程序代写案例-WFNS0001
时间:2022-06-22
WFNS0001
Numeracy Skills for
Everyday Life
WORKBOOK
Week 2
Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 46

Week 2 – Topic List


Lesson 1
2.1.1 Algebra operations, Expand/Factorise, Simplify Fractions 47
2.1.2 Linear and Quadratic Equations 50
2.1.3 Problem solving 54

Lesson 2
2.2.1 Linear patterns 59
2.2.2 Coordinate Geometry – plotting points and lines 60
2.2.3 Distance, Midpoint, Gradient, Parallel and Perpendicular Lines 62

Lesson 3
2.3 Applications to coordinates geometry 67


Lesson 4
2.4 Simultaneous Equations 69


Lesson 5
2.5.1 Applications of Simultaneous Equations – Breakeven analysis 73
2.5.2 Problem solving using simultaneous equations 75


ANSWERS 194









Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 47

Week 2 – Lesson 1

2.1.1 Basic Operations with Algebra
Algebra is a system of notation used in Mathematics and other Sciences, where letters (or pronumerals, or
variables) are used to represent numbers/things which can vary.

Just like in any new area of study, there are certain rules that have been developed for Algebra and its use.
Some of the rules are:

NOTATION RULES:
• 4w means 4 × (in algebra, we do not write the multiplication sign)



means ÷ (in algebra, we use fractions for division, rather than using the division sign)
• We write 3w not w3 (number first, then letters)
• 1w is always written as w (and -1w is written as -w)
• × (−) × × = − (sign first, then pronumerals in alphabetical order)
• + = ; − =



= × =
ARITHMETIC RULE RULES FOR EXPONENTS
• b aba
c c
  = 
 
m n m na a a +× = ;
m
m n
n
a a
a
−=

a
ac
b bc
 
 
  = ; a acb b
c
=
 
 
 
( )nm m na a ×=
• a c ad bc
b d bd
+
+ = ; a b a b
c c c
+
= + ( )n n nab a b= ;
n n
n
a a
b b
  = 
 

• a c ad bc
b d bd
+
+ = 1n na a
− = ;
n na b
b a

   =   
   


• a b b a
c d d c
− −
=
− −
0 1a =

a
adb
c bc
d
=
Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 48

E.g. 1
Simplify the following:
(a) 3 + 4 − 4 + 5 (b) 3 + 42 − 3 + 4 (c) 1622
123
(d) (−63)2
Solution:
(a) 3 + 4 − 4 + 5 = 3 − 4 + 4 + 5 = − + 9
(b) 3 + 42 − 3 + 4 = 3 + 4 + 42 − 3 = 7 + 42 − 3
(c) 16
22
123
= 4
3

(d) (−63)2 = (−63)(−63) = 3626

Expanding and factorizing algebraic expressions

Expanding an expression involves removing the brackets by multiplying each term inside the brackets by the
term outside the brackets.
( + ) = + and ( + )( + ) = + + + )
OR:
And

E.g. 2

Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 49
Factorising an expression involves finding the factors that when multiplied together would equal the initial
expression. Factorising is the opposite process to expanding.

E.g. 3: (a) 32 + 12 = 3( + 4) (b) 5( − 3) + 9( − 3) = ( − 3)(5 + 9)



Simplifying algebraic fractions

When dividing algebraic terms, write as a fraction first. Remember, cancelling/simplifying can ONLY occur
when the algebraic expressions are factorised (the terms of the expression are in a product not in a sum).

E.g. 4: (a)
8
3
× 6
5
= 16
5
; (b) 6
2
5
× 10
92
× 7

= 28
3
;
(c)
20(2−)
52
× 7
42(2−) = 223





















Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 50

2.1.2 Linear and Quadratic equations
Linear Equations in one variable
A few points to note:
• A linear equation is called a polynomial of degree 1.
• An equation has a LHS (left hand side) and a RHS (right hand side) that are equal (like a balanced
scale).
LHS = RHS

https://www.istockphoto.com/th/photo/simple-balance-scale-3d-rendering-gm537276346-95152075

• To solve for the unknown variable, you must isolate the variable
(usually on the LHS).
• When isolating the variable, the concept of opposite operations should be used. (‘+’ on one side
becomes ‘-‘ on the other side, ‘x’ opposite to ‘÷’, opposite to √ ).
This is equivalent to: ‘whatever you do to an equation, do the SAME thing to BOTH sides of that
equation.’
• A linear equation can have one solution, no solution or an infinity of solutions.

