Solve the following linear equations. a 7x 4 = 17 bc THINK WRITE a Write the equation.a7x417

(1)

1

VCE

co

covverage

erage

Areas of study

Unit 1 • Functions and graphs

• Algebra

In this

cha

chapter

pter

1A Solving linear equations

1B Rearrangement and

substitution

1C Gradient of a straight line

1D Equations of the form

y

=

mx

+

c

1E Sketching linear graphs

using intercepts

1F Simultaneous equations

1G Perpendicular lines

1H Formula for finding the

equation of a straight line

1I Distance between two

points

(2)

Solving linear equations

Throughout the Maths Methods course, the skill of being able to rearrange or solve equations is often called upon.

A linear equation is one that involves a pronumeral 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 , x2, x3, x−1, x−2 and so on.

For example, y= 7x− 3, ax+by=c and x+ 1 = 9 are all linear equations. Note that y= is not a linear equation, as the is really x−1. To isolate a particular pronumeral — known as ‘making x (or whatever the pronumeral is) the subject’, we focus on the pronumeral by ‘undoing’ other terms and operations. In doing so, remember to do the same 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, in this section some other types of equation will be included for skills practice.

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

x 1 2

---8 3

---1

x

--- 1---_x

Solve the following linear equations.

a 7x− 4 = 17 b c

THINK WRITE

a Write the equation. a 7x − 4= 17

Add 4 to both sides. 7x = 21

Divide both sides by 7. x =

x= 3

b Write the equation. _b + 5 = 1

Subtract 5 from both sides. _{= −4}

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

Add 2 to both sides. 3x = −14

c Write the equation. _c ₌ 6

Divide both sides by 2. = 3

Add 1 to both sides. ₌ 4

Multiply both sides by 5. 4x = 20

x = 5 3x–2

4

---+5 = 1 2 4x 5 ---–1

 

  ₌ ₆

1 2

3 21

7

---1 3x–2

4

---2 3x–2

4

---3 4

5 –14

3

---1 2 4x

5 ---–1

 

 

2 4x

5 ---–1

3 4x

5

---4

5 20

4

---1

(3)

C h a p t e r 1 L i n e a r f u n c t i o n s

3

When an equation has pronumerals (for example, x’s) on both sides, at some stage they must be gathered together on the same side of the equation.

Solve:

a 4x− 3 = 3(6 −x) b c

THINK WRITE

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

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

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

7x − 3 = 18

Add 3 to both sides. 7x = 21

x = 3

b Write the equation. _b ₌

Find the lowest common denominator for all three terms. Here, we use 6.

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

=

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

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

Expand all brackets. 27x + 9 = 26x + 14

Collect x’s on the LHS and numbers on the RHS. 27x − 26x = 14 − 9

Simplify and solve. x = 5

c Write the equation. c =

Find the lowest common denominator for all three terms. Here, we use 20.

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

=

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

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

Collect x’s on the RHS and numbers on the LHS. 15 − 126 − 20 = 18x + 5x

Simplify and solve.

−131 = 23x − =

= 9x+3

2

--- 13x+7 3

---= 3–x

4

--- 9(x+7) 10 ---+1 = 1 2 3 4 5 21 7

---1 9x+3

2

--- 13x+7 3

---2

3 3 9x( +3)

6

--- 2 13x( +7) 6 ---4 5 6 7

1 3–x

4

--- 9(x+7) 10 ---+1

2

3 5 3( –x)

20

--- 2×9(x+7) 20 --- 20 20 ---+ 4 5 6 7 131 23 --- x x 131 23 ---–

2

(4)

Using a graphics calculator

Solving linear equations using solver

Press , then continue to press until

0:Solver is highlighted. Select it by pressing .

Press to move to the equation line.

Enter the equation to be solved in the form 0 = . . .; in this case:

Press , the green key and [SOLVE].

Solving linear equations graphically

1. Press and enter the equation. 2. Press . Select 6:ZStandard to

obtain a graph (of ‘standard’ dimensions).

3. If you cannot see where the graph

cuts the x-axis, press ,

select 3:Zoom Out and press .

5. Press to return to the

graph, then [CALC] 2:zero.

MATH _▼

ENTER ▼

2(x–6) 7 ---+8

▼ ALPHA

Y= ZOOM

ZOOM

ENTER

GRAPH 2nd

4. The axis divisions may not be suitable or may be cluttering the window. If so, change them

by pressing and selecting new

values for Xscl and Yscl.

WINDOW

6. To calculate where the line crosses the

x-axis: (a) move the cursor to the left of where the graph cuts

the axis and press ,

(b) move the cursor to the right of where the

graph cuts the axis and press ,

(c) move the cursor closer to the crossing point

and press again.

ENTER

ENTER

(5)

5

Solving linear equations

1 Solve the following linear equations.

2 Solve: 3 Solve: 4 Solve: 5 Solve: 6 Solve: 7 Solve:

a 3x− 19 = −13 b −6x+ 11 = 59 c 8x− 23 = −47

d 15 − 2x= 1 e 4x+ 25 = −7 f 63 − 7x= 21

g −9x+ 21 = 3 h 45x+ 72 = −423 i 9x+ 19 = −2

j 15 − 6x= 2

a b c

d e f

g h

a b c

d e f

g h

a b c

d e f

a 2x− 9 = 3(2x− 11) b 5x+ 6 = 2(3x+ 4) c 7x− 1 = 17(3x− 13) d 5x+ 9 = −4(x+ 9) e x+ 11 = 2(x+ 12) f 5 − 2x= 3(3 −x) g 3x− 7 = 2(35 − 2x) h 16 − 4x= 7(1 −x)

a b c

d e f

a b c

d e f

remember

Linear equations can be solved by rearranging to make the pronumeral the subject.

