20 x

y

− 1–2

b 321

−1−2−3−4−5

−2 20 x

y

c 321

−1−2−3−4−5

−1−2 1 20 x

y d 54321

−1−2−3−4−5

−1 1 2 3 4 5 60 x

y

e 54321

−1−2−3−4−5

−1 1 2 3 40 x

y f 987654321

−1−2−3−4−5−6−7−8−9

−3 −2 −1 1 2 30 x

y

diGital doCdoc-9701

linear graphs


9 We10 Sketch graphs of the following linear equations, showing x- and y-intercepts.a y = 6x + 18 b y = 3x − 21c y = −2x − 3 d y = 10 − 5xe y = −9x + 30 f y = 2(x − 8)

10 We11 Sketch graphs for each of the following equations.a 2x + 3y = 6 b 4x + 5y = 20c 6x − 3y = −18 d 7x − 5y = 35

11 Sketch the graph for each equation.a 6x + 7y + 42 = 0 b 5x − 2y + 20 = 0c −3x + 4y − 16 = 0 d y − 3x + 6 = 0

12 We12 Sketch the graph for each equation.a x + y = 0 b x − y = 0c 2x + y = 0

13 mC Which of the following is in the form ax + by = c?a 2x − 3y − 1 = 0 B 2x + 3y + 1 = 0C 2x + 3y = x d 2x + 3y = 1

e y = 23 x − 1

14 mC The x- and y-intercepts for the equation 2y = −3x + 12 are (respectively):a 2 and 3 B −3 and 12C −4 and 6 d −4 and −6e 4 and 6

15 mC Which of the following has a y-intercept of −3?a y = −3x − 3 B y = −3x + 3C x + 3y = 9 d x − 3y + 9 = 0e 3x + y + 9 = 0

1e Simultaneous equationsSimultaneous equations are groups of equations containing two or more variables. In this section, we look at pairs of linear equations involving the variables x and y. Each equation, as we have learned in previous sections, may be represented by a linear graph that is true for many x- and y-values. If the graphs intersect (when wouldn’t they?), the values of x and y at the intersection are those that make both equations true.

Graphical solutionFinding the point of intersection of two straight lines can be done graphically; however, the accuracy of the graph determines the accuracy of the solution.

Consequently, using a calculator to solve the equations graphically is more reliable than reading the solution from a hand-drawn graph.

Worked example 13

Use a CAS calculator to solve the following simultaneous equations graphically.y = −3x + 5 4x − 7y + 8 = 0

think Write

1 Using a CAS calculator, make y the subject of the second equation.Complete the entry line as:solve (4x − 7y + 8 = 0, y)

solve (4x − 7y + 8 = 0, y)

Result: = +y

x4( 2)

7

Both graphs havethe same x- andy-values here.

y

x


2 On a Graphs page, complete the function entry lines as:f 1(x) = −3x + 5

f xx

2( )4 ( 2)

7= × +

Use the calculator to fi nd the intersection point.

f 1(x) = −3x + 5

f 2(x) = +x4( 2)

7

3 Write the answer. Solving y = −3x + 5 and 4x − 7y + 8 = 0simultaneously produces the solution (1.08, 1.76).

algebraic solutionIt is possible to solve simultaneous equations without graphs, that is, algebraically. The methods of substitution and elimination taught in earlier years may be used.

Worked example 14

Use the substitution method to solve the following simultaneous equations.y = 2 x + 5−x + 3y = 25

think Write

1 Write down and label the equations. y = 2x + 5 [1] −x + 3y = 25 [2]

2 Substitute [1] into [2] and label the resulting equation [3].

−x + 3(2x + 5) = 25 [3]

3 Solve [3] for x and label the solution as [4]. −x + 6x + 15 = 255x + 15 = 25

5x = 10 x = 2 [4]

4 Use the solution to solve for y and label the solution as [5].

Substitute [4] into [1].y = 2(2) + 5y = 4 + 5y = 9 [5]

5 State the complete answer. Solution: (2, 9)

6 Optional check: substitute [4] and [5] into [2] to check that these values for x and y make [2] true.

Check in [2].LHS = −x + 3y

= −2 + 3(9) = 25 = RHS ✓

Worked example 15

Use the elimination method to solve these simultaneous equations.2 x + 9y = −55 x − 2y − 12 = 0

think Write

1 Write down and label the equations. 2x + 9y = −5 [1]5x − 2y − 12 = 0 [2]

2 Rearrange [2] so it is in a similar form to [1]. Call this [3]. Write down [1] again.

5x − 2y = 12 [3] 2x + 9y = −5 [1]


3 Obtain 10x in both [1] and [3]. 2 × [3] 10x − 4y = 24 [4]5 × [1] 10x + 45y = −25 [5]

4 Eliminate x as shown. [5] − [4]: 49y = −49

5 Solve for y. =−

y49

49 y = −1 [6]

6 Substitute [6] into [1] to fi nd x. Substitute [6] into [1].2x + 9 (−1) = −5

2x − 9 = −52x = −5 + 92x = 4

x = 2 [7]

7 State the solution. Solution: (2, −1)

8 Again, [6] and [7] may be checked in [2] if desired. Check in [2].LHS = 5x − 2y − 12

= 5(2) − 2(−1) − 12= 10 + 2 − 12= 0= RHS ✓

Worked example 16

Two shoppers buy the following at a fruit shop, paying the amounts given. What was the cost of each apple and each banana?Shopper 1: 4 apples and 3 bananas for $2.59Shopper 2: 6 apples and 5 bananas for $4.11

think Write

1 Decide on variable names for the unknown quantities.

Let a = cost of an apple (in cents).Let b = cost of a banana (in cents).

