1.1

Mathematical Skills for Physics What is this? VCE Physics does not have any pre-requisites and it‟s not all maths! But as with all subjects, some prior skills and knowledge are assumed before commencing the Physics course. As you are probably new to the DECV we need to identify any gaps in your mathematical background which might hinder you. We can then plan how to overcome them early. The aim is that we can plan your studies with us, not to stop you!

OK – What do I do? Below is a series of questions to complete, covering some common skills. Most areas you will have covered before, but some may be new, so don‟t panic! Our aim is to find what you do know and areas we need to cover with you. 1. 2. 3. 4.

5.

Answer the questions. Check the answers yourself. (You do not need to send these to your teacher.) The answer sheet has details of how to get the answer and further advice. For some topics, further explanations and examples are given after the answer sheet. You may want to revisit particular topics, or read it all as a general revision. Fill in the VCE Physics Skills Evaluation at the end and return the Week 1 submission sheet at the end of the week. If you have any concern about your preparedness or the content of the course, please contact your teacher.

We hope you will find this helpful, and contact the DECV if you have concerns about our courses, the enrolment process or how we can work together to make your study a worthwhile and successful experience.

1.2

Questions (You may use a calculator if necessary.) Find the answer for the following: 1. (a) −45.1  2.1 (b) −3 + 5 (c) 5  0.004 (d) −98 − 586

2.

3. 4.

(e)

46−19  2

(f)

45 – (− 30)

(g) (h) (i) (j) (k)

−25.6  −3 83 + (− 123) 350 + 450 ÷ 9 – 4 × 5 1 2  3 5 2 4  3 5

3 as a decimal 16 a) to 3 decimal places b) to 2 decimal places c) to 2 significant figures 3 Convert into a percentage 4 Write 16% as a number in decimals. Write

5.

Find the answer to: 42 (a) (b) 3  24 (c) 4  32 20  11 (d)

6.

Scientific notation expresses numbers as a combination of a simple number with a part to the power of 10. e.g. 3.2 × 104 Express these in scientific notation: (a) 1245 (b) 0.005

7.

Rearrange these equations to find the value of y (a) y +4 = 21 (b) 162y = 7 (c) 4 – 3(y + 2) = 7 (d) 8 – 3y = 9 + 5y 6 (e)  12 y

8.

Make „a‟ the subject of the following: b c (a) a a b d (b) c 1 1 1   (c) a b c

1.3

9.

If v = 100, a = 10, s = 450, find the value of „u‟ using the following equation: v2 = u2 + 2as

10.

If s = 300, a = 10, u = 20, find the values of „t‟ using the following equation: s = u t + ½ at2

11.

Analyse the following graph:

(a) (b) (c) (d) (e) (f)

12.

What does the horizontal axis show? What is the velocity when the time is 40 seconds? When does the object have a velocity of 300 m/sec? When is the velocity greatest? Find the gradient or slope of the line from the start to 15 seconds? The area under this graph has the shape of a triangle. Find the area under the graph. A rectangular sheet of paper is 21 cm wide and 30cm long. (a) Find the area of the rectangle in cm2 and then convert it to m2.

13.

(b) If I stack those sheets in a pile 10 cm high, what is the volume of the pile? The area of a circle is A  r 2 . What number does the symbol  stand for?

14.

A right angled triangle has short sides of length 6m and 8m. How many metres is the longest side?

15.

Check the prefixes list on page 2.5 to help you answer the questions below. Convert: a) b) c) d)

16.

100km/hr to m/s 3.5 m/s to km/hr 4GHz to Hz 23 nanometres to metres

How many seconds are there in 365 days?

1.4

17.

Find the following: (a) sin 60o (b) cos 45o

18.



The angle above is 30o. If F = 20 what is the size of a and of b? 19.

Using the right angle triangle above, if the length of a = 3, b= 4, F= 5, what is the angle ɵ (in degrees)?

20.