remember

1A

WORKED Example

1a

EXCEL Spreadshe_e_t Equation

solver

GC pr_ogr

am Equation

solver 3x–1

4

--- = 5 4x+2

11

--- = –2 2x+8

3 --- = 6 5x+20

7

--- = –5 12–3x

3

--- = 5 10–x

4

--- = –2 11–3x

7

--- = 6 6x+13

2

--- = –4

W WORKEDORKED E Examplexample

1b

Ma_th_c a_d

Equation solver

4x–6 3

---–7 = 3 2x–7

5

---+11 = 8 12–3x

3

---–5 = 6 12–9x

2

---+7 = –5 7x+4 3

---–8 = –9 1–x

2

---+17 = 20

x+16 5

---–4 = 0 19–x

4 –

---+3 = –1

1c ₂ 3x

5 ---–1

 

  ₌ ₁₀ ₃ 5x

6 ---+4

 

  ₌ ₂₇ ₄ 2x

3 ---+5

 

 

– = 4

7 8 x

4 ---–

 

  ₌ ₇₇ _{5 8} 2x

7 ---–

 

 

– = –20 6 20x

9 ---+15

 

  ₌ _–₁₅₀

WORKED Example 2a W WORKEDORKED E Examplexample

2b x+2

6

--- x–5 3

---= x+11

3

--- 2(x+14) 9

---= x–1

12

--- 3x–17 8

---=

2x–4 5

--- 11–x 2

---= 4x+66

3

--- 13–3x 4

---= x+10

9

--- 2 7( –3x) 5

---=

WORKED Example

2c 6x+7

5

--- 5x+1 4 ---+1

= 2x+29

3

--- x+44 8

---= +2 7x–9

2

--- 5x+3 4 ---+6

=

9x+28 5

--- 7x–6 2 ---–5

= 7x–9

9

--- 21–x 3 ---–18

= 17–x

2

--- 49+2x 5 ---+5

(6)

Rearrangement and substitution

When there is more than one pronumeral involved in an equation, we may rearrange to make a particular pronumeral the subject using the same rules of equation solving described in the previous section.

Rearrange each of the following to make the pronumeral in red the subject.

a 6x+ 8y− 48 = 0 y b s=ut+ at2 u c T= k

THINK WRITE

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

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

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

Divide both sides by 8. y =

Cancel if possible. Here, divide the top and bottom of the fraction by 2.

y =

Other ways of representing the answer are shown

opposite. or y = − x

= 6 − x

=− x + 6

b Write the equation. b s = ut + at2

Subtract at2 from both sides. s − at2 = ut

Multiply both sides by 2. 2s − at2 = 2ut

Divide both sides by 2t. ₌_u

Write the equation with the desired pronumeral on the left.

u =

c Write the equation. c T =

Divide both sides by 2π. ₌

Square both sides. =

Turn both sides upside down.

Note: This can be done only if both sides are fractions. =

Multiply both sides by m. = k

Write the equation with the desired pronumeral on the left. k =

1 2 --- m k ----2π 1 2 3

4 48–6x

8

---5 24–3x

4 ---6 24 4 --- 3 4 ---3 4 ---3 4 ---1 1 2

---2 1₂--- 1

2

---3

4 2s–at2

2t

---5 2s at

2 – 2t ---1 m k ----2π 2 T 2π --- m k ----3 T 2π ---   2 m

k

----4 2π

T ---   2 k

m

----5 m 2π

T ---   2

6 m 2π

T ---   2

3

(7)

7

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

a The formula for converting temperature in degrees Fahrenheit (F) (which is the system used in the USA) to degrees Celsius (C) is C= .

i Make F the subject.

ii What is the temperature in Fahrenheit when the temperature measured in degrees Celsius is 21°C?

b The area (A) of a circle is given by A=πr2, where r is the radius. Find the value of r

when A= 20 cm2.

THINK WRITE

a i Write the equation. a i C =

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

Divide both sides by 5. ₌

Add 32 to both sides. = F

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

F =

or F =

a ii Replace C with 21. Note that 9C

means 9 ×C.

a ii F =

Evaluate F. F = + 32

F = 37.8 + 32 F = 69.8 so 21°C = 69.8°F

b Write the equation. b A =πr2

Solve for r (make r the subject) as follows.

Divide both sides by π. =

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

r =

Substitute A = 20 into the new formula. If A = 20,

r =

Evaluate r. = 2.523 cm

5(F–32) 9

---1 5(F–32)

9

---2

3 9C

5

--- F–32

4 9C

5 ---+32

5 9C

5 ---+32 9C+160

5

---1 9×21

5 ---+32

2

189 5

---1 2

3 ---A_π r2

4 A

π

---5

20

π

---6

4

(8)

Computer algebra systems

Software applications such as Mathcad,

DERIVETM_{, MathView and the TI–89 and TI–92}

graphics calculators are able to isolate a pronumeral in an equation. The screen at right is from a TI–92 graphics calculator.

Notice that the second ‘argument’ of the solve function is k, the pronumeral being solved for.

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

The final velocity v m/s of an object that begins with velocity u m/s and accelerates at

a m/s2 over a distance of s m is given by the equation . a Find the value of v when u= 16, a= 2 and s= 60.

b Rearrange the given equation to make s the subject.

c Find the distance travelled by an object which begins with a velocity of 10 m/s, and has a final velocity of 4 m/s while accelerating at −1 m/s2.

THINK WRITE

a Write the given equation. a

Substitute u= 15, a= 2 and s = 60. Simplify and evaluate v.