2 Write equations involving these variables. Work in terms of cents.

4a + 3b = 259 [1] 6a + 5b = 411 [2]

3 Choose a variable to eliminate, in this case b. 5 × [1]: 20a + 15b = 1295 [3] 3 × [2]: 18a + 15b = 1233 [4]

4 Find [3] − [4] and solve for a. [3] − [4]: 2a = 62a = 31 [5]

5 Solve for b. Substitute [5] into [1]. 4 × 31 + 3b = 259 124 + 3b = 259 3b = 135

b = 45 [6]

6 State the answer using [5] and [6] as a guide. The cost of an apple is 31 cents, and the cost of a banana is 45 cents.

exercise 1e Simultaneous equations 1 We13 Use a CAS calculator to solve the following simultaneous equations.

a y = −2x, y = −4x − 6 b y = 3x − 5, y = 20

c y = 3x + 5, y = 7x − 4 d −3x + y = −4, y = 6x + 5

e y = 10x + 1, 2x + y = −6 f 9x + y = 17, x + y = 14

diGital doCSdoc-9702Simultaneous linear equations — graphical method


2 We14 Use the substitution method to solve the following simultaneously.a y = 3x + 1, y = 2x + 2 b y = 5x + 5, y = −x − 19

c y = x + 2, 3x − 4y = −1 d y = −2x + 3, −5x + 2y = 1

e −4x − 3y = 2, y = −6x + 7 f y = 10 − x, 2x + 7y = 5

3 We15 Use the elimination method to solve these simultaneous equations.a 9x + 10y = 153, 3x − y = 12 b 7x − 11y = −13, x + y = 11

c 6x − 2y = 10, 2x + 5y = −8 d −3x + y = 8, 4x + 2y = 21

e 7y − x = 11, x + y = 10 f x − 11y = −15, y + 6x = 9

4 We16 At the conclusion of a tour of Wonky Willy’s confectionery factory, Nutrina buys 10 choc balls and 8 fizz wizzers for $4.30, and her friend purchases 6 choc balls and 9 fizz wizzers for $4.05. Determine the cost of each type of lolly.

5 The sum of two whole numbers, x and y, is 41. The difference between them is 3. Write two equations involving x and y and solve them to find the numbers.

6 A farmer counts emus and sheep in a paddock, and notes there are 57 animals and 196 feet. Assuming no animal amputees, how many of each animal are there?

7 A sports store supplies 24 basketballs and 16 cricket balls to one school for $275.60, and delivers 12 basketballs and 32 cricket balls to another school for $211. If delivery is free, how much did the supplier charge for each type of ball?

8 A businessperson hires a stretch limousine for 2 days and a sedan for 3 days while on an interstate trip. If the total car hire cost of the trip was $675, and the limousine cost triple the price of the sedan, find the cost per day of the limousine.

9 mC A manufacturing plant produces square and circular metal panels in fixed sizes. If the mass of a square panel is 13 kg and that of a circular panel is 22 kg, how many of each panel are there in a truck loaded with 65 panels of total mass 1205 kg? The equations to solve are:a 13s + 22c = 1205, s + c = 65 B 22s + 13c = 1205, s + c = 65C 13s + 22c = 65, s + c = 1205 d 22s + 13c = 65, s + c = 1205e 13s + 22c = 1205, s + c = 35

10 mC Which of the following is a solution of 11x + 2y = −121 and 10x + 12y = −222?a (11, 2) B (−121, −222) C (10, 12) d (−9, −11) e (6, 10)

1F Finding the equation of a straight lineConsider a general linear graph containing the particular points (x1, y1), (x2, y2) and the general point (x, y) (which could be any point).

Using the fi rst two of these points in the formula for gra dient, we have

my yx x2 1

2 1=

−−

[1]

Using the fi rst point and the general point in the same formula yields

my y

x x1

1= −

− [2]

Putting [2] = [1] gives y y

x x

y y

x x1

1

2 1

2 1

−−

= −−

, which may be rearranged to

− = −

−

−y yy y

x xx x( )1

2 1

2 11 [3]

Since my yx x2 1

2 1=

−− , equation [3] may be written

y − y1 = m(x − x1) [4]

This last formula may be used to fi nd the equation of a straight line when two points are given or when the gradient and only one point are given. When two points are given, the gradient m may be found

fi rst using my yx x2 1

2 1=

−− and substituted into the formula y − y1 = m(x − x1) along with one of the points.

y

x

(x1, y1)

(x2, y2)

(x, y)


Worked example 17

Find the equation of the line having gradient 34 that passes through (7, 11).

Express your answer in the forms i ax + by + c = 0 and ii y = mx + c.

think Write

1 As one point and the gradient are known, use the formula y − y1 = m(x − x1).

y − y1 = m(x − x1)

2 List the given information. m = 34

(x1, y1) (7, 11)

3 Substitute for all variables except x and y. y − 11 = 34 (x − 7)

4 Simplify, expressing in the formax + by + c = 0.

i 4y − 44 = 3(x − 7) 4y − 44 = 3x − 21

3x − 4y + 23 = 0

5 Express your answer in the form y = mx + c.

ii 3x + 23 = 4y

y x34

234

= +

Worked example 18

Find the equation of the straight line containing the points (2, −5) and (−3, 1).Express your answer in the forms i ax + by + c = 0 and ii y = mx + c.

think Write

1 Write down the points so they match the variables in the formula.

(x1, y1) (x2, y2)(2, −5) (−3, 1)

2 As two points are known, first use the

formula my y

x x2 1

2 1= −

− to find m.

my y

x x

1 5

3 26

5

2 1

2 1

6

5

= −−

= −−

=

=

−

−

−

−

3 Write the formula y − y1 = m(x − x1). y − y1 = m(x − x1)

4 Substitute the calculated gradient m 65

=−

and the first point (x1, y1) = (2, −5). Leave x and y as they are.

− = −− −y x5 ( 2)6

5

5 Simplify and express in the two forms required. + = −−y x5 ( 2)65

−5y − 25 = 6(x − 2)−5y − 25 = 6x − 12

−5y = 6x + 13 i So 6x + 5y + 13 = 0, or

ii y x65

13

5= −

−

perpendicular linesPerpendicular lines are lines that meet at right angles. The gradients of two perpendic ular lines, when multiplied together, equal −1. Stated mathematically:

m1 × m2 = −1


Another way to write this relationship is m m1

12

=−

. This type of relationship is known as a negative

reciprocal. For example, the negative reciprocal of 23 is 3

2

−. So, two lines are perpendicular if their

gradients are negative reciprocals.

exercise 1F Finding the equation of a straight line 1 Copy and complete the table below.

Equation 1f1(x)

Gradient of equation 1

(m1)Equation 2

f2(x)

Gradient of equation 2

(m2) m1 × m2

a y = 2x + 1 y = −12x

b y = 3x − 4 y = −13x + 2

c y = 14x + 6 y = −4x − 9

d y = 25x − 3 y =

−52x + 4

e y = − 97 x y = 7

9 x + 1

2 Sketch f1(x) and f2(x) for part a above using a calculator or using graph paper with identical scales on each axis to produce accurate graphs. What do you notice about the graphs?