Find the answer to the following: (a) (b) 1010 103 (c) 3108  4 105 (d) 7 106  2 104

1.5

Answers Below are answers to the “test” questions. Check the answers and explanations. If you are still having difficulties, note the entry in the “T” column. After the answers we’ve included a “Further Details” section with more explanations about each topic. T A

Question 1 (a)

Answer 94.71

Details The rules for multiplying and dividing numbers are: If the signs are the same, the answer is positive If the signs are different, the answer is negative. So 2  3 = 6 and 2  3 = 6 but 2  3 = 6 and 2  –3 = 6.

A

(b)

2

You can think of these sums as moving along a “number line”, with a starting point at zero. Positive numbers are to the right and negative numbers to the left. The -3 moves you to the left by 3 from zero. Adding 5 moves you to the right by 5. This will send you pass the starting point (0) and to the right by 2. So the finishing point is +2. Another way to think of it is that you take away the numbers and the sign of the bigger one wins!

A

(c)

0.02 5 x 0.004 = 0.02 Normally we don‟t show the extra zeros at the end of a decimal (like 0.020), but any zeros between the decimal point and the other digits must be shown. That is 0.02 or 0.020 are alright but not 0.2 or 2.

A

(d)

684

The 98 moves you to the left by 98 from zero. Subtracting 586 moves you further to the left by 586. So the finishing point is – (98 + 586).

A

(e)

8

The order of the operations is important. You should do them in the following order: brackets multiplication and division (from left to right) addition and subtraction (from left to right) So in this sum, you should start with the multiplication, then the addition 46 – 19  2 = 46 – 38 =8 More examples: 15 –6  3 = 13 not 3 12 + 3  (6 + 4) = 12 + 3  24 = 12 + 72 = 84 not 94.

1.6 A

(f)

75

When there is two negative side by side, the overall is a positive, therefore the sum becomes 45 + 30 = 75.

A

(g)

76.8

Negative number multiply by another negative number gives you a positive number. 25.6  3 = 76.8

A

(h)

40

  + = , therefore the sum becomes 83 – 123 = 40.

A

(i)

380

BODMAS Do 450 ÷9 first & you get 50 then do 5  4 = 20 So now the sum looks like this: 350 + 50 – 20 = 380

B

(j)

11 15

When you multiply fractions you multiply the top numbers (numerators) together and the bottom numbers (denominators) together. When adding or subtracting fractions the denominators must be the same. We can multiply or divide denominator by any number as long as we do the same to the top. Here the denominators are 3 and 5. We can make the 3 into 15 by multiplying by 5 and can multiply the 5 by 3 to get 15. 1 1 5 5    3 3 5 15 2 2 3 6    5 5 3 15 Finally you add the numerators together. 1 2 5 6 11 So     3 5 15 15 15 You use the same method for subtraction.

B

C

(k)

2

(a)

8 15

0.188

Multiply the top numbers and multiply the bottom numbers. 2 4 2 4 8    3 5 3  5 15 Dividing by a fraction is the same as multiplying by its inverse or “reciprocal”. 2 4 2 7 14 For example,     5 7 5 4 20 We could reduce this further by dividing both 14 and 7 20, to give 10 On a calculator you punch 3 16 = 0.1875 Some calculator have an “ENTER” key instead of “=”.

1.7 D

(b)

0.19

When rounded off, if the first digit being pruned is 5 or more you go up; less than 5 goes down.

(c)

0.19

See page 2.13 (week 2) for more rules on significant figures To get a percentage you turn the fraction into a decimal and multiply by 100. 3 So  0.75 4 0.75 = 0.75  100% = 75%

C

3

75%

C

4

0.16

D

5

D

(a)

(b)

16

To go from a percentage to a decimal, divide by 100. 16% = 16  100 = 0.16 When a number is multiplied by itself, we write it as the number and an index. For example: 2  2  2 = 23 and 3333=34

48

42 = 4  4 = 16 3  24 =32222 = 48

D

(c)

13

D

(d)

3

Note: It was the 2 that was raised to a power of 4, not 32. If the question was (3  2)4 the answer would be (3  2)4 = 64 = 6  6  6  6 = 1296 The 32 is really a multiplication, so we do this before the addition. 4 + 32 = 4 + 9 = 13 This is the square root of 20 , so we have to do that sum first.