Final velocity is ±4 m/s.

b Begin with the given equation. _b _v₌

Square both sides. v2= u2− 2as

Subtract u2 from both sides. v2− u2 = −2as

Divide both sides by −2a. ₌_s

Reverse so that s is given on the left. _s₌

c Match the pronumerals with the given

information.

c u = 10 v = 4 a = −1

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

s =

Substitute the values given in step 1. ₌

Simplify and evaluate. =

= = −42

Explain the answer in words. The object travels 42 m in the opposite

direction to its initial direction.

v = u2_–₂_as

1 v = u2–2as

2 ₌ 162_–₂_×₂_×₆₀

3 = 256–240

16 =

4 ± =

1 u2_–_2as

2 3

4 v2–u2

2a –

---5 v2–u2

2a –

---1

2 v2–u2

2a –

---3 42–102

2×–1 –

---4 16–100

2

---84 –

2

---5

5

(9)

9

Rearrangement and

substitution

1 Each of the following is a real equation used in business, mathematics, physics or another area of science. Make the pronumeral shown in red the subject in each case.

2 Find the value of the subject (the first mentioned pronumeral), given the values of the other pronumerals.

a W=Fd F= 10, d= 5.6

b P= 2l+ 2w l= 6, w= 9

c A= bh b= 10, h= 16

d s= 12, a= 4, b= 9, c= 11

e R₁= 50, R₂= 100

f k= 60, d= 15

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

h n= 3, λ= 2.8

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

j ω= 2, r= 1.6, y= 1

a A=L+P P b A=lw l

c t d C= 2πr r

e R f V= πr3h r

g E=I₄R₄−I₁R₁+I₂R₂−I₃R₃ I₃ h R₁=R₂(1 +αθ) α

i E=αθ+βθ2 β j r

k φ l V₁N₂=V₂N₁ V₂

m pV=nRT n n s=ut+ at2 a

o Fd= mv2− mu2 v p r

q U _r γ

s S= 2w(l+h) + 2lh w _t S= 2πr2+ 2πrH H

1B

remember

Equations may be rearranged by applying the same rules as those used to solve equations.

remember

Ma_th cad Rearranging

equations

W WORKEDORKED E Examplexample

3

v d

t

---=

A PRT

100

---= 1

3

---F kQq

r2

---=

E nφ t

---=

1 2

---1 2

--- 1 2

--- _F µI

2

2πr

---=

f₂ V f1 V+U

---= v = γrT

4

Ma_th cad Substitution

1 2

---A = s s( –a)(s–b)(s–c) R_T R1R2

R₁+R₂

---=

I k

d2

---=

D n 1

2

---–

( )λ

=

(10)

3 Make the pronumeral in red the subject, and find its value using the given information.

4 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 net force, F, measured in Newton (N) acting on a mass m kilograms (kg) is found using the equation F=ma, where a is the acceleration of the mass measured in m/s2.

a Find the net force required to accelerate a 2.5 kg rock at the rate of 4 m/s2. b Make a the subject of the equation.

c Find the acceleration produced by a 700 N force acting on a 65 kg person.

6 The area of a trapezium (figure A) is given by

, 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, find an equation for determining side a in terms of the area A and side b.

c Find a in figure C.

a A= l2 l A= 60

b V= πr3 r V= 1000

c F=mg−kv2 v F= 250, m= 60, g= 10, k= 0.1

d v=u+at a v= 25, u= 0, t= 6

e S=πr(r+h) h S= 120, r= 2, π= 3.14

f T= 2π l T= 4, g= 9.8, π= 3.14

g d f= 2, l= 15

h H=U+PV V H= 26, U= 4.5, P= 2

i c K= 6.9, α= 0.05

j u H_i= 34, H₀= 4, v= 40

W WORKEDORKED E Examplexample

4

4 3

---l g

---f l 2_–_d2

4l

---=

K cα 2

l–α

---=

H_i H₀

--- v

u

---=

5

A a+b

2

--- 

 _h =

a

h

b

Area

A

Figure A

9 m 16 m 21 m

Figure B

62 cm

50 cm

a

Area = 2000 cm2

(11)

11

7 The size of a 2-year investment account with a particular bank is given by

, 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.

8 The length of a circular curve is given by the formula

where L= length of curve, π= 3.14,

r= radius of curve and θ is the angle of the curve.

a Determine, to the nearest

degree, the angle required for a curve of radius 15 m and length 40 m.

b Find the length of the curved

edge of the fan at right.

9 The length, L, of metal strip

required to make the bracket at right is given by .

a What length is required to make a bracket for which

a= 25 cm, b= 35 cm and w= 5 mm?

b Find the value of w (in cm) necessary to make the bracket at right from a 60 cm length of metal.

10 The object and image positions for a lens of focal length f are related by the formula , 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?

A D 1 r

100

---+

 

 2

=

L

r

θ

L πrθ

180

---=

123

ο

8 cm

a

b

w

L a b w

2

----+ ----+ =

40 cm

17 cm

1

u

--- 1

v

---+ 1

f

(12)

Qualifications: Nil (No formal qualifications)

Employer: Self-employed

Company website: http://home.netc.net.au/~rick/

I grew up on the family vineyard and was encouraged to develop an interest in wine-making at a very early age. Thus, I was familiar with the workings of a vineyard from nearly day one. A usual day for me is to crush the previous day’s grape harvest, add yeast to establish the correct fermentation and then I am out to the vineyard for picking the next block of grapevines. Part of my job is to look after customers by serving tastings of the wines. I then fill, cork and label more wine for the sales room. The wine fermentations are checked and any wine that has completed its fermentation needs the skins to be pressed off.