3 Repeat question 2 for graphs b to e. Do you notice anything special about each pair of graphs?

4 Find the gradient of a line perpendicular to another line that has a gradient of:

a 4 b −9 c 17

d 89

−e 7

2f 1.

5 Find the gradient of a line that is perpendicular to the line with equation:

a y = −5x + 2 b y = x − 1 c y = 23 x + 1

d y x 276

= −− e 2x + y = 5 f 3x − 4y = 7.

6 We17 Find the equation of a straight line having the gradient given and passing through the point listed. Express your answer in the forms i ax + by + c = 0 and ii y = mx + c.a (1, 2) gradient 3 b (4, 1) gradient 5c (3, −2) gradient −1 d (−3, 2) gradient 1

2

7 We18 Find the equation of the line containing each pair of points. Express your answers in the forms i ax + by + c = 0 and ii y = mx + c.a (5, 2) (3, 1) b (1, 1) (5, 5) c (6, 3) (8, 2) d (2, −2) (0, 1)

8 Find the equation of the line passing through (3, −3) that makes an angle of 45° with the positive x-axis.

9 Find the equation of the line containing (7, −2) that makes an angle of 71.565° with the positive x-axis.

10 Find the equation of the line (in y = mx + c form) that:a is perpendicular to the line with equation y = 3x + 1, passing through (−3, 6)b is parallel to the line with equation y = 2

5x − 9, passing through (4, −7)

c is parallel to the line with equation 3x + 6y = 8, passing through (2, 2)d is perpendicular to the line with equation −6x + 7y − 2 = 0, passing through (4, 0)e has gradient 2, passing through the intersection of the lines with equations y = 3x − 5 and y = −2x + 5

f has gradient 34

−, passing through the intersection of the lines with equations x + 4y = −14 and

−5x + 2y = 4.

11 Find the equation of the line that passes through the point of intersection of the lines whose equations are 7x − 3y − 19 = 0 and 3x + 2y + 5 = 0, given that the required line is parallel to the line with equation −5x − 2y = 3.

12 Find the equation of a line containing the intersection of the lines with equations y = −3x + 4 and 5x − 3y + 40 = 0 that:

a has a gradient of 67 b is perpendicular to the line with gradient

23

c passes through the point (−1, 9) d is parallel with the line joining (−8, 5) and (0, 4).

diGital doCdoc-9703

SkillSHEET 1.2reciprocals and

negative reciprocals

diGital doCdoc-9704

equation of a straight line


13 A line passes through the points (−8, −5), (4, −3) and (a, 12). Find the value of a.

14 The points (2, 7) and (6, 9) lie on the same straight line. Does the point (4, 8) also lie on this line?

15 The height of a particular young pine tree is found to increase in a linear manner each month in the first year after planting. Find an equation connecting height with time in months after planting, using the information supplied in the diagram below.

34 cm

After 2 months

52 cm

After 5 months

1G distance between two points and midpoint of a segmentThe distance, d, between any two points on the Cartesian plane may be found using Pythagoras’ theorem applied to a right-angled triangle as shown at right.

Using Pythagoras’ theorem: c2 = a2 + b2 or c a b2 2= + andreplacing c with d, a with (x2 − x1) and b with (y2 − y1),

we have d x x y y( ) ( )2 12

2 12= − + − .

Worked example 19

Find the distance between the points (−3, 7) and (5, −2) correct to 3 decimal places.

think Write

1 Match up (−3, 7) and (5, −2) with (x1, y1) and (x2, y2).

(x1, y1) (x2, y2)(−3, 7) (5, −2)

2 Substitute into the formula for d and simplify.

= − + −

= − + −

= +

= +

=

− −

−

d x x y y( ) ( )

(5 3) ( 2 7)

(8) ( 9)

64 81

145

2 12

2 12

2 2

2 2

= 12.042, correct to 3 decimal places

diGital doCdoc-9705WorkSHEET 1.2

interaCtiVitYint-0260distance between two points and midpoint of a segment

y2

y1

d

x1 x2

y

x

(y2 − y1)

(x2, y2)

(x2 − x1)(x1, y1)



midpoint of a segmentThe middle or midpoint M (xm, ym) of a segment joining two general points A (x1, y1) and B (x2, y2) is shown on the Cartesian plane below.

y

x

M (xm, ym)

y2 − ym

x2 − xm

ym − y1

xm − x1

A (x1, y1)

B (x2, y2)

E

DC

Consider the triangles ACM and MEB.

∠MAC = ∠BME (since AM and BM have the same slope)∠CMA = ∠EMB (since ∠ACM and ∠MEB are both 90°)AM = BM (given, as M is the midpoint)Therefore ΔACM ≡ ΔMEB.Since the horizontal and vertical sides of the two triangles must be equal, we have:

xm − x1 = x2 − xm and ym − y1 = y2 − ym

Simplifying these,

2xm = x2 + x1 2ym = y2 + y1

x

x x

2m

2 1= −

y

y y

2m

2 1= +

So the point M has coordinates + +

x x y y

2,

21 2 1 2 .

Worked example 20

Find the midpoint of the segment joining (5, 9) and (−3, 11).

think Write

1 Match (5, 9) and (−3, 11) with (x1, y1) and (x2, y2). (x1, y1) (x2, y2)(5, 9) (−3, 11)

2 Substitute values into the formula for M and simplify. = + +

= + +

=

=

−

x x y yM

2,

2

5 3

2,9 11

2

2

2,20

2(1,10)

1 2 1 2

exercise 1G distance between two points and midpoint of a segment 1 We19 Find the distance between each of the following pairs of points.

a (4, 5) and (1, 1) b (7, 14) and (15, 8) c (2, 4) and (2, 3) d (12, 8) and (10, 8)

2 Calculate the distance between each of the pairs of points below, accurate to 3 decimal places.a (−14, 10) and (−8, 14) b (6, −7) and (13, 6) c (−11, 1) and (2, 2)

3 Find the distance between each of the following pairs of points in terms of the given variable(s).a (a, 1), (2, 3) b (5, b), (0, 6) c (c, 2), (4, c) d (d, 2d), (1, 5)

diGital doCdoc-9706

distance between two points


4 Two hikers are about to hike from A to B (shown on the map below). How far is it from A to B ‘as the crow flies’, that is, in a straight line?