E.g. 1
(a) 6 + 7 = 13 + 5 (b) 5 + 2 = 2 + 6 (c) 2( + 1) = 2 + 2
6 − 5 = 13 − 7 2 − 2 = 6 − 5 2 + 2 = 2 + 2
= 6 0 = 1() True for any value of x
One solution No solution Infinity of solutions










Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 51

Quadratic Equations and methods of solving

General form of a quadratic equation: + + =

Note: An expression is called ‘quadratic’ because the highest power of the variable is 2. A quadratic is also
referred to as a polynomial of degree 2.

Methods for solving quadratic equations are:
• Factorising
• The quadratic formula

FACTORISATION
Expanding (product to sum)
(x + 3)(x + 4) = x² + 4x + 3x + 12
Factorising (sum to product)

Why factorise?
• For equations – easier to solve (if × = ⇒ = or = !!!!!)
• For fractions – to be able to simplify
E.g. 2: ( − 5)( + 2) = 0 ⇒ ( − 5) = 0; ( + 2) = 0 ∴ = 5; = −2 very easy! 

Some ways of factorising a quadratic expression are:
 Common factor
 Difference of two squares formula
 Monic and non-monic trinomials (three terms) (not covered in this course)

Common factor: + = ⇔ ( + ) = ⇒ = ; = −




E.g. 3 (a) 62 − 3 = 3(2 − 1) (b) −6 − 182 = −6(1 + 3) (c) 22 + 3 = (2 + 3)

(d) (2 − 1) − 3(2 − 1) = (2 − 1)( − 3) (e) 6( − 1) + 3( − 1) = 3( − 1)(2 + 1)


Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 52



Difference of two squares: − = ⇒ = ; = −

By expanding the product of a sum of two numbers with the difference of the same numbers, we notice a
pattern: ( + )( − ) = 2 − + − 2 = 2 − 2.
We can therefore use this pattern every time we have a difference of two squares:
2 − 2 = ( + )( − )
The formula can be applied when solving quadratic equations in this form:
 2 − 2 = 0 ⇔ ( + )( − ) = 0 ⇒ = ; = −
OR
 2 − 2 = 0 ⇔ 2 = 2 ⇒ = ±√2 ∴ = ; = −

E.g. 4 2 − 16 = 0 ⇔ ( + 4)( − 4) = 0 ⇒ = 4; = −4 OR 2 = 16 ⇒ = ±√16 ∴ = 4; = −4

E.g. 5 ( − 3)2 − 16 = 0 ⇔ ( − 3)2 = 16 ⇒ − 3 = ±√16 ⇒ − 3 = ±4 hence = 7; = −1

Q1: What are the solutions of equation: 2 + 2 = 0?
Your answer:


Q2: Expand: ( + )2 = ( + )( + ) =? then ( − )2 = ( − )( − ) =?.
What can you notice? Write your findings in the answer.
Your answer:










Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 53
QUADRATIC FORMULA
The Quadratic formula for solving the equation 2 + + = 0 is:

= −±√

; = − ( is called the discriminant)

∆< −
∆> −
∆= −

E.g. 6 Solve using the quadratic formula and leave your answer in exact form.
Solution:

1. 32 − 7 + 12 = 0 = 3; = −7; = 12
= (−7)2 − 4 × 3 × 12 = −95 < 0 ⇒ no real solutions

2. 2 − 7 + 5 = 0 = 1; = −7; = 5
= (−7)2 − 4 × 1 × 5 = 29 > 0 ⇒ two distinct real solutions
1,2 = −(−7)±√292×1 = 7±√292 ⇒ 1 = 7+√292 ; 2 = 7−√292

3. −42 + 12 − 9 = 0 = −4; = 12; = −9
= 122 − 4 × (−4) × (−9) = 0 ⇒ two identical solutions
1 = 2 = −122×(−4) = −12−8 = 32











Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 54

2.1.3 Problem solving using equations

In real life, equations are useful in solving various problems. The problem-solving exercise is in fact a
translation process where the problem given in English is being translated into Mathematical sentences
(equations), then solved using the methods learned/studied. Some of the steps to follow when solving
word problems are given below:


Note:
The key information in the problem is not necessarily only numerical. There are key words that
can make a significant difference to the solution if not understood correctly.
Not every word in the given problem is directly relevant to the problem-solving process. Some
words are there to describe the situation, for a better understanding of the real-life problem, like
setting the ‘scene’ in a story. The meaning of these words can be at times derived from the
context or at times be left out without impacting the ‘translation’ to the Mathematical process or
the overall understanding of the problem.