Mathematics is very important in winemaking as it is essential that quantities are calculated accurately. For example, to calculate the amount of pure alcohol to add to a wine to fortify it to 18.5% by volume, the Pearson Square formula is used.

whereX = litres of fortifying spirit required

V = litres of wine to be fortified

C = final alcohol strength of the wine in % by volume

A = alcohol strength of the wine before fortification

B = alcohol strength of fortifying spirit

If we have 5000 litres of wine at 10% alcohol by volume and our fortifying spirit is 95% alcohol by volume then the litres of spirit that are required is:

= 555.5.

That is, 555.5 litres of spirit is required. It is necessary to know the exact capacity of the mixing tank where the fortification is carried out. If the tank is cylinder shaped, then the formula is a straightforward volume of a cylinder calculation. Then, once the number of cubic centimetres is known, it can be converted into litres. A wooden dipstick is constructed with either centimetre marks and the number of litres per centimetre calculated or marked in actual litres for that particular tank. The exact amount of spirit to add to a wine can be calculated using maths. This can save a lot of money and reduce wastage.

Questions

1. What are the two ways in which a wooden dipstick can be marked?

2. Rick uses skills in substituting into an equation to calculate the amount of pure alcohol needed to fortify a wine. What other areas of mathematics are useful to him? 3. Although Rick has no formal qualifications,

are there any courses for vignerons? If so, what are they? Where else could you gain skills for working at a vineyard?

Testing the sugar content of the fermenting wine with an hydrometer

to measure the specific gravity uses a French graduation called

degrees Baum/e. 1 baum/e = 1.008 SG

X V C – A( ) B–C

---=

5000 18.5( –10) 95–18.5

---= 42500 76.5

---R I C K M O ---R ---R I S — V i g n e r o n

(13)

13

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

Gradient of a straight line

The gradient of a line describes its slope or steepness. You may recall from previous studies the following types of gradient for straight lines.

The gradient may be calculated using the formula

gradient or . These terms are

illustrated at left. Greater

+ gradient

Positive gradients

y

x

Negative gradient

y

x

Zero gradient

y

x

Infinite gradient

y

x

y

x

(x₁, y₁) (x₂, y₂)

Run Rise

m rise

run

---= m y2–y1 x₂–x₁

---=

Here are two examples of where gradient can affect our everyday lives. Can you think of others?

(14)

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.

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

THINK WRITE

Identify the rise and run. rise = 14, run = 2 Calculate .

y

x

–2 14

1

2 m rise

run

---= m 14

2 ---=

7 =

6

WORKED

E

xample

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

THINK WRITE

Use the formula .

Match up the terms in the formula with the values given.

(x₁, y₁) (x₂, y₂)

(3, −6) (−1, 8)

Substitute the given values.

Simplify.

Cancel if possible.

1 m y2–y1

x₂–x₁

---= m y2–y1

x₂–x₁ ---=

2

3 m 8––6

−1–3 ---=

4 14

4 – ---=

5 –7

2 ---=

7

WORKED

E

xample

y

x

rise run

tan = rise = m

run

θ

(15)

15

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

b Find the gradient of the line shown. Express your answer to 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.

b The angle given is not the one between

the graph and the positive direction of the

x-axis. Calculate the required angle θ.

a m = tanθ = tan 40° = 0.839

b θ= 180°− 60° = 120°

Use m= tanθ to calculate m to 2 decimal places.

m = tanθ = tan 120° = −1.73

y

x

60°

1 y

x

60° θ

2

8

WORKED

E

xample

remember

The gradient (m) of a straight line may be calculated using the following formulas:

.

where (x₁, y₁) and (x₂, y₂) are points on the line.

m = tanθ where θ is the angle the line makes with the positive direction of the

x-axis.

m rise

run

---=

y

x

Rise Run

m y2–y1 x₂–x₁

---=

(16)

Gradient of a straight line

1 Calculate the gradient of each of the following linear graphs using the intercepts shown.

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

3 Calculate the gradient of the line joining each pair of points.

a b c

d e f

g h

a (2, 4) and (10, 20) b (4, 4) and (6, 14)

c (10, 4) and (3, 32) d (5, 31) and (−7, 25)

e (7, 2) and (12, −28) f (−3, 2) and (42, 17)

g (0, −30) and (−8, −31) h (−11, −25) and (0, −3)

i (217, 4) and (19, 4) j (3, −3) and (−45, 21)

k (1, 32) and (67, −100) l (−2, −5) and (0, 0).

a b c

1C

W WORKEDORKED E Examplexample

6

–3

y

x

5

–1

y

x

1 –4

y

x

2

6

y

x

12 3

y

x

7 7

y

x

32

–10

y

x

100

45

y

x

7

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x y

(1, 3) (5, 5)

x y

(1, 6)

(6,1)

x y

(3, –2)

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17

4 Find the gradient of the line joining each pair of points.

d e f

g h

a b c

d e f

g h x y (–2, –2) (–7, –10) x y (2, 2) (–2, –2) x y (–3, 8) (–2, –4) x y (4, 3) (–5, 3) x y (6, 9) (6, –3) W WORKEDORKED E Examplexample

7 5 4 3 2 1 –1 –2 –3 –4 –5

–5 –4 –3 –2 –10 1 2 3 4 5 x y 5 4 3 2 1 –1 –2 –3 –4 –5

–5 –4 –3 –2 –10 1 2 3 4 5 x y 6 5 4 3 2 1 –1 –2 –3 –4 –5

–5 –4 –3 –2 –10 1 2 3 4 5 x y 5 4 3 2 1 –1 –2 –3 –4 –5

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5 Find the gradient of each line below.