N

Grid spacing : 1 km S

EW

B (E7, N4)

Lake Phillios

100 m

200 m

50 m

A (W12, S5)

300 m

200 m100 m

5 Using the coordinates shown on the aerial photo of the golf course, calculate (to the nearest metre):a the horizontal distance travelled

by the golf ball for the shot down the fairway

b the horizontal distance that needs to be covered in the next shot to reach the point labelled A in the bunker.

6 We20 Find the midpoint of the segment joining each of the following pairs of points.

a (1, 3) and (3, 5) b (6, 4) and (4, −2) c (2, 3) and (12, 1) d (6, 3) and (10, 15)

7 Find the midpoint of the segment joining each of the following pairs of points.a (7, −2) and (−4, 13) b (0, 22) and (−6, −29) c (−15, 8) and (−4, 11) d (−3, 40) and (0, −27)

8 Find the coordinates of the midpoint of each of the following pairs of points, in terms of a variable or variables where appropriate.a (2a, a) and (6a, 5a) b (5, 3c) and (11, 3c) c (3f, 5) and (g, −1)

9 Find the value of a in each question below so that the point M is the midpoint of the segment joining points A and B.

a A (−2, a), B (−6, 5), M (−4, 5) b A (a, 0), B (7, 3), M (8, 32)

y (in metres)

x (in metres)

A (320, 148)(225, 96)

(80, –64)

diGital doCdoc-9707midpoint of a segment


10 A fun-run course is drawn (not to scale) at right. If drink stations D1, D2 and D3 are to be placed at the middle of each straight section, give the map coordinates of each drink station.

11 Find the equation of a line that has a gradient of 5 and passes through the midpoint of the segment joining (−1, − 7) and (3, 3).

12 Find the equation of a line parallel to the line with equation 9x − 3y = 5 that passes through the midpoint of the segment connecting (0, −4) and (−2, 10).

1h linear modellingMany real-life applications, such as fees charged for services, cost of manufacturing or running a business, patterns in nature, sporting records and so on, follow linear relation ships. These relationships may take the form of a linear equation; for example, F = 50 + 30t may be used by a tradesperson to calculate her fee (in dollars) for t hours of work.

Here, F is the fee in dollars, and t the time in hours. The 50 represents an initial fee for simply turning up, while the 30t is the amount charged for the time spent on the job.

For example, if t = 2 hours, 30t = 60, so the total charge for the work would be $(50 + 60) = $110.Equations like F = 50 + 30t are sometimes referred to as ‘linear models’, a common form of which is:

Total cost = Fixed cost + Cost per unit × Number of units.

This is, of course, equivalent to y = mx + c. y = c + mx.

Worked example 21

A generator company charges a $200 delivery fee, and a rental fee of $1500 per day.a Find an expression relating total charge to the number of days for which the generator is hired.b Sketch a graph of the relationship.c What would be the charge for 4 weeks of rental?

think Write/draW

a 1 Define convenient variables. a Let T = total charge (in dollars) and n = number of days the generator is hired.

2 The fixed cost is $200, and the cost per unit is $1500. (c = 200, m = 1500)

T = 200 + 1500n

b 1 Find the vertical intercept (when n = 0). b If n = 0, T = 200

2 The total cost rises $1500 each day, so the graph must show this.

1 2

3200

200

1700

T ($)

n (Days)

c 1 After 4 weeks, n = 28. Substitute this in the equation from part a.

c If n = 28T = 200 + 1500 × 28

= 200 + 42 000 = 42 200

2 Write the answer in words. After 4 weeks, the total cost is $42 200.

y

x

(−4.5, 5)(1.5, 3.5)

(13, −8)(3, −7)

(1.5, −2)START/FINISH

Of�cial tent

D1

D2

D3Coordinatesare in kilometres.


Worked example 22

‘Rent-a-Chef’ provides food cooked and served by a qualifi ed chef at parties. The company charges $120 as a booking fee, and an additional $30 per hour. Another company, ‘Greased Lightning’, provides fast food served by two students at a cost of $65 per hour, with no booking fee. Under what conditions would it be cheapest to hire ‘Greased Lightning’?

think Write

1 Defi ne convenient variables. Let C = cost (total) in dollars and t = time in hours.

2 Write an equation for the cost of hiring both organisations.

Rent-a-Chef C = 120 + 30t [1]Greased Lightning C = 65t [2]

3 Use simultaneous equations to fi nd when the cost is the same with each group.

Put [1] = [2]120 + 30t = 65t

120 = 35t

t120

35=

= 3.4 hours

4 At 3.4 hours, the cost is the same. Since Greased Lightning has the higher per hour cost, after 3.4 hours, they would be more expensive.

It is cheaper to hire Greased Lightning for food preparation and service of less than 3.4 hours (3 hours and 26 minutes) duration.

Notes1. 0.4 hours = 0.4 × 60 minutes = 24 minutes.2. An alternative approach would be to use a CAS calculator and fi nd the point at which the

two graphs crossed.

exercise 1h linear modelling 1 We21 The cost of hiring a floodlit

tennis court consists of a booking fee and an hourly rate.a Use the photo to write an equation

for the total hire cost in terms of the hourly rate.

b Sketch a graph of the relationship.c What would be the charge for

3 hours?

2 A singing telegram service charges a $60 appearance fee, and $8 per minute sung.a Write an equation for the total cost

of a singing telegram in terms of the number of minutes sung.

b Sketch a graph of the relationship.c What would be the charge for a 5-minute singing telegram?

3 Colleen delivers junk mail and is paid $32 to traverse a particular route, and a further 10 cents per leaflet delivered.a Write an equation for the total payment she receives.b Sketch a graph of the relationship expressed in a.c What would be Colleen’s pay if she delivers 1650 leafl ets along the route?

4 A pay-TV salesperson receives $300 per week plus $20 for every household he signs up to have pay-TV connected. How much does the salesperson receive for a week in which he signs up 33 households?

5 We22 A computer firm, SuperComputers Inc., offers a back-up plan covering the ongoing service and troubleshooting of its systems after sale. The cost of signing up for the service plan is $215, and there is an


diGital doCdoc-9708Simultaneous linear equations


hourly rate of $65 for the serviceperson’s time. Purchasers not signing up for the plan are charged a flat rate of $150 per hour for service. Would it be advisable to sign up for the service plan if you expected to need 3 hours of service assistance during the life of a computer purchased from Super Computers Inc?

6 Two amusement parks show the following information for school-age tourists in a promotional brochure. After how many rides does an excursion to Fun World become the cheaper option for the same number of rides?