It is important to try to ‘visualise’ the real-life problem for a better interpretation of the answer
once the problem has been solved. Placing yourself ‘in the situation’ and becoming the ‘main
character’ in the problem will make it even more real .
Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 55





















https://ccmit.mit.edu/problem-solving/

1. Find the dimensions of a rectangle if the length is triple the width and the area is 75 cm2

Solution: = 3
× = 75 ⇒ 3 × = 75 ⇔ 32 = 75 ⇒ 2 = 25 ∴ = ±5

= −5 rejected (cannot have a negative width)
∴ = 5, = 15

2. The sum of the squares of two positive consecutive whole numbers is 313. Find the numbers.
Solution: two consecutive numbers can be written: , +

Therefore 2 + ( + 1)2 = 313 ⇔ 22 + 2 + 1 = 313

22 + 2 − 312 = 0/÷ 2 ⇒ 2 + − 156 = 0 ⇒ = −1±√625
2
= −1±25
2


= −13 < 0 rejected; = 12 ⇒ the numbers are 12 and 13.



Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 56


Exercises – Lesson 2.1.1

1. Simplify the following:
(a) 3 + 8 (b) 9 − 4 (c) 14ℎ + 6 − 12ℎ (d) 4 + + 4 − 3
(e) 14 + 2 × 7 − 8 (f) −2(4 − 7) + 2 − 5

2. Multiply/divide the following:

(a) −62 × 5 (b) 45 × 32 (c)3045 ÷ 1522 (d)(3)2

(e) (5)0 (f) 3 × 4 (g) (−32)(4)(−2)2 (h)(3)4 ÷ (2)3
(i)
4
3
× 6
8
(j)3

× 5
9
× 6ℎ
10


3. Expand then simplify:
(a) 5(2 + 4) + 3 (b) 6 − 4(2 − 3) (c) 9 − (3 − 2 + 5) (d) 2( + 3) + 4( − 1)
(e) 6(22 + 3) + 2( + 1) − 3(2 + 4) (f) ( + 7)(2 − 3) (g) ( + + 6)(2 − )
(h)( + 5)(2 − 3) − 2( + 4) + ( − 2)( − 3)

4. Factorise:
(a) 82 + 24 − 16 (b)7 − 49 (c) 12 + 152 (d) 22 − 7
(e)( + 2) + ( + 2) (f)4(3 − 5) + 2(3 − 5) (g)22 − 3 + 2 − 3
(h) 5 + 15 + 22 + 6


5. Factorise the following expressions:
(a) 2 − 36 (b) 2 − 92 (c) 492 − 2 (d) 92 − 162












Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 57
Exercises – Lesson 2.1.2

1. Solve the following equations:
(a) 6 + 7 = 13 + 7 (b) 13 − 4 = 1 − (c) −14 + 6 + 7 − 2 = 1 + 5
(d) 7 − 3 = 3 + 6 (e) −10 + + 4 − 5 = 7 − 5 (f) −8 + 4(1 + 5) = −6 − 14

(g) 7(5 − 4) − 1 = 14 − 8 (h) 8 + 4(4 − 3) = 4(6 − 4) + 4
(i) −3( − 1) + 8( − 3) = 6 + 7 − 5

2. The length of a rectangle is four metres longer than its breadth x. Write the perimeter and the area as a
function of x.


3. Niel is 6 years older than Sean, who is h years old, and Nikki is twice Sean’s age. The sum of their ages is 86.
Find the equation that shows this information, then solve to find their ages.

4. The sum of 3 consecutive numbers is 117. Find the largest of the numbers.

5. 3 consecutive odd numbers add to give 405. Find the smallest of the numbers.

6. I am thinking of a number that if multiplied by 6, and the result is added to 38, gives an answer of 230.
What number am I thinking of?