6 Find the gradient (accurate to 3 decimal places) of a line making the angle given with the positive x-axis.

7 Find the gradient of each line below. Give answers to 2 decimal places.

8 Which of these lines has: a a non-zero positive gradient?

b a negative gradient?

c a zero gradient?

d an undefined gradient?

a b c

d e f

g h

a 50° b 72° c 10° d −30°

e 150° f 0° g 45° h 89°

a b c d

5 4 3 2 1 –1 –2 –3 –4 –5

–5 –4 –3 –2 –10 1 2 3 4 5 x y 5 4 3 2 1 –1 –2 –3 –4 –5

–5 –4 –3 –2 –10 1 2 3 4 5 x y 6 5 4 3 2 1 –1 –2 –3 –4 –5

–5 –4 –3 –2 –10 1 2 3 4 5 x y 6 5 4 3 2 1 –1 –2 –3 –4 –5 –6 –5

–6 –4 –3 –2 –10 1 2 3 4 5 6x y 5 4 3 2 1 –1 –2 –3 –4 –5

–5 –4 –3 –2 –10 1 2 3 4 5 x y 5 4 3 2 1 –1 –2 –3 –4 –5

–5 –4 –3 –2 –10 1 2 3 4 5 x y 6 5 4 3 2 1 –1 –2 –3 –4 –5 –6 –5

–6 –4 –3 –2 –10 1 2 3 4 5 6x y 6 5 4 3 2 1 –1 –2 –3 –4 –5 –6 –5

–6 –4 –3 –2 –10 1 2 3 4 5 6x y

W WORKEDORKED E Examplexample

8a

8b y x 43° y x 69° y x 28° y x 15° 5 4 3 2 1 –1 –2 –3 –4 –5

–5 –4 –3 –2 –10 1 2 3 4 5 x y

A

C

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19

9

10 Arrange the following in order from smallest to largest gradient. The same scale has been used to draw each graph.

11 Sketch two different graphs that have:

a the same gradient

b zero gradients

c different negative gradients.

12 Burghar plots the coordinates of a proposed driveway on a plan which is shown

below. What is the gradient of the proposed driveway? a Which of the following graphs has a

gradient of −2?

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

c Which of the following lines has a gradient of ?

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

a b c d e

m

multiple choiceultiple choice

5 4 3 2 1 –1 –2 –3 –4 –5

–5 –4 –3 –2 –10 1 2 3 4 5 x y A C E B _D 5 4 3 2 1 –1 –2 –3 –4 –5

–5 –4 –3 –2 –10 1 2 3 4 5 x y A C E B D 1 2 ---5 4 3 2 1 –1 –2 –3 –4 –5

–5 –4 –3 –2 –10 1 2 3 4 5 x y A C E B D 2 5 ---5 4 3 2 1 –1 –2 –3 –4 –5

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

14 A passenger jet takes off along the flight path shown below. What is the gradient of the path?

15 Find 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: 2

b (2, 1) and (8, a) gradient: 5

c (0, 4) and (a, −11) gradient: 3

d (a, 5) and (5, 1) gradient: −2

16 For safety considerations, wheelchair ramps are constructed under regulated specifi-cations. One regulation requires that the maximum gradient of a ramp exceeding 1200 mm in length is to be .

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

requirements?

b Does a ramp with gradient meet

the specifications?

c A 16 cm high ramp needs to

be built. Find the hori-zontal length of the ramp required to meet the specifications.

0.85 m

15 m

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150 m 110 m

500 m Runway

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1.1

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---C h a p t e r 1 L i n e a r f u n c t i o n s

21

Equations of the form y

=

mx

+

c

A common form for linear equations is y=mx+c, sometimes referred to as ‘y=’ form.

The following exercise aims to demonstrate the effect of altering m and c. A graphics calculator would be useful, but is not essential.

Equations of the form

y

=

mx

+

c

1 Use a graphics calculator or other method to sketch graphs of the following on the same set of axes.

2 What is the effect on the graph of the number in front of the x (the ‘x-coefficient’ or ‘gradient’)?

5 What is the effect on the graph of the number at the end of the equation (the

‘y-intercept’)?

6 State the gradient for each of the following equations.

7 State the y-intercept for each of the equations in the previous question.

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

a y=x+ 1 b y=x+ 2 c y=x+ 3 d y=x− 4

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

a y= 5x+ 7 b y= 6x− 4 c y= −9x+ 1 d y= 2x− 13

e y= −8x− 5 f y=x+ 2 g y= −x− 10 h y= 5x+ 0

i y= 3x j y= 0x+ 17 k y= 2 l y= 0

y = mx + c

y

x

y

x

Gradient

Gradient m

y-intercept

x-intercept

These lines have identical gradients (equal m values)

remember

1. The general equation for a straight line of gradient m and y-intercept c is

y=mx+c. 2. Lines with the same gradient (m) are parallel.

remember

1D

Ma_th cad Linear graphs

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8 Write the equation of a line having the following properties (where m= gradient and

c=y-intercept).

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

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

11

Which of the following lines is parallel to (that is, has the same gradient as)

y= −4x− 7?

A 5 − 4y= 13 B x= −4y− 7 C 4x+y+ 7 = 0 D y= 4x− 7 E y= 5x− 8

12

Which of the following lines has the same y-intercept as y= 18x− 2?

A y= 2 − 18x B y− 18x− 2 = 0 C 3x+ 7y= −2

D 14 − 7y− 2x= 0 E 2x= −14 − 7y

13 Write three equations starting with y=, that each have a gradient of −7. 14 Write three equations starting with y=, that each have a y-intercept of −6.

15 Write three equations that have the same gradient as 3y+ 5x= 17. Write them in the same form as the equation in this question.

16 Write three equations that have the same y-intercept as 3y + 5x= 17. Write them in the same form as the equation in this question.