Aqua World $8.00 entry$2.50 per ride

$12 entry$1.50 per ride

7 A telephone company, Opus, offers calls to Biddelonia for a connection fee of $14, and thereafter $1 per minute. Its rival, Elstra, offers calls for $2 per minute (no connection fee) to the same country.a Compare the cost of a 10-minute call to Biddelonia using each company.b At what point would it be cheaper to use Opus?

8 It costs you $6 to get into a taxi (the ‘flagfall’), and $1.50 per kilometre if you use PinkCabs, while NoTop taxis charge $8 flagfall, and $1.20 per kilometre.a How much would it cost with each company to travel 15 kilometres in one of its cabs?b When would it cost the same to use both companies?

9 Medirank, a health insurance company, charges $860 per year (for a single person) and requires customers to pay the first $100 of any hospital visit. HAB, on the other hand, charges an annual fee of $560 and requires its members to pay the first $150 of any hospital visit. Determine the number of hospital visits in a year for which the cost of health services is the same whichever company insures you.

10 Nifty is a car hire firm that charges insurance of $135 and $50 per day car hire. A competitor, Savus, simply charges $65 per day and offers ‘free’ insurance. You are planning a holiday, and would prefer to use Savus. Under what conditions (days hired) could you justify this choice?


SummarySolving linear equations and inequations

• Do the same to both sides and remember inverse operations + and −, × and ÷, √ and 2. • Aim to get a single variable by itself.• Solve inequations the same way as equations, keeping the original inequality sign at each step,

unless multiplying or dividing by a negative number.

rearrangement and substitution

• ‘Make x the subject’ means manipulate into the form ‘x = …’.• ‘Substitute’ means to replace a variable with a value.

Gradient of a straight line

• my y

x x2 1

2 1= −

− m = tan (θ)

where θ is the angle the line makes with the positive direction of the x-axis.

Sketching linear functions

• The general equation for a straight line of gradient m and y-intercept c is y = mx + c.• Lines with the same gradient (m) are parallel.• To fi nd the y-intercept, let x = 0 and fi nd y.• To fi nd the x-intercept, let y = 0 and fi nd x.• If y = 0 when x = 0, substitute another x-value (for example x = 1) to fi nd another point

on the line.• Join two points and/or intercepts with a straight line.

Simultaneous equations

• Simultaneous equations can be solved with a calculator.• For equations of the form y = ax + b, y = cx + d, consider using substitution.• For equations of the form ax + by = c, dx + ey = f, consider using elimination.

Finding the equation of a straight line

• Formulas for fi nding the equation of a straight line:

y = (mx + c) − = −−

−y yy y

x xx x( )1

2 1

2 11 y − y1 = m(x − x1)

• For perpendicular lines, m1 × m2 = −1.

distance between two points and midpoint of a segment

• d x x y y( ) ( )2 12

2 12= − + −

• = + +

x x y yM

2,

21 2 1 2

linear modelling • Total cost (y) = fi xed cost (c) + cost per unit (m) × number of units (x)• y = c + mx


Chapter review 1 Solve the equation

x x3(5 4)

7

6(4 3)

5

−=

+.

2 Find the value of x where x

x37 8

104 9

+

= − .

3 Solve the inequality x x3( 4)

4

1

2

+≤

−−.

4 The following formula may be used to study planetary motion.

GmM

R

m R

T

42

2

2

π=

Make T the subject of the equation.

5 Using Pythagoras’ theorem, find the length of the hypotenuse of the triangle shown. Express your answer:a in surd formb to 3 decimal places.

6 Calculate the gradient of each of the following lines.

a y

x

6

−8

b

x

y654321

−1−2−3

−5 −4 −3 −2 −1 1 2 3 4 5 60

c

x

y1 gridsquare = 1 unit

d y

x

(−4, −10)

(−12, −3)

7 Find the gradient of the line joining (−7, 15) and (2, −6).

8 Find the gradient of the line shown.y

x77°

9 State the gradient of the line below.y

x5

Short anSWer

6

9

c


10 State the gradient and y-intercept (in that order) for each of the following.

a y = 3x − 7 b 5x + 3y = 30 c 2x − 4y − 8 = 0

11 Find the equation for a linear graph having gradient 25 and y-intercept −3.

12 Sketch graphs of the following, showing intercepts.

a y = −3x + 24 b −x + 8y = 40 c 9x − 7y − 63 = 0 d y + 6x = 0

13 Solve y = 3x + 10 and y = −2x − 15 graphically.a Sketch the solution on a set of axes. b State the solution (point of intersection).

14 Solve y = −3x, y = 6x − 15 using substitution.

15 Use the method of elimination to solve 4x − 7y = 21, −2x + y = 6.

16 A piggybank contains 67 coins. If there are only one- and two-dollar coins in the piggybank, and there are 25 more one-dollar coins than two-dollar coins, how many of each type are there?

17 Find the gradient of a line perpendicular to 3x − 9y = 7.

18 Find the equation of the line containing (−4, 8) and (3, 1).

19 Find the equation of the line having gradient 67

− that passes through (1, 4).

20 Find the equation of the line perpendicular to y = 14x − 5 that passes through (−8, 6).

21 The distance between (2, −7) and (a, −2) is 41 units. Find the value of a if it is positive.

22 Show that the triangle with vertices (3, 7), (3, 3) and (6, 3) is a right-angled triangle.

23 The midpoint of the line joining (k, 2h) and (9k, 6h + 2) is (20, −11). Find k and h.

24 The washing machine repair company ‘Washed out’ charges $75 to come to your house, as well as an hourly charge of $65, calculated to the nearest half hour.a Write an equation that may be used to calculate the cost of any service call if the

time taken by the repairer is known.b Sketch a graph of the relationship between repair cost and time taken to do a repair.c How much would it cost to have a repair done that takes 3 1

2 hours?

mUltipleChoiCe 1 The first step in solving

x7 23

399

− = would be to:a add 23 to both sides B divide both sides by 3C divide both sides by 7 d multiply both sides by 3e multiply both sides by 7

2 x = −5 is a solution to the equation:

a 3x + 7 = −8 B 2x − 7 = −5 C x 25

65

+ =

d 2(x + 3) = 10 e x5

945=

−

3 The solution to x

x14

32( 2)