7. Tom earns four times Chris’ wage and Chris earns $17 more than Dawn. How many dollars does Tom earn
if their combined wages are $163?

8. The two equal sides of an isosceles triangle are given as (x+2)cm and the third is (x+6)cm. If the perimeter
of the triangle is 385cm, find the value of x.

9. A caramel chocolate is 11cents dearer than a milk chocolate. 3 milk chocolates and 2 caramel chocolates
cost 57 cents. What is the cost of a caramel chocolate in cents?















Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 58

Exercises – Lesson 2.1.3

1. Solve the following equations using the formulas found for ( + )2; ( − )2:
(a) 2 + 6 + 9 = 0 (d) 2 − 18 + 81 = 0 (e) 42 + 12 + 9 = 0 (h) 42 − 36 + 81 = 0

2. Solve the following equations:
(a) 2 − 121 = 0 (c) 42 − 36 = 0 (e) ( − 5)2 = 9 (g) 2
9
− 100 = 0

3. Solve the following equations using the quadratic formula:
(a) 2 + 6 + 8 = 0 (b) 2 + 8 + 12 = 0 (e) 22 − 11 = −5 (f) 2 − 5 + 6 = 0

4. Solve the following equations:
(a) 2 + 3 + 2 = 0 (e) 2 + 2 − 3 = 0 (f) 2 − 5 + 6 = 0 (g) 2 − 4 − 12 = 0

5. Solve the following equations:
(a) 22 + 13 + 15 = 0 (b) 72 + 11 − 6 = 0 (c) 32 − 2 − 1 = 0 (d) 22 − 5 + 3 = 0

6. The product of two consecutive integers is 1122. Find the numbers.


8. The sum of the squares of two numbers is 40. If the numbers differ by 4, find the numbers.


9. A ladder 4m long is placed against a wall so that it reaches 2m further up the wall than the foot of the
ladder is from the base of the wall. How far is the foot of the ladder from the base of the wall?
10. For what value of ‘k’ does the equation + + = have:
(a) no real solutions; (b) one double real solution; (c) two real solutions


12. A garden measuring 12m by 16m is to have a pedestrian pathway installed all
around it, increasing the total area to 285m2. What will be the width of the
pathway?









Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 59
Week 2 – Lesson 2

2.2.1- Linear Functions – patterns

A number pattern that increases or decreases by the same amount for each new term is called a linear
pattern.

Which of the following patterns are linear?
5, -2, -9, -16, … 4, 8, 16, 32, … 1, 2, 3, 5, …
6, 10, 14, 18, … 3, 13, 23, 33, … 24, 12, 6, 3, …

Consider the following matchstick pattern sequence:



Pattern Number (n) 1 2 3 4 … n … 140
No of Matchsticks (m) 6 11 16 21 … 5n+1 … 701

How?
• For each new pattern, we need extra 5 matches ( n goes up by 1, m = 5n).
• If n = 1, m = 6, therefore m = 5n+1 (adjusted value)

If a linear number pattern is graphed on a number plane, the graph is a straight line. That is
why the pattern is called ‘linear’. The graph below illustrates the number pattern in the
example, using the values from the table.



Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 60
2.2.2 – Coordinate Geometry

Coordinate Geometry provides the link between Algebra and Euclidean Geometry through graphs of
lines and curves in the coordinate plane (or Cartesian plane).

The invention of Cartesian coordinates in the 17th century by René Descartes (Latinized name: Cartesius),
revolutionized mathematics.

René Descartes was a French philosopher and mathematician, famous also for his saying:
“I think, therefore I am”.

The Coordinate plane has four quadrants and each point in the plane has two coordinates: x = horizontal
coordinate and y = vertical coordinate (x,y).




• To graph a linear number pattern, we need a minimum of two points in the coordinate plane.
• All the points on a line satisfy a linear relationship between variables x and y.
• Two points in the coordinate plane determine a unique line, with a unique relationship between
the x and y coordinates.




Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 61

E.g. 1: Given the pattern below find the linear relationship between the two variables.

19,16,13,10,7,4,…

Solution:
The pattern can be presented in a table:
x 1 2 3 4 5 6
y 19 16 13 10 7 4





= −3 + 22


E.g. 2: Given the linear relationship = 4 − 3, graph the line on the coordinate plane.