20  11  9  3 Again you need to see which numbers are under the square root symbol. If only 20 was under it the answer would be 20  11  4.47  11  6.53 D

6

(a)

1.245  103

1245 = 1.245  1000 = 1.245  103 A way of looking at it is you moved the decimal point 3 places to the left, so it is 103. If the number is less than 1 you need to move the decimal point to the right, giving a negative index. In this course we usually express numbers greater than 9,999 and less than 0.01 in scientific notation.

D

(b)

5  10-3

0.005 = 5  1000 = 5  103 = 5  10-3 You had to move the decimal point 3 places to the right, so the index of 10 is –3.

1.8 E

7

(a)

y = 17

To rearrange the equations you need to get y on its own on one side. The rules are: 1. You can do any operation, as long as you do the same thing to both sides 2. The order of the operations is important. You should do them in the opposite to the normal order: addition and subtraction (from left to right) multiplication and division (from left to right) brackets 3. Use the “inverse” of the operations eg. to get rid of the +4, subtract 4 from both sides. y + 4 = 21 y + 4 – 4 = 21 – 4 y + 0 = 17

E

(b)

4.5

16 – 2y = 7 First get rid of the 16 by subtracting 16 from both sides 16 – 2y – 16 = 7 – 16 2y = 9 y is multiplied by –2 so opposite operation means divide by –2 2 y 9  2 2 9 y   4.5 2

E

(c)

3

E

(d)

 0.125

4 – 3(y+2) = 7 Start by getting rid of the 4 4  3(y + 2) –4 = 7 – 4  3(y + 2) = 3 Divide through by –3 to undo the multiplication 3( y  2) 3  3 3 y  2  1 Now subtract 2 from both sides y+2–2=12 y = 3 8 – 3y = 9 + 5y Get all the “y‟s” on one side. Add 3y to both sides. 8 – 3y + 3y = 9 + 5y +3y 8 = 9 + 8y 8 – 9 = 9 + 8y –9 1 = 8y 1 y 8 y  0.125

1.9 E

(e)

0.5

6  12 y 6  y  12  y y 6  12  y 6 y 12 y  0.5

y is the denominator, so we multiply by„y‟ both sides

(OR we can swap y and 12) G

8

(a)

a

G

(b)

a=cd+b

G

(c)

a

b c

bc bc

b a We want “a” on its own. So multiply both sides by a and then divide both sides by c. ac = b b a c The shortcut is to swap ‘a‟ and „c‟. a b d c To get rid of the “bottom line” (or denominator), multiply both sides by c a b d c  c c d c  a b Now add b to each. b + dc = a c

1 1 1   a b c To combine the fraction on the right hand side, make the denominators bc. 1 1 c 1 b   a bc cb 1 c b   a bc bc The terms on the right have the same denominator, so we can combine them.

1.10

1 cb  a bc (c  b )  a 1 bc bc  (c  b)  a bc a cb F

9

31.62

Substitute given values into corresponding terms in the equation you will get: 1002 = u2 + 2 (10)(450) 10000 = u2 + 9000 10000−9000 = u2 + 9000 -9000 u2 = 1000 therefore = 31.62 √

F

6 and 10

10

Substitute the given values into the equation, you get: 300 = 20t + ½ (10) t2 5t2 + 20t – 300 =0 The formula for solving quadratic equation is that if ax2 + bx +c = 0

x

b  b 2  4ac 2a

So in this case, a  5, b  20, c  -300 t 

20  202  4(5)(300) 2(5)

20  6400 10 20  80 20  80  or 10 10 Therefore t = 6 and 10 

11

(a)

Time (in seconds)

(b)

About 150 m/sec

On a graph you normally have 2 axes, one “horizontal” (across the page) and the other “vertical” (up and down the page). Each should have a scale, a name and usually the units. Here time is measured in seconds and velocity in m/s. The horizontal axis represents time. To find the velocity at 40 seconds: Go to 40 on the time axis and draw a vertical line up to where it cuts the graph. From that point go horizontally to cut the vertical “velocity” axis, and work out the value.

1.11

H

(c)

About 13s and 27.5s

Find 300 on the velocity axis and draw a horizontal line across to the graph. It hits the graph at two points, drop a vertical line down to the axis. This cuts the time axis at about 13 seconds and at about 27.5 seconds.