+ ≥ + is:

a x ≤ 10 B x 25

≥ C x 25

≤ d x 25

≤ −e x 6

5≤

−

4 When c2 = a2 + b2 is rearranged to make a the subject, the equation becomes:

a c a b2 2= + B a2 = b2 + c2 C a c b2 2 2= −

d a c b2 2= − e a = b + c

5 Which values, when substituted into K mv12

2= , give a value for K of 4?

a m = 2, v = 4 B m = 4, v = 2 C m = 8, v = 12

d m = 8, v = 1 e m = 1, v = 16

6 Using the equation P = m1v1 + m2v2, if P = 10, m1 = 2, m2 = 6 and v1 = 4, v2 would equal:

a 13

B 12

C 1 d 2 e 3


7 The line shown has a gradient of:a −6B −3C −2d 2e 6

3

8 The gradient of the line shown at right is 3. The value of a must be:a −2 B −1 C 5d 7 e 11

9 The gradient of the line joining (−1, 0) and (4, −10) is:a −4 B −2 C 2

d 4 e 5

10 Which of the graphs below has a gradient of 67?

a y

x

7

−6

B y

x

6

−7

C y

x

6

7

d y

x−7

−6

e y

x

−7

6

11 The gradient of the line with equation y x 167

= − is:

a −1 B 67 C

76 d 6 e 7

12 The y-intercept of the line with equation y x12 23

= + is:

a 23 B

32 C 2 d 3 e 12

13 The gradient and y-intercept (in that order) of the line with equation 2x − 3y = 7 are:

a 2 and −3 B 2 and 7 C 23

− and 7

d and23

73

−e 3 and 7

2−

14 Which of the following could be the graph of y = 2x + c?

a y

x

−c

B y

x

c

C y

xc

y

x3

6

y

x

(2, −5)

(6, a)


d y

x−c

e y

x

c

15 The equation of the line shown at right is:a 2x − 5y = 1 B 2x − y = 4C 15x + 6y = −30 d 10y − x = −2

e x y

2 51− =

16 To solve the equations 2x + y = 5 and 3x − 6y = 12 graphically on a calculator, you would enter the equations in the function entry line as:a f 1(x) = 2x + y and f 2(x) = 3x − 6yB f 1(x) = 5 and f 2(x) = 12C f 1(x) = 5 − 2x and f 2(x) = 12 + 6yd f 1(x) = 2x + 5 and f 2(x) = 3x + 12e f 1(x) = −2x + 5 and f 2(x) = x

2 − 2

17 Which of the following would be the most effective way to solve the following equations simultaneously?

y = 2x − 13 [1] y = 7x + 2 [2]

a Multiply [1] by 2 and [2] by 13 and add the newly formed equations.B Multiply [1] by 7 and put it equal to [2].C Multiply [2] by 2 and put it equal to [1].d Multiply [1] by 2 and [2] by 7 and subtract the newly formed equations.e Put [1] equal to [2].

18 The gradient of a line perpendicular to a line with a gradient of 7 is:

a 17

− B −7 C 71

−

d 7 e 71

19 The gradient of a line perpendicular to yx7 5

16= +−

is:

a 17

−B

17 C

167

d 516 e

716

−

20 The equation of the line containing (1, −2) and (2, −3) could be expressed as:

a y − 2 = x − 1 B y + 2 = 1 − x C y + 3 = x − 1

d x − 2y = −3 e 3x − 5y = 1 21 A line with equation y − 7 = 5(x − 1) has:

a gradient 5 and contains the point (7, 1) B gradient −7 and contains the point (−1, −7)C gradient 5 and contains the point (1, 7) d gradient −5 and contains the point (1, −7)e gradient −1 and contains the point (5, 7)

22 The distance between (4, 3) and (−2, 1) is equal to:

a (−2 − 4)2 + (1 − 3)2 B (4 3) ( 2 1)2 2− + −− C (4 3 ) ( 2 1 )2 2 2 2+ − −−

d (4 2) (3 1)2 2+ + − e ( 2 4) (1 3)2 2− − −−

y

x

−5

2


23 The midpoint of the segment joining (11, −3) and (−5, 17) is:

a ( 5 , 1 )12

12

− − B (3, 7) C (6, 14)

d ( 2 ,8 )12

12

− e (4, 6)

24 Bote lives 5 kilometres from the nearest post office. At noon one day he begins cycling (from home) at 20 kilometres per hour in a constant direction away from the post office. At t hours after Bote begins cycling, the distance, D km, that he is from the post office is given by:a D = 5t B D = 20t C D = 5t + 20

d D = 20t + 5 e D = 20t − 5

25 The linear function f D R f x x: , ( ) 6 2→ = − has range [−4, 12]. The domain D is:

a [−3, 5] B [−5, 3] C R

d [−14, 18] e [−18, 14]

extended reSponSe

1 The graph at right is a rough sketch of three points on a section of sheet metal that are to be drilled by a programmed robotic drilling arm. Any deviation from a straight path, no matter how slight, means the arm must be programmed for more than one direction. The coordinates marked are correct. Will the robotic arm be able to move in one direction only to drill all three holes?

2 Points A, B and C have the coordinates (1, 6), (0, 0) and (−2, 2). Find the coordinates for a point D so that the four points form a parallelogram.

3 Consider the points (−4, −2), (6, 2), (4, −1) and (0, −7).a Find the coordinates of the midpoints of each side of the quadrilateral formed by the points.b Show that the shape formed by the midpoints is a parallelogram.c Repeat parts a and b for a different set of starting points.d What can you conjecture based on your answers?

4 The cost of a parachuting course consists of a charge of $250, which covers equipment hire and tuition, and a further expense of $55 per jump.a Express the total cost, C, as a function of j, the number of jumps.b How many jumps could a person doing the course afford if she was prepared to spend up to

$1000?

5 A physics student conducts an experiment to find out how much a spring stretches when various weights are hung from it. Her results are shown in the table below.

Length of spring (cm) Force applied (N)

4 0

7 10

12 20

16.5 30

20.5 40

25 50

a What is the natural or ‘unstretched’ length of the spring?b Plot a graph of the student’s results.c Draw a straight line through the points that best describes the data.d Select two points on the line and use them to fi t a linear equation to the line.

y

x

6

8 14 20

12

16

A

B

C


A second student conducts the same experiment on a similar spring. His results are shown below.

Length of spring (cm) Force applied (N)

5 0

10 10

16 20

21 30

24 40

28 50

e On the same set of axes you used in part b, plot the results of the second experiment and join the points with a line of best fi t.

f Write an equation that describes the relationship between the force applied and the length of the second spring.