Solution: To graph a line we need a minimum of two points:

Let = 0 ∴ = 4 × 0 − 3 = −3
= 1 ∴ = 4 × 1 − 3 = 1





E.g. 3: Determine if point (8,-4) lies on the line of equation − 2 = 5.

Solution: Substitute = 8, = −4 in the equation: 8 − 2 × (−4) = 5 ⇒ 16 = 5() ∴the point is
not on the line.


x 0 1
y -3 1
-3 -3 -3 -3 -3
Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 62

2.2.3 - Distance, Midpoint, Gradient

The distance between two points of given coordinates can be found by applying Pythagoras’ Theorem.


NEW CENTURY MATHS 10

The coordinates of the midpoint (in the middle) of the interval between two given points are the
averages of the coordinates of the two points.


NEW CENTURY MATHS 10

The gradient m of a line is the slope of the line. =





NEW CENTURY MATHS 10



Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 63




E.g. 1: A(-8,5) and B(3,-2) are two points in the coordinate plane. Find:
(a) The distance between A and B.
(b) The midpoint between A and B.
(c) The gradient of line AB.
Solution:
(a) = �(−8 − 3)2 + �5 − (−2)�2 = �(−11)2 + (7)2 = √170 ≃ 13.04
(b) �−8+3
2
, 5−2
2
� = �−5
2
, 3
2

(c) = 5−(−2)
−8−3
= 7
−11
= − 7
11
downward slope

Note: In this example we considered (2,2), (1,1)and substituted in the relevant formulas.
Therefore: 2 = −8,2 = 5 and 1 = 3,1 = −2. The answer would be the same if the two would
be swapped.

E.g. 2: Prove that the points A(-3,-2), B(0,7) and C(3,1) form an isosceles triangle.
Solution:
= �(−− ) + (− − ) = √ + = √
= �( − ) + ( − ) = √ + = √
= �(− − ) + (− − ) = √ + = √ = ∴△ is isosceles.
Ascending line (m>0) Descending line (m<0)
Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 64
• The Two-Point Formula: 1 2 1
1 2 1
y y y y
x x x x
− −
=
− −
, where 1 1( , )x y and 2 2( , )x y are two points
on the line.

(this formula is obtained by substituting the gradient into the point-gradient form)

Linear Relations and Graphs

Linear Relations can be expressed in different forms:

• The General Form: + + = 0, where , , are integers and is positive

(all the terms to the left of the equality, 0 only on the right; sometimes this form is neater and more
convenient to use).



• The Gradient-Intercept Form: = + , where is the gradient and is the y-
intercept.


• The Point-Gradient Form: − 1 = ( − 1), where is the gradient of a line which
passes through the point(1,1).









E.g. 1: Find the gradient and the y-intercept of the lines with the equations:
(a) = 8 − 4 (b) = 3+1
4
(c) − 2 = 9
Solution: Write the equations in the gradient-intercept form = + .
(a) = −4 + 8 implies = −4 and = 8
(b) = 3
4
+ 1
4
implies = 3
4
and = 1
4

(c) = 2 + 9 implies = 2 and = 9

E.g. 2: Write equation of the line:
(a) with gradient 6 and the y-intercept -4
(b) with gradient -3 and passing through the point of coordinates (1,-2)
(c) passing through the points of coordinates (1,1) and (0,3)
Solution:
(a) = 6; = −4 ⇒ = 6 − 4 (using the gradient-intercept form)
(b) Using the point-gradient form: − (−2) = −3( − 1) ⇔ = −3 + 1
(c) Using the two-point formula:
−1
−1
= 3−1
0−1

−1
−1
= 2
−1
⇔ = −2 + 3 (when (1,1) is (1,1) and (0,3) is (2,2))
OR −3
−0
= 1−3
1−0

−3

= −2
1
⇔ = −2 + 3 (when (0,3) is (1,1) and (1,1) is (2,2))
SAME answer!
Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 65

Parallel and perpendicular lines

Two lines 1 and 2 have the gradients 1 and 2 respectively.
The following statements are true:

IF ∥ THEN = OR IF = THEN ∥


IF ⊥ THEN × = − OR IF × = − THEN ⊥


E.g. 3: Determine if the following pairs of gradients represent parallel lines, perpendicular lines or neither:
(a) 1 = 3,2 = −3 (b) 1 = −2,2 = 12 (c) 1 = 1,2 = 1 (d) 1 = 23 ,2 = 32

Solution:
(a) neither (b) perpendicular (1 × 2 = −2 × 12 = −1) (c) parallel (1 = 2 = 1) (d) neither

E.g. 4: A line joins A(4,-1) to B(2,3) and M is the middle of AB. Find:
(a) the coordinates of M;
(b) the equation of the perpendicular on AB in M;
(c) the gradient of any line parallel to AB.