(d)

15 s

The highest velocity reached was 400. This was at about 15 seconds.

(e)

26.7 ms-2

The gradient is how steep the line is. To find this you: choose 2 points on the line find how much the vertical value has changed find how much the horizontal value has changed divide the vertical change by the horizontal change So here at t =15, v=400 at t = 0, v=0 so the gradient =

2.7 12

(f)

11000 m

(a)

630 cm2 Or 0.063m2.

(b)

6300 cm3

point (15,400) point (0,0)

rise 400  0 400    26.7 m/s2. run 15  0 15

Area of a triangle = 0.5  length of base  height Area = 0.5  55 (seconds)  400 (metres/second) = 11000 m. The area under a velocity-time graph represents distance travelled. Area of rectangle = length  width So area = 21 cm  30 cm = 630 cm2 1m2=100 × 100 cm2 =10000cm2, therefore 1cm2 = 10-4m2. To convert cm2 to m2, you multiply 630 with 10-4 630 × 10-4 = 6.3 × 10-2 = 0.063m2 OR 0.21 m× 0.3m = 0.063m2. Volume of a cube = length  width  height So here volume = 21cm  30cm  10cm = 6300 cm3

13

3.1416

 stands for pi which has a value of approximately 3.1416.

14

10

For a right-angled triangle (one with an angle of 90 degrees), the length of the sides obey Pythagoras‟ Theorem. The longest side (c) is related to the other sides (a and b) by: c2 = a2 + b2 So here c2 = 62 + 82 c2 = 36 + 64= 100 c = 10

1.12 15

(a)

27.7ms-1

100km = 100 000m 1hour =60 mins × 60 seconds = 3600seconds Therefore

will be

=27.7m/s. 2.6

(b)

12.6km/h

See page 2.6, to convert m/s to km/h, you multiply by 3.6. So, 3.5 × 3.6 =12.6km/h

2.5

(c)

4 × 109 Hz

2.5

(d)

2.3 × 10-8 m

See page 2.5, to convert from giga hertz to hertz, you multiply by 109 so 4GHz = 4 × 109 Hz See page 2.5, to convert from nano metres to metres, you multiply by 10-9 so 23nm = 2.3 × 10-8 m.

16

31536000 s

1 year has 365 days 1 day has 24 hours 1 hour has 60 mins 1 min has 60 seconds Therefore 1 year = 365 x 24 x 60 x 60= 31536000 s

17

18

(a)

0.866

(b)

0.7071 a=17.32 b= 10

sin 60o = 0.866 (make sure your calculator is in DEG mode rather than RADIANS). You can check, in degree mode, sin 30ᵒ = 0.5 cos 45o = 0.7071 For right angled triangles, the sin of an angle equals the length of the opposite side divided by the longest side (called the hypotenuse). The cosine is the adjacent side divided by the longest. So in this case: b sin  = so b = F sin  (sin = opposite/hypotenuse) F b = F sin  = 20 sin (30o) = 20  0.5 = 10

a so a = F cos  (cos = adjacent/hypotenuse) F a = F cos  = 20 cos (30o) = 20  0.866 = 17.32 cos  =

1.13 53.1ᶱ

19

We can use either sin, cos or tan to solve this question since the 3 sides are given. If you use sin, b sin   F therefore  = sin-1 (4/5) = 53.1ᶱ If you use cos, a cos   F therefore  = cos-1 (3/5) = 53.1ᶱ

I

20

(a)

108

(b)

107

If you use tan, (tan = opposite/adjacent) b tan  = a therefore  = tan-1 (4/3) = 53.1ᶱ You can use calculator to solve this question OR apply the law of indices. When multiplying two or more numbers or pronumerals with the same base, keep the base and add the powers together. am  an  am  n Therefore When dividing two or more numbers or pronumerals with the same base, keep the base and subtract the powers. am  an  am - n 1010  10103  107 103

(c)

1.2 104

We group the SAME base together to apply the indices laws and multiply the numbers without common base as per normal. We can rewrite: 3108  4 105   3 4  108( 5) 12 103

(d)