17 State the equation for each of the following graphs.

a b

a m= 2, c= 7 b m=−3, c= 1 c m= 5, c=−2

d m= 0, c= 3 e m= 1, c= 0 f m= , c=−5

g m= , c= h m=− , c=− i y-intercept 12, gradient −2

j y-intercept −3, gradient

a y= 9 + 3x b y= −42 + 7x c y= 12 − 4x

d y= −35 − 5x e y− 3x= 10 f y+ 6x= 24

g y+ 16x= −15 h y+ 9x+ 1 = 0 i y−x+ 23 = 0

j y+ 18 − 4x= 0

a 2y= 8x+ 10 b 3y= 12x− 24 c −5y= −20x+ 30

d −y= 3x− 1 e 16 − 4y= 8x f 22 − 2y= −6x

g 21x+ 3y= −27 h −10x+ 5y= 25 i 6y+ 3x= −18

j −11y− 2x= 66 k 8x+ 3y− 2 = 0 l −3x− 4y+ 13 = 0

m 15 − 6y+x= 0 n 2y+ 7 + 5x= 0

1 2 ---2

3 --- 1

3

--- 3

4

--- 1 2 ---5

2

---m

multiple choiceultiple choice

m

multiple choiceultiple choice

G

Cpr ogram

Guess the equation

6 5 4 3 2 1 –1 –2

–1 0 1 2 x

y

– 1 – 2

3 2 1 –1 –2 –3 –4 –5

–2 0 2 x

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23

c d

e f

18 a A set of axes is placed as shown

over the Leaning Tower of Pisa. Find the equation of the line representing the sloped left wall of the Tower.

b What angle does the Tower make

with the ground?

19 State the gradient and y-intercept (in that order) for the following.

a y=ax+b

b ax+by=c

c ax+by+c= 0

d 2y= 4kx− 6h

3 2 1 –1 –2 –3 –4 –5

–2 –1 0 1 2 x

y

5 4 3 2 1 –1 –2 –3 –4 –5

–1 0 1 2 3 4 5 6 x y

5 4 3 2 1 –1 –2 –3 –4 –5

–1 0 1 2 3 4 x y

9 8 7 6 5 4 3 2 1 –1 –2 –3 –4 –5 –6 –7 –8 –9

–3 –2 –1 0 1 2 3x y

y

5.2 m

55.8 m

x Wor

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Sketching linear graphs using intercepts

To 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. So, rather than construct a table of values and plot several coordinates, it is quicker to mark only the points where a line cuts (or intersects) an axis. These points are called x- and y-intercepts. The x-intercept occurs when y= 0, while the y-intercept occurs when x= 0. This information is the basis for the approach to sketching illustrated in the following example.

y

x

Sketch the graph of y= −6x+ 15, showing x- and y-intercepts.

THINK WRITE

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

If x = 0 y = −6 × 0 + 15 y = 15

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

If y =0 0 = −6x + 15 6x = 15

x = x = or 1

Mark the intercepts on a set of axes.

Note that = (a little over 1 ).

Join the intercepts with a straight line.

1

2

15 6

---5 3 --- 2

3

---3

5 3

--- 1.66˙ 1

2

---y

x

5 – 3

15

4

9

WORKED

E

xample

Sketch the graph of 3x− 2y= 12.

THINK WRITE

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

If x =0 3 × 0 − 2y = 12 −2y = 12

y = y = −6

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

If y =0 3x − 2 × 0 = 12 3x = 12 x = 4

1

12 2 –

---2

10

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25

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=kx or ax+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.)

THINK WRITE

Mark the intercepts on a set of axes. Join the intercepts with a straight line.

3 y

x

–6 4

4

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

THINK WRITE

Try substituting x= 0 to find the

y-intercept.

If x =0 4 × 0 − 3y = 0 −3y = 0 y = 0

Note that the graph passes through (0, 0). There is no point substituting

y= 0, as we know we’ll get x= 0. Substitute another x-value. In this example we use x= 1.

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

Plot the points (0, 0) and (1, ) on a set of axes. Note that is 1 , which is a little less than 1 .

1

2

3

4 3

---4 4

3 ---4

3 --- 1

3 ---1

2

---y

x

4 – 3

(1, )

11

WORKED

E

xample

remember

To sketch a linear graph:

1. Let x= 0 and find the y-intercept. 2. Let y= 0 and find the x-intercept.

3. If (0, 0) is an intercept, find another point on the line by substituting x= 1 (or any other convenient non-zero value).

4. Mark and join the intercepts with a straight line.

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Sketching linear graphs using

intercepts

1 Sketch graphs of the following linear equations, showing x- and y-intercepts.

2 Sketch graphs for each of the following.

3 Sketch the graph for each equation.

4 Sketch the graph for each equation.

5

Which of the following is in the form ax + by = c? (One or more answers.)

6

The x- and y-intercepts for the equation 2y= −3x+ 12 are (respectively):

7

Which of the following has a y-intercept of −3?

To find x- and y-intercepts using a graphics calculator, proceed as follows:

1. Enter the equation for y in the Y= menu.

2. Press and then or adjust

settings to obtain a view of the graph that includes both intercepts.

3. To find the x-intercept, press [CALC] and select 2:Zero. Scroll to the left of the x-intercept

and press . Scroll to the right of the

x-intercept and press . Press

again.