Solution:
(a) = 4+22 = 3; = −1+32 = 1,∴ (3,1)
(b) = 3−(−1)2−4 = −2 ⇒ ⊥ = 12 ( × ⊥ = −1); M on the line, therefore, using the point
gradient formula, the equation of the perpendicular on AB in M is:
− 1 = 12 ( − 3) ⇔ = 12 − 12
(c) = −2 ⇒ ∥ = −2 (same gradient)


E.g. 5: Which of the following lines is perpendicular to =
3
− 2 ?
(i) = 3 − 2 (ii) = −3 + 2 (iii) 3 + − 7 = 0

Solution:
= 13 (i) = 3 (ii) = −3 (iii) = −3 Therefore (ii) and (iii) are perpendicular on the
given line (−3 × 1
3
= −1)





Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 66
Exercises – Lesson 2.2.3


1. Find the formula for each of the following tables of values:

(a) x 0 1 2 3
y 7 9 11 13

(b) a 1 2 3 4
b 16 12 8 4

(c) x 0 2 4 6
y 5 9 13 17

2. Determine if the point (1, -2) lies on the lines of equations:
(a) = − 3 (b) = −2 (c) + = 1 (d) 2 − − 4 = 0

3. Given the points A(2,-1), B(3,4), C(-2,2), D(-5,-6), answer the following questions:
(a) What quadrants do the points belong to? Find those points on the number plane.
(b) Find the exact value of the length of the intervals AB, BC, CD, AD.
(c) Find the coordinates of the midpoints of the intervals AB, BC, CD.
(d) Find the gradients of AC, BD, AD and specify if the lines are ascending or descending.

4. Find the coordinates of the centre of the circle if the endpoints of a diameter are:
(a) A(-2,9), B(8,5) (b) E(3,11), F(9,9) (c) C(-5,6), D(7,-10)

5. If the midpoint of (a,b) and (9,9) is (6,2), what are the values of a and b?

6. Arrange each of the following equations in the form y = mx + c, and hence write down the gradient and y -
intercept. Sketch the line that represents each equation.
(a) 3y = 2x+ 9 (b) y + 3x + 6 = 0 (c) 3y - 7x = 10
(d) -12x= 7y +14 (e) 12 y = 9x (f) y + 9 = 0

7. Find the equation of the straight line that passes through the points (2, 3) and (-1, 4).

8. For each of the following, find the gradient of AB and PQ and then determine whether AB is parallel or
perpendicular to PQ.

(a)
(b)
(c)
(d)

9. A straight line passes through the point (3, 4) and is perpendicular to the line joining the points
(-2, 3) and (1, 2). Find the equation of the line, giving the answer in the form ax + by + c = 0.



A B P Q
(4,5) (8,6) (9,1) (8,5)
(-5,0) (-2,9) (1,7) (5,19)
(-2,11) (2,3) (1,9) (-3,17)
(1,3) (-2,1) (5,5) (3,8)

Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 67

Week 2 – Lesson 3

2.3 Applications of Coordinate Geometry

Coordinate Geometry can be used to prove different properties of geometric figures like triangles,
quadrilaterals, and circles as well as to prove various conjectures.

Also, by using the coordinate system, we can represent real world and mathematical problems by
graphing points in the coordinate plane and interpret coordinate values of points in the context of the
situation.

E.g.1: The points A(3,6), B(-6,3) and O(0,0) are the vertices of triangle and M is the middle of AB.
(a) Show that triangle ABC is isosceles.
(b) Find the coordinates of M.
(c) Prove that OM is perpendicular to AB.
(In an isosceles triangle, the perpendicular on the base is also a median.)