3.5 1010

This can be further simplified to 1.2 104 We group the SAME base together to apply the indices laws and divide the numbers without common base as per normal. 7 106  2 104   7  2  106 4  3.5 106 4  3.5 1010

1.14

Further Details Below are more details about the mathematical processes used in Physics. You may use it as a general review, but particularly read sections for topics in which you had difficulties. After each section there are a few extra examples to try and the answers on page 21. {Extracts below come from "Maths Quest" by Nolan et al from Jacaranda}

Topic A: Dealing with numbers When solving a mathematical expression that has several operations, the order you do them is important. 1. brackets must be evaluated first, then 2. multiplication and division (in order from left to right), then 3. addition and subtraction (from left to right). Inside brackets the same order applies. Evaluate the following: (a) 3 + 8  7 – 4

(b) (7 + 2  3)  (6  2 – 11)

(a) There are 3 operations here: +  and -. According to the rules, do multiplication first then working left to right we do + then (b) We start with the brackets. In both the highest priority is the multiplication Then the other operations in the brackets. As no more brackets, now do the division. Further examples (answers at the end of the week): (1) 13 – 6  2 (2) (56  2 – 20  2

3+87–4 = 3 + 56 – 4 = 59 – 4 = 55 (7 + 2  3)  (6  2 – 11) = (7 + 6)  (12 – 11) = 11  1 = 11

(3)

12 – (6 + 4)

Topic B: Fractions 3 1 2 Evaluate 2   1 4 5 3 Write the expression Change mixed numbers into improper fractions The lowest number that is a factor of 4, 5 and 3 is 60. Change the denominator to 60 by multiplying the top and bottom numbers of the first fraction by 15, the second by 12 and the third by 20. Change into a mixed number

3 1 2 2  1 4 5 3 11 1 5 =   4 5 3 11  15 1  12 5  20    4  15 5  12 3  20 165  12  100  60 77  60 17 1 60

1.15 The final answer should be given in the simplest form. When multiplying and dividing fractions, mixed numbers need to be changed into improper fractions first. To multiply 2 or more fractions, multiply the numerators together (the top numbers) and multiply the denominators (bottom numbers) together. To divide a fraction, turn it upside down and multiply. (eg. dividing by 3

same effect as multiplying by

2

2 3

has the

.

Also check to see if you can cancel numbers before doing the multiplication. eg. 2



3

6 7



2 1 3

Calculate



23



7

(a)

2



1

1 3 2



2 7

4



4 7

(b)

5

7 8

1

3 4

a Write the expression

1 4 3  2 5 7 4   2 5 7 2   1 5

Change the mixed number into an improper fraction We can cancel by dividing the 4 in the numerator by 2 and the 2 in the denominator by 2. Multiply the numerators together and the denominators together. Change into a mixed number.

14 5 4 2 5 7 3 1 8 4 7 7   8 4 7 4   8 7 1 7 4 1   4  2 1 7 1 1   2 1 1  2 

b Write the expression Change the mixed number into an improper fraction Turn the second fraction upside down and multiply instead of divide. The sevens in the numerator and denominator can be cancelled. Also they can be further reduced by dividing by 4. Multiply the numerators together and the denominators together

Further examples (answers at the end of the week): (a) (b) 1 2 1 2 1  1  2 3 2 3 (c)

1 2 1  2 3

(d)

7 2  8 4

1.16

Topic C: Percentages The term “per cent” means “per hundred”. For instance 29% means 29 parts out of 100. Any percentage can be changed into a fraction by dividing it by 100. Write the following percentages as: (i) common fractions 3 (a) 12% (b) 10.3% (c) 76 % 8 a(i) To show 12% as a fraction, put it over a hundred and remove the % sign Simplify (ii) To write it as a decimal, divide it by 100 (that is move the decimal point 2 places to the left) and remove the % sign. b(i) Put 10.3 over 100 and remove the % sign Multiply the numerator and denominator by 10 to get rid of the decimal point. (ii) Divide the 10.3 by 100 (moving the decimal point 2 places to the left) and remove the % sign. c(i) Change the fractional part into a decimal Put this over 100 and remove the % sign Get rid of the decimal point by multiplying the top and bottom by 1000 Simplify (ii) Change the fractional part into a decimal Divide by 100 by moving the decimal point 2 places to the left, and drop the % sign. Further examples (answers at the end of the week ): (1) 23% as a decimal 2 (2) as decimal and percentage 5 (3) 0.16 as a percentage

(ii) decimals.