4. To find the y-intercept, press [CALC] and select 1:Value. Enter X = 0 and press .

a y= 6x+ 18 b y= 3x− 21 c y= 5x+ 12

d y= −2x− 3 e y= 10 − 5x f y= 1 −x

g y= −9x+ 30 h y= 2(x− 8)

a 2x+ 3y = 6 b 4x+ 5y= 20 c −5x+ 8y= 10

d 6x− 3y= −18 e 7x− 5y= 35 f 8y−x= 4

g x−y= 2 h −2x+ 11 = 6y

a 6x+ 7y+ 42 = 0 b 5x− 2y+ 20 = 0 c −3x + 4y− 16 = 0

d y− 3x+ 6 = 0 e 9x+ 18 − 2y= 0

a x+y= 0 b x−y= 0 c 2x+y= 0

A2x− 3y− 1 = 0 B 2x+ 3y+ 1 = 0 C 2x+ 3y=x D2x+ 3y= 1 E y= x− 1

A 2 and 3 B −3 and 12 C −4 and 6 D −4 and −6 E 4 and 6

Ay= −3x− 3 B y= −3x+ 3 C x+ 3y= 9 Dx− 3y+ 9 = 0 E 3x+ y+ 9 = 0

1E

WORKED

Example

9

Mathc ad

Gradient of a straight

line WWORKEDORKED E Examplexample

10

Cabr

i Geometry

11

m

2 3

---m

m

Graphics Calculator

tip!

Finding

x

-

and

y

-

intercepts

GRAPH ZOOM WINDOW

2nd ENTER

ENTER ENTER

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27

Simultaneous equations

Simultaneous equations are groups of equations containing two or more pronumerals. In this section, we look at pairs of linear equations involving the pronumerals x and y. Each equation, as we have learnt 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 solution

The following example shows how a graphics calculator may be used to solve simultaneous equations graphically. Hand drawn sketch graphs may be used if graphics calculators are not available. Refer to earlier sections if you need reminding how to draw linear sketch graphs, and make sure you use a consistent and accurate scale.

Both graphs have the same x- and

y-values here.

y

x

Use a graphics calculator to solve the following simultaneous equations and sketch the screen from which the solution was obtained.

y= −3x+ 5 4x− 7y+ 8 = 0

THINK WRITE/DISPLAY

Convert the second equation into

Y= form so it may be entered in the

Y= menu of the graphics calculator.

4x – 7y + 8 = 0 4x + 8 = 7y

7y = 4x + 8 y = x +

Press . At Y1=, enter (-)3X+5. At Y2=, enter 4/7X+8/7.

Press , and adjust the WINDOW

settings if required, to encompass the point of intersection of the two graphs.

Press [CALC]. Select 5:intersect

and then press three times to

find the coordinates of the intersection.

Sketch the screen.

1

4 7 --- 8

7

---2 Y=

3 GRAPH

4 2nd

ENTER

5

12

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Algebraic solution

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

Use the substitution method to solve the following simultaneously. −2x+ 7y= 8

y= 3x− 1

THINK WRITE

Write down and label the equations. −2x + 7y = 8 [1]

y = 3x − 1 [2]

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

Substitute [2] into [1].

−2x + 7(3x − 1) = 8 [3]

Solve [3] for x and label the solution as equation [4].

−2x + 21x − 7 = 8 19x − 7 = 8 19x = 15

x = [4]

Use the solution to solve for y. Substitute [4] into [2].

y =3 − 1 = − 1

= −

= or 1 [5]

State the complete answer. Solution: ,

Optional double-check: substitute equations [4] and [5] into [1] to check that these values for x and y make [1] true.

Check: In [1], LHS = −2x + 7y

= −2 + 7 = − + = = 8 = RHS ✓

1

2

3

15 19

---4

15 19

---   

45 19 ---45 19 --- 19

19 ---26 19 --- 7

19

---5 15

19

---  26

19 ---_

6

15 19

---    26

19

---   

30 19 --- 182

19 ---152

19

---13

WORKED

E

xample

Use the elimination method to solve these simultaneous equations. 2x+ 9y=−5

5x− 2y− 12 = 0

THINK WRITE

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

5x − 2y − 12 =0 [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]

1

2

14

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29

THINK WRITE

Obtain 10x in both [1] and [3] as explained opposite.

2 × [3]: 10x − 4y = 24 [4] 5 × [1]: 10x + 45y = −25 [5]

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

Solve for y.

y =−

y = −1 [6]

Substitute [6] into [1] to find x. Substitute [6] into [1].

2x + 9 (−1) = −5 2x − 9 = −5

2x = −5 + 9 2x = 4

x = 2 [7]

State the solution. Solution: (2, −1)

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 ✓

3

4

5 49

49

---6

7 8

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 $2.59 Shopper 2: 6 apples and 5 bananas for $4.11

THINK WRITE

Decide on pronumeral names for the unknown quantities.

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

Write equations involving these pronumerals. Work in terms of cents.

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

Choose a pronumeral to eliminate. In this case b.

5 × [1]: 20a + 15b = 1295 [3] 3 × [2]: 18a + 15b = 1233 [4]

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

a = 31 [5]

Solve for b. Substitute [5] into [1].

4 × 31 + 3b = 259 124 + 3b = 259 3b = 135

b = 45 [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.

1

2

3

4

5

6

15

WORKED

E

xample

remember

1. With a graphics calculator, express as Y1 = and Y2 =and find the intersection

using [CALC] and 5:intersect.

2. If form is y=ax+b, y=cx+d consider using substitution. 3. If form is ax+by=c, dx+ey= f consider using elimination.

2nd

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Simultaneous equations

1 Use a graphics calculator to solve the following simultaneous equations, and sketch the screen from which the solution was obtained.