Solution:

(a)

= �(3 − 0)2 + (6 − 0)2 = √45
= ⇒△

(b)

(c) = 92−0−3
2
−0
= −3; = 6−33−(−6) = 13
× = −3 × 13 = −1 ⇒ ⊥














2 2( 6 0) (3 0) 45OB = − − + − =
3 6 6 3 3 9, ,
2 2 2 2
M M− +   ⇒ −   
   
Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 68
Exercises – Lesson 2.3


1. Prove that A(0,1), B(3,4), C(5,2) form a right triangle in two ways:
(a) using gradients of lines
(b) using distances and Pythagoras’ Theorem.

2. Show that the points A(4,3), B(0,2) and C(-4,1) are collinear (lie on the same line).

3. The points A(-2,-3), B(-5,2), C(0,4) and D(3,-1) are the vertices of a quadrilateral. Prove that:
(a) ABCD is a parallelogram (2 pairs of parallel sides);
(b) the diagonals AC and BD bisect each other;
(c) the opposite sides are equal (AB=CD and BC=AD).
(In a parallelogram, the opposites sides are equal and the diagonals bisect each other.)

4. The points A(-4,5), B(1,4), C(2,-1) and D(-3,0) are the vertices of a quadrilateral. Prove that:
(a) ABCD is a rhombus (parallelogram with two adjacent sides equal);
(b) the diagonals AC and BD are perpendicular;
(The diagonals of a rhombus are perpendicular.)

5. A(-3,-6), B(1,10) and C(7,2) are the vertices of a triangle.
(a) Find the coordinates of the midpoint M, of AB.
(b) Find the coordinates of the midpoint N, of AC.
(c) Show that the length of MN is half of BC.
(In a triangle, the line joining the midpoints of two sides is half the length of and parallel to the third side.)

6. A The coordinates of a lighthouse are L(2,0). A boat A is sailing towards the lighthouse on a straight line of
equation 1: 1
2
l y x= − + . A second boat B was anchored at point B(0,-4).
(a) If boat B starts moving towards the lighthouse, find the equation k of the line that connects B to L.
(b) How are the directions of movement of the two boats (k and l) compared to each other?
(c) Prove that boat A will reach the lighthouse by keeping its course l.
(d) Find the exact distance BL between the lighthouse and the second boat.













Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 69
Week 2 – Lesson 4

2.4 Simultaneous Equations

Solving simultaneous equations means finding solutions that satisfy all the equations in the set.
Simultaneous equations can only be solved when the number of unknowns equal the number of
equations in the given set.

In this unit, only sets of two equations with two unknowns will be considered.

When considering two linear equations simultaneously, the following situations can occur:

 The two lines meet (intersect) in a point (unique) one solution
 The lines never meet (are parallel) no solution
 The lines overlap (co-incident) infinite solutions (all the points on the line)

E.g. 1: + = −3
3 − = −1 A: one solution the point of coordinates (-1,-2)
E.g. 2: + 2 = 5
+ 2 = 9 A: no solution (the lines are parallel and distinct)
E.g. 3: + = 3
2 + 2 = 6 A: infinite solutions (the lines are identical)


Solving Simultaneous Equations

Simultaneous equations can be solved using different methods. Three of those methods are:
1. Graphical method – graph the lines and find the point of intersection.
2. Elimination method – eliminate one variable to reduce the set to one equation with one variable
then solve.
3. Substitution method – express one variable as a function of the other and reduce the set to one
equation with one variable, then solve.

Let’s consider two linear equations and solve them simultaneously using the three methods listed:
+ = −3 and 3 − = −1





Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 70

1. GRAPHICAL METHOD
+ = − − = −
x 0 1
y -3 -4

Graph the two lines and find the point of intersection.

From the graph, the lines intersect at point of
coordinates (-1, -2)

Therefore, the solution of the simultaneous
equations is:
= −, = −




2. ELIMINATION METHOD
+ = −3 (1) The aim is to eliminate one variable, x or y. In this case, y is easier to 3 − = −1 (2) eliminate as the coefficients of y, in both equations, are equal in
absolute value.