12 100 3 = 25

12% =

12 % = 0.12

10.3 100 103 = 1000

10. 3 % =

10.3% = 0.103

3 76 % = 76.373% 8 76.375 = 100 76375 = 100000 =

611 800

3 76 % = 76.375% 8 = 0.76375

1.17

Topic D: Positive and Negative Numbers When multiplying or dividing two numbers the sign of the answer is: First no. + + -

Second no. + +

Answer + + -

"Like" numbers give a positive answer; "Unlike" numbers give a negative answer. Evaluate (a) – 18  (–6)  5 (b) (–3)3 (a) Write the expression As the operations are all multiplication or division, they should be done from left to right. So divide –18 by –6. As both are negative, the answer is positive. Multiply 3 by 5. Since both are positive, the answer is positive. (b) Write the expression Multiplying two negative numbers gives a positive. Multiplying a negative and a positive gives a negative.

– 18  (–6)  5 =35

= 15 (–3)3 =(–3)  (–3)  (–3) = 9  (–3) = – 27

Evaluate (a) 5 – 11 (b) –3 + (–6) (c) 6 – (–2) (d) –5 + 12 (a) Write the expression 5 – 11 = – 6 As the signs are opposite, find the difference. The result takes the sign of the larger one. (b) Write the expression –3 + (–6) = – 9 As the numbers have the same sign, add them. the result takes the sign of the larger one. (c) Write the expression 6 – (–2) = 6 + 2 =8 Subtracting a negative number is the same as adding the positive equivalent. (d) Write the expression –5 + 12 As the signs are opposite, find the =7 difference. The result takes the sign of the larger one. Further examples (answers at the end of the week): (1) 13–43 (2) –12 (2(–3)) (3)

20–15

Key Points: 1. Order of operations is (a) brackets (b)  and  from left to right (c) + and–from left to right Inside the brackets you should use the same rules on order.

1.18 2.

For fractions (a) Change mixed numbers to improper fractions (eg 16/3) before any calculations. (b) Change fractions to the lowest common denominator before adding or subtracting. (c) To divide by a fraction, turn it upside down and multiply. (d) Check whether you can cancel before multiplying. (e) Give the answer in its simplest form, and change improper fractions to mixed numbers.

3.

Positive and Negative Numbers When multiplying or dividing two numbers the sign of the answer is: First no. + + -

Second no. + +

Answer + + -

"Like" numbers give a positive answer; "Unlike" numbers give a negative answer.

Topic E: Equations Equations are used extensively in Physics They can have one unknown quantity, denoted by a symbol (eg. x, y, ) or more than one(eg. 5 = y + 4).. We need to be able to rearrange them so that an unknown quantity is on its own on one side of the equation (y = 5 - 4 = 1). A key is that in an equation the two sides are equal. They will stay equal as long as any change made to the left side is matched by the same change on the right side. To solve an equation you perform a number of "inverse operations" to both sides of the equation until the value of the unknown is found. When solving them, the order of operations (BODMAS) is reversed. This means you do subtraction and addition first, then multiplication and division, and brackets last. Solve the following equations: x (a) 2x – 3 = 4 (b)  2  8 6 (a) Write down the equation

(c) 10 

Think of the operations performed on x in the correct order, and their inverse. To solve the equation we will start at the bottom of the inverse column and go up. Solve the equation by making x the subject. Add 3 to both sides Divide both sides by 2

3x 5 2 (a)

2x – 3 = 4 Operation Inverse 2 2 –3 +3

2x –3 = 4 2x – 3 + 3 = 4 + 3 2x = 7 x = 3.5

1.19 (b) Write down the equation

(b)

Think of the operations performed on x in the correct order, and their inverse. To solve the equation we will start at the bottom of the inverse column and go up.

x 2 6

Solve the equation by making x the subject. Subtract 10 from both sides

Multiply both sides by 2

Divide both sides by 3

=8

x 22 = 8 – 2 6 x =6 6 x =66 = 36

Multiply both sides by 6

Think of the operations performed on x in the correct order, and their inverse. To solve the equation we will start at the bottom of the inverse column and go up.