The gradients of graphs such as the ones you have drawn give an indication of the stiffness of a spring. The greater the gradient, the harder it is to stretch the spring. The lower the gradient, the easier it is to stretch the spring.g Comment on the stiffness of the two springs investigated by the

students.h Is it likely that these two springs will ever be the same length at

a given force; that is, is it likely that the intersection of the two graphs could ever really happen? Explain your answer.

diGital doCdoc-9709Test YourselfChapter 1


ICT activitiesChapter openerdiGital doC

• 10 Quick Questions doc-9695: Warm up with ten quick questions on linear functions (page 1)

1a Solving linear equations and inequationsdiGital doC

• doc-9696: Use trial and error to balance an equation and hence solve for the unknown variable (page 3)

1B rearrangement and substitutiondiGital doC

• Career profile doc-9697: Learn how a vigneron uses substitution and other areas of mathematics in his work (page 8)

1C Gradient of a straight linediGital doCS

• doc-9698: Calculate the gradient between two given points (page 10)• SkillSHEET 1.1 doc-9699: Practise using the gradient to find the value

of a parameter (page 11)

1d Sketching linear functionsdiGital doCS

• WorkSHEET 1.1 doc-9700: Solve linear equations, use substitution and calculate gradients of straight line graphs (page 12)

• doc-9701: Investigate the effect of changing the gradient and y-intercept for a linear graph (page 14)

tUtorial• We 12 eles-1404: Watch a tutorial on how to sketch a linear

graph by hand (page 13)

1e Simultaneous equationsdiGital doC

• doc-9702: Use the graphical method to solve simultaneous linear equations (page 17)

1F Finding the equation of a straight linediGital doCS

• SkillSHEET 1.2 doc-9703: Practise writing positive and negative reciprocals of rational numbers (page 20)

• doc-9704: Investigate the equations of straight lines (page 20)• WorkSHEET 1.2 doc-9705: Sketch linear graphs, determine equations

of linear graphs and application problems (page 21)

1G distance between two points and midpoint of a segmentinteraCtiVitY

• Distance between two points and midpoint of a segment int-0260: Consolidate your understanding of how to calculate the distance between two points and the midpoint of a segment (page 21)

tUtorial• We 19 eles-1405: Watch how to calculate the distance between

two points on a Cartesian plane (page 21)

diGital doCS• doc-9706: Investigate the distance between two points (page 22)• doc-9707: Investigate the midpoint of a segment (page 23)

1h linear modellingdiGital doC

• doc-9708: Investigate simultaneous linear equations (page 25)

tUtorial• We 22 eles-1406: Watch how to apply linear algebra skills to

determine the conditions for which it would be cheapest to hire a catering company (page 25)

Chapter reviewdiGital doC

• Test Yourself doc-9709: Take the end-of-chapter test to test your progress (page 33)

To access eBookPLUS activities, log on to www.jacplus.com.au


Answers CHAPTER 1

linear FUnCtionS exercise 1a Solving linear equations and inequations

1 a 2 b −8 c 7

3

−

d 7 e −1 f 9 g −1 h 10 i −12 2 a 6 b 5 c −13 d 112

9 e 12 f −5

g −9 h 76

59 i 3

j −4 k −9 l −7

3 a ≥x 13

6 b x31

3≤

− c x > 3

d x < −3 e x ≥ 11 f x 3

2≥

exercise 1B rearrangement and substitution

1 a P = A − L b lA

w=

c td

v= d r

C

2π=

e E

2β αθ

θ= −

f rkQq

F=

g vFd mu

m

2 2

= + h

v

rT

2

γ =

i wS lh

l h

2

2( )= −

+ j H

S r

r

2

2

2ππ

= −

2 a 0.267 b 350 c 7 d 13 100 e 2.498

3 a l A, 7.746= b rV3

4, 6.2043

π=

c = −a

v u

t, 4.167 d

π=

l g

T

2, 3.972

2

e cK(1 ), 2622

2

αα

= −

4 a 42 cm

b wP

l wP l

2or

2

2= − = −

c 40 mm

5 a 240 m2

b aA

hb a

A bh

h

2or

2= − = − c 18 cm

6 a $1123.60

b = −

= −

rA

D

A D

D100 1 100

c 41.4%

7 a fuv

u v=

+ b =

−u

f v

v f

c 150 cm

8 b = 2

9 h25

π= cm

exercise 1C Gradient of a straight line 1 a 2 b 1

3

−

2 a 2 b 5 c −4 d 1

2

3 a 1.192 b 3.078 c 0.176 d −0.577 e −0.577 f 0 g 1 h 57.290 4 a 0.93 b 2.61 c −0.53 d −3.73 5 a D b C c A d B 6 a B b E 7 2

17 8 17

300

−

9 a 4 b 31 c −5 d 3 10 a No b Yes c 224 cm

exercise 1d Sketching linear functions

1 a–e

x

y

y = x

y = 2xy = −2x

y = −x

y = 3x

2 The higher the number, the steeper the graph. Positive values make the graph slope up when moving (or tracing) to the right; negative values make the graph slope down when moving to the right.

3 a–d

x

1

−4

y

4

y = x − 4

y = x + 1

y = x + 2

y = x + 3

4 a–d

x0

1

6

−5

−7

y

1−1

y = 2x + 1

y = 2x − 7

y = 3x − 5

y = −3x + 6

5 The number is where the graph cuts the y-axis (hence the name ‘y-intercept’).

6 a y = 2x + 7 b y = −3x + 1 c y = 5x − 2 d y x23

13= +

e y x3412= −

− f y = −2x + 12

7 a 4, 5 b 4, −8 c −3, 1 d −2, 4 e −7, −9 f 2, 5

g ,2

11

− −6 h

−,8

3

2

3

i ,16

5

2 j

− −,5

2

72

8 a y = 4x + 2 b y = 5

2

−x − 5

c y x 24

3= − d y = 5

6

−x + 5

e y = 2x − 1 f y = −5x

9 a y

x−3

18

b y

x

−21

7

c y

x

−3

3 – 2−

d y

x

10

2

e y

x

30

10 — 3

f y

x8

−16

10 a y

x

2

3

b y

x

4

5

c y

x−3

6


d y

x

−7

5

11 a y

x

−6

−7

b y

x

10

−4

c y

x

4

16 — 3−

d y

x2

−6

12 a y

x

(1, −1)

b y

x

(1, 1)

c y

x

(1, −2)