2 Use the method of substitution to solve the following simultaneously.

3 Use the elimination method to solve these simultaneous equations.

4 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

A manufacturing plant produces fixed size square and circular metal panels. 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:

10

Which of the following is a solution of 11x+ 2y= −121 and 10x+ 12y= −222? a y= −2x, y= −4x− 6 b y= 4x, y= 3x− 5

c y= 3x− 5, y= 20 d y= −2x− 4, y= −5x+ 5

e y= 3x+ 5, y= 7x− 4 f y= −5x+ 12, y= 2x− 1 g −3x+y= −4, y= 6x+ 5 h 2x−y− 11 = 0, y= −4x+ 8 i y= 10x+ 1, 2x+y= −6 j 9x+y= 17, x+y= 14

a y= 3x+ 1, y= 2x+ 2 b y= −3x, y= 4x+ 14 c y= 5x+ 5, y= −x− 19 d y= −4x− 3, y= 3x− 24 e y=x+ 2, 3x− 4y= −1 f y= 3x− 6, 2x+ y= 9 g y= −2x+ 3, −5x+ 2y= 1 h 6x−y= 8, y=x+ 4 i −4x− 3y= 2, y= −6x+ 7 j y= 10 −x, 2x+ 7y= 5

a 9x+ 10y= 153, 3x−y= 12 b 2x+ 4y= 8, 7x+y= −37 c 7x− 11y= −13, x+y= 11 d 6x− 9y= 51, −6x+ 11y= −49 e 6x− 2y= 10, 2x+ 5y= −8 f 2x+ 7y= 16, 3x− 6y= 2 g −3x+y= 8, 4x+ 2y= 21 h 8x+ 3y= −9, 4x−y= −3 i 7y−x= 11, x+y= 10 j x− 11y= −15, y+ 6x= 9

A 13s+ 22c= 1205, s+c= 65 B 22s+ 13c= 1205, s+c= 65 C 13s+ 22c= 65, s+c= 1205 D 22s+ 13c= 65, s+c= 1205 E 13s+ 22c= 1205, s+c= 35

A (11, 2) B (−121, −222) C (10, 12) D (−9, −11) E (6, 10)

1F

W WORKEDORKED E Examplexample

12

Mathca

d

E

XCEL Sp

readsh_eet

Simultaneous linear equations — graphical method

Mathca

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Geometry

Simultaneous linear equations

13

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Simultaneous linear equations E

XCEL Sp

readsh_eet

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15

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31

Using matrices to solve

simultaneous equations

The steps below show how to solve the following set of three simultaneous equations. They may be easily modified to handle any number of equations.

2x+ 3y+ 4z= 21 [1]

x – 3y – z= 0 [2]

6x + 7z = 44 [3]

In matrix notation, the above set may be written as follows:

or AX = B for short, where

A =

The solution to the set of equations may be found by calculating

A–1B.

A–1 is the ‘inverse’ of A, and may be found using the graphics calculator. The following steps show how to use matrices on the graphics calculator to solve the given set of equations.

1. Press and select EDIT and

1:[A]. Overtype the dimensions with

3 × 3 if necessary, and press .

Move the cursor to the various positions within the matrix and enter the required

values (press after each one).

2. Now press and select EDIT and

2:[B]. Overtype the dimensions with 3 × 1,

then press .

3. Fill in the required values.

4. Press [QUIT] to return to the home screen.

5. Press and select NAMES and 1:[A].

6. Press , and , and select

NAME and 2:[B].

7. Press to find the solution matrix,

which gives the solution for x, y and z

from top to bottom, that is, in this example, x= 5, y= 1 and z= 2.

2 3 4

1 –3 –1

6 0 7

x y z

= 21

0 44

2 3 4

1 –3 –1

6 0 7

, X = x y z

and B = 21

0 44

MATRIX

ENTER

ENTER MATRIX ENTER

2nd

MATRIX

x–1 X MATRIX

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Perpendicular lines

The reciprocal of a value is equal to one divided by the value. Using algebra, the reciprocal of is

and the reciprocal of a is .

The reciprocal of a fraction is an ‘upside down version’ of the fraction. For example, the reciprocal of is . The reciprocal of 7 (which may be thought of as ) is .

A negative reciprocal of a value is the opposite sign, and the reciprocal of the value. The negative reciprocal of is − . The negative reciprocal of − is .

Perpendicular lines

The equations below are in pairs where each has the negative reciprocal gradient of the other.

1 Copy and complete the table below. You may use a graphics calculator to help you if one is available.

2 Sketch Y₁ and Y₂ for part a above using a graphics calculator (make the scale of the

axes equal by pressing and selecting 5:ZSquare) or using graph paper with

identical scales on each axis to produce accurate graphs. What do you notice about the graphs? (Come back to this question later if you’re not sure.)

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 which has a gradient of:

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

Equation 1 (Y1)

Gradient of Equation 1

(m1)

Equation 2 (Y2)

Gradient of Equation 2

(m2) m1×m2

a y = 2x+ 1 y=− x

b y = 3x− 4 y=− x+ 2

c y = x+ 6 y=−4x− 9

d y = x− 3 y = − x+ 4

e y = – x y= x+ 1

a 4 b −9 c d − e f 1

a y= −5x+ 2 b y=x− 1 c y= x+ 1

d y=− x− 2 e 2x+y= 5 f 3x− 4y= 7

a b

--- 1 a

b

---÷ 1 b

a ---× b a ---= =

1÷a 1

a ---= 2 3 --- 3 2 --- 7 1 --- 1 7 ---4 9 --- 9 4 --- 9 5 --- 5 9

---1G

Skill

SHEET

1.2

1 2 ---1 3 ---1 4 ---2 5 --- 5 2 ---9 7 --- 7 9 ---ZOOM 1 7 --- 8 9 --- 7 2 ---2 3 ---7 6

---remember

If two graphs have negative reciprocal gradients, (m₁×m₂= −1), they are

perpendicular; if they are perpendicular, then they have negative reciprocal gradients.

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33

Formula for finding the equation of a

straight line

Consider 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 first two of these points in the formula for gradient, we have

[1] Using the first and last point in the same formula yields

[2]

Putting [2