=8

Operation Inverse 6 6 +2 -2

Solve the equation by making x the subject. Subtract 2 from both sides

(c) Write down the equation

x 2 6

(c)

10 

3x 2

Operation (-3) 2 +10

10  10 

3x 2

3x  10 2 3x 2 3x 2 2 3x x

=5 Inverse (-3) 2 -10

=5 = 5 – 10 = -5 = -5  2 = -10  10 = 3 1  3 3

You can check your answer by substituting it back into the original equation.

1.20 If the unknown appears more than once, you need to move all of them across to the one side, and the rest of the terms on the other. Solve for x in the equation 2x – 4 = 4x + 6 2x – 4 2x – 4x – 4 –2x – 4 –2x – 4 + 4 -2x 2 x 2 x

Write down the equation Get all the “x”s on the left hand side by subtracting 4x from both sides. Add 4 to both sides Divide both sides by –2 Simplify

= 4x + 6 = 4x – 4x + 6 =6 =6+4 = 10 10 = 2 = -5

If the expression contains brackets, these should be dealt with first. However sometimes both sides of the equation can be divided by a number to simplify the arithmetic. Solve for x in the equation (a) 2(x + 5) = 3(2x – 6) (a)

Write down the equation Expand the brackets. Get all x‟s on the left side by subtracting 6x from both sides. Subtract –10 from both sides. Divide both sides by –4

(b)

Write down the equation To simplify it, we can divide both sides by 4 Take away 6 from both sides Divide both sides by 2

(b) 4(6 + 2x) = 20 = 3(2x – 6) = 6x – 18 = 6x – 18 – 6x = – 18 = – 18 – 10 = –28  28 = 4 =7 = 20 = 20  4 =5 6 + 2x – 6 = 5 – 6 2x = – 1 x = –1 2 1    0.5 2

2(x + 5) 2x + 10 2x + 10 – 6x 10 – 4x 10 – 4x – 10 –4x 4 x 4 x 4(6 + 2x) 6 + 2x

If an equation has a fraction, we should first remove the denominators by multiplying each term of the equation by the lowest common denominator.

x2 x  5 3 2 x2 x = 5 3 2 The LCD will be 6

Find the value of x which will make the following true: Write down the equation Find the lowest common denominator Multiply each term by the LCD Simplify both sides Expand the bracket

( x  2) x  6  5 6   6 3 2 2(x+2) = 30 –3x 2x + 4 = 30 – 3x

1.21 Move the x‟s to the left by adding 3x Subtract 4 from both sides Divide both sides by 5

2x + 4 + 3x = 30 – 3x + 3x 5x + 4 = 30 5x = 30 – 4 = 26 26 x 6 1  5  5.2 6

Further examples (answers at the end of the week ): (1) 2x + 5 = 16 (2) 5 – x = x +7 (3) 5x – 3.5x = (x+2)

Topic F: Substitution A formula is an equation that shows the relationship between 2 or more variables. A change in one variable (eg. rainfall this month) will change another (eg. amount of water in reservoirs). If all but one value in a formula is known then we can get the value of the unknown variable by simply substituting the known values into the formula. If the formula for the conversion of temperature from Celcius (C) to degrees 9 Fahrenheit (F) is given by F  C  32 find the value of F when: 5 (a) C = 35 (b) C = –10 (a) Write down the formula for the conversion 9 F  C  32 of temperature 5 Substitute 35 in place of C in the formula Evaluate Answer the question including the correct unit. (b) Write down the formula Since C=-10, substitute -10 in place of C in the formula. Evaluate Answer the question and include the appropriate unit.

If C

= 35 F  95  35  32  63  32

 95 35C is equivalent to 95F 9 F  C  32 5 9   10  32 5 = –18+32 =14 –10C = 14F

Further examples (answers are the end of the week): To rent a car costs $50 a day plus 10 cents a kilometer. What is the total cost if you travel (1) 100km or (2) 500km.