13 D 14 E 15 A exercise 1e Simultaneous equations 1 a

x

1

−1

y

1−1

y = −4x − 6

(−3, 6)

y = −2x

b

x

20

−5

y

5−5

y = 20

y = 3x − 5

253( ), 20

c

x

y

9 ,4

474( )

y = 3x + 5

y = 7x − 4

d

x

5

0

y

(−3, −13)

y = −4 + 3xy = 6x + 5

e

x

5

−5

y

5

y = 10x + 1

y = −6 − 2x712

296( ),− −

f

x

14

yy = 17 − 9x

y = 14 − x

38

1098

( )

179

,

17

2 a (1, 4) b (−4, −15) c (−7, −5) d (5

9, 179

)

e (23

14,

20

7

−) f (13, −3)

3 a (7, 9) b (6, 5)

c (1, −2) d (1

2,19

2)

e (59

8,21

8) f (

84

67,99

67)

4 15 cents and 35 cents 5 22 and 19 6 16 emus, 41 sheep 7 Basketballs $9.45, cricket balls $3.05 8 Limousine $225 (sedan $75) 9 A 10 D

exercise 1F Finding the equation of a straight line 1 a 2,

1

2

−, −1 b 3,

1

3

−, −1

c 1

4, −4, −1 d

2

5,5

2

−, −1

e 9

7

−,7

9 , −1

2 They are perpendicular. 3 They are perpendicular.

4 a 1

4

− b

1

9 c −7

d 9

8 e

2

7

− f −1

5 a 1

5 b −1 c

3

2

−

d 6

7 e

1

2 f 4

3

−

6 a i 3x − y − 1 = 0 ii y = 3x − 1 b i 5x − y − 19 = 0 ii y = 5x − 19 c i x + y − 1 = 0 ii y = −x + 1

d i x − 2y + 7 = 0 ii y x1272= +

7 a i x − 2y − 1 = 0 ii y x1212= −

b i x − y = 0 ii y = x

c i x + 2y − 12 = 0 ii y = 1

2

− x + 6

d i 3x + 2y − 2 = 0 ii y = 3

2

− x + 1

8 y = x − 6 9 y = 3x − 23

10 a y = 1

3

−x + 5 b y x25

435= −

c y = 1

2

−x + 3 d y x76

143= +

−

e y = 2x − 3 f y x3492= −

−

11 y x5232= −

−

12 a y x67827= + b y x 73

2= +−

c y = −x + 8 d = +−y x8

394

13 94 14 Yes 15 H = 22 + 6t

exercise 1G distance between two points and midpoint of a segment 1 a 5 b 10 c 1 d 2 2 a 7.211 b 14.765 c 13.038

3 a a a4 82 − + b b b12 612 − +

c c c2( 6 10)2 − + d d d5 22 262 − +

4 21.024 km 5 a 216 b 108 6 a (2, 4) b (5, 1) c (7, 2) d (8, 9)

7 a (1 , 5 )1

2

1

2 b ( 3, 3 )12

− −

c ( 9 , 9 )1

2

1

2− d ( 1 , 6 )1

2

1

2−

8 a (4a, 3a) b (8, 3c) c f g3

2, 2

+

9 a 5 b 9 10 D1 (−1.5, 4.25), D2 (−1.5, 1.5), D3 (8, −7.5) 11 y = 5x − 7 12 y = 3x + 6


exercise 1h linear modelling 1 a C = 25 + 5t

b

1 2

35

2530

Cost ($)

Time (h)

c $40 2 a C = 60 + 8m

b

1 2

76

6068

Cost ($)

Time (min)

c $100 3 a P = 32 + 0.1n b

10 20

34

3233

Payment ($)

Numberof lea�ets

c $197

4 $960 5 Yes ($410 compared to $450) 6 After 4 rides 7 a Opus $24, Elstra $20 b After 14 minutes 8 a PinkCabs $28.50, NoTop $26 b After 6.7 km (6

2

3 km)

9 6 visits 10 Savus would be cheaper for up to 9 days

hire.

Chapter reVieWShort anSWer

1 −2 2 6 3 x ≥ −14

4 TR

GMR

R

GM

4or 2

2 3π π=

5 a 15 b 3.873

6 a 3

4 b

7

11

−

c 5

11 d

7

8

−

7 7

3

− 8 4.331

9 Undefi ned

10 a 3, −7 b 5

3

−, 10

c 1

2, −2

11 y x 32

5= −

12 a y

x

24

8

b y

x−40

5

c y

x

−9

7

d y

x

(1, −6)

13 a

x

5

−5

y

−5

f (x) = 3x + 10

f (x) = −2x − 15

(−5, −5)

b (−5, −5)

14 ( , 5)5

3− 15 ( , )63

10335

− −

16 21 two-dollar and 46 one-dollar coins 17 −3 18 y = −x + 4

19 y x67347= +

−

20 y = −4x − 26 21 6

22 Teacher to check. 23 k = 4, h = −3 24 a C = 75 + 65t b

1 2

205

75

140

C ($)

t (h)

c $302.50

mUltiple ChoiCe

1 D 2 A 3 C 4 D 5 D 6 A 7 C 8 D 9 B 10 B 11 B 12 A 13 D 14 B 15 E 16 E 17 E 18 A 19 C 20 B 21 C 22 D 23 B 24 D 25 A

extended reSponSe

1 No, the points are not collinear. This may be shown by calculating gradients or equations for lines joining different pairs of points.

2 (−1, 8) 3 a Midpoints: (1, 0), (5,

1

2), (2, −4), (−2,

9

2

−)

d The midpoints of any quadrilateral form a parallelogram.

4 a C = 250 + 55j b 13 jumps 5 a 4 cm b, c, e

60

50

40

30

20

10

Length of spring (cm)

Forc

e ap

plie

d to

spr

ing

(N)

Student 1Student 2

5 10 15 20 25 300

d Answers will vary. One possible answer

is y x30

13

95

13= − .

f Answers will vary. One possible answer is y = 2x − 10.

g The fi rst spring is stiffer than the second, as the gradient of its graph is greater than that of the second spring.

h The graphs intersect at the point (−8.75, −27.5). It is not possible for the springs to have a negative length, so this point is not achievable.

Year 11 Methods Chapter 1

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Transcript of Year 11 Methods Chapter 1