1.22

Topic G: Transposing Formulae. In transposing formulae we really follow the same rules as when rearranging an equation. Again we do "inverse operations" until we get an unknown alone on one side of the formula, and we must do the same thing to both sides. The only difference is that in the end we don't get a specific value of the variable, but rather it expressed in terms of other variables. Transpose the formulae to make y the subject. by  cy 4 (a) 4x = 2y – 3 (b) A   y 2 (c) P  3 d a Write down the formula Add 3 to both sides of the equation

(d) m 

= 2y – 3 = 2y – 3 + 3 = 2y 2y  2 =y 4 A   y2 3 3A = 4 y2

4x 4x + 3 4x + 3 4x 3  2 2 2 x  1.5

To get y on its own divide both sides by 2 Simplify b Write down the formula Multiply both sides by 3

3A  y2 4

Divide by 4 Take square root of each side

y

c Write down the formula

Divide both sides by bc

y

Express the following with y as the subject. (1) 2x – 3y = 6 2( y  3) b (2) 5 (3) 5(3  y) = 6(f + 5)

Pd (b  c)

m  pq  ry m2= pq – ry 2 m – pq = – ry

D Write down the formula

Further examples (answers at the end of the week):

3A 4

P = Pd = by – cy Pd = (b – c) y

Multiply by d

Square both sides Subtract pq from both sides Divide both sides by –r

pq  ry

y=

1.23

Topic H: Gradient of a Graph The slope of a line is called the gradient, m. It can be calculated by choosing 2 points on the line, then finding the difference between the two y-values (y2-y1) and dividing it by the difference between the x-values (x2-x1). These are also called the “rise” and “run”. y y So the gradient m  2 1 x2  x1 Find the gradient of the line passing through the points (-1,4) and (2,10)

Pick two points and write down their coordinates.

Let (1, 4) be (x1, y1) and (2, 10) be (x2, y2).

Calculate the gradient using m 

y2  y1 x2  x1

10  4 2  (1) 6  2 1  2 So the gradient of the line is 2 m

Further examples (answers at the end of the week): Find the gradients of lines that go through the following points (1) (1, 2) and (5, 14) (2) (4, 6) and (7, 21)

Topic I: Multiplication and division of numbers with power of 10 When multiplying two or more numbers or pronumerals with the same base, keep the base and add the powers together. am  an  am  n When dividing two or more numbers or pronumerals with the same base, keep the base and subtract the powers. am  a m - n or a m  a n  a m - n n a

1.24 Find the values of: (a) 107 104 (b) 7.5 1011  9 1019 (c) 6.7 103  7 104 (d) 34 103  0.001 (a) The numbers both have a base of 10. As 107 104  1074  1011 they are being divided, subtract the indices. (b) We have normal numbers and some with a 7.5 1011  9 1019 base 10, so rearrange them   7.5  9   1011  1019  Divide the normal parts in the usual way. Because we are dividing, for the base 10 parts subtract the indices. (Remember that subtracting a negative is the same as adding a positive!)

 67.5 1011 19

If possible divide or multiply by 10 to get it in standard form.

 6.75109

 67.5 101119  67.5 108

6.7 103  7 104

(c) We have normal numbers and some with a base 10, so rearrange them

  6.7  7   103 104 

Multiply the normal parts in the usual way. Because we are multiplying, for the base 10 parts add the indices.

 48.9 103 4

If possible divide or multiply by 10 to get it in standard form.

 4.89 102

 48.9 101

34 103  0.001

(d) There are two ways you can approach this question. You can group the numbers and multiply like above and then convert the answer into standard notation OR convert 0.001 into common base 10 and apply the indices law.

  34  0.001  103  0.034 103  3.4 103 2  3.4  101  34 OR 34 103  0.001

  34 103   1103   34 10

3  3

 34 100  34 1  34

Answers to Further Examples: Topic A Topic B Topic C Topic D Topic E Topic F Topic G Topic H

10 5 6 0.23 -30 5.5 $60

y 3

72 1

2 x2 3

0.4, 40% 2 -1 $100 y  2.5b  3 9

2

1 4 16% 5 4 2

y= 1.2f3

1

3 8

1.25