(1) + (2): + 3 = −3 − 1 ⇒ 4 = −4 ∴ = −1
Substitute x in equation (1): −1 + = −3 ∴ = −2
Solution: point of coordinates (-1, -2)

3. SUBSTITUTION METHOD
+ = −3 (1) Express one variable as a function of the other using one equation then
substitute it into the second equation. 3 − = −1 (2)

Substitute y from (1): = − − 3 into (2): 3 − (− − 3) = −1 then solve for x 3 + + 3 = −1 ⇒ 4 = −1 − 3 ⇒ 4 = −4 ∴ = −1
Solution: point of coordinates (-1, -2)

x 0 1
y 1 4
Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 71

A few points to note:

• The Graphical method gives a visual representation of the solution but is very limiting as it is
based on measurements (involve error). The solution can only be clearly determined, if the
coordinates are whole numbers of reasonable value, otherwise the solution will be approximate.

Graphing often takes more time than the other methods. This method is suited for break-even
analysis, that will be studied separately.

• The Elimination and Substitution methods are algebraical methods that will always find the
accurate solution for the simultaneous equations.

• The Substitution method is best used when one (or both) of the equations is already solved for
one of the variables. Also, it works well when one of the variables has a coefficient of 1.

• The Elimination method is best used when the equations are in standard form (ax+by=c) and
when all of the variables have a coefficient other than 1.

• Always check your answer by substituting it into both equations to make sure it is correct!

























Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 72

Exercises – Lesson 2.4

1. Check if the point of coordinates (1, -2) is a solution of the given sets of simultaneous equations:
(a) + = 3 (b) − = 3 (c) 2 − = 7 (d) 4 − 3 = 10
−2 + = −4 4 + = 2 5 + 3 = −1 − + 5 = −11

2. Solve using the graphical method:
(a) = + 3 (b) − = 4
= 3 − 5 + = 2

3. Solve using the elimination method:
(a) − + = 7 (b) 3 + = 13
+ 3 = 5 + = 7
4. Solve using the substitution method:
(a) −2 + = 2 (b) − 3 = 2
3 + 5 = 10 4 + 5 = 25

5. Solve using the method of your choice:
(a) − − 2 = −4 (b) −2 + = 1
1
4
+ 3 = 1 − 2 = −5

6. Solve the following sets of simultaneous equations:
= 2 + 1
= 2 − 2 − 20

(solving a set of simultaneous equations (one linear and one quadratic) is similar to finding the points of
intersection of a line (linear) with a parabola (quadratic))












Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 73

Week 2 – Lesson 5

2.5.1 Applications of Simultaneous Equations – Breakeven analysis

In Algebra, the breakeven point is the point where two linear functions intersect, or the solution of two
linear equations solved simultaneously.
In Marketing/Business/Accounting, the breakeven point represents the point where the revenue equals
the cost, or where a product neither makes a profit nor incurs a loss.

The purpose of the breakeven analysis is to calculate the amount of sales required for the revenue to
equal cost.

In very simple terms:
Cost = fixed cost (rent, machinery, etc.) + variable cost (cost per unit: material)
Revenue = income from the product (units sold)
Profit = Revenue – Cost Loss = Cost - Revenue

E.g. Beverly is the accountant in charge of a large furniture factory’s production line. She wants to
measure how many units they will have to produce and sell to cover their expenses (to breakeven). How
many units do they have to produce to make a profit of $150,000?
The following costs apply:
Total fixed cost: $50,000
Variable cost per couch: $300
Sale price per couch: $500

Solution: Let x = number of couches produced and y = money paid/received
- the cost function: () = = 50,000 + 300
- the revenue function: () = = 500
- breakeven point: () = () ⇒ 500 = 50,000 + 300 (equivalent to solving the
simultaneous equations above, algebraically by substitution)

Therefore: = 250 couches needed to have a cost = revenue of: = $125,000
To make a $150,000 profit: () − () = 150,000 500 − (50,000 + 300) = 150,000 ⇒ 200 = 200,000 ∴ = 1,000 couches




Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 74


Graphical solution:
cost function: = 50,000 + 300 revenue function: = 500
x 0 100
y 50,000 80,000







• For < 250 ⇒Revenue < Cost (the factory makes a loss)
• For = 250 ⇒Revenue = Cost = $125,000
• For > 250 ⇒Revenue > Cost (the factory makes a profit)
• For = 1000 ⇒Profit = $150,000

A few points to note:

• The Breakeven analysis is a simple planning aid rather than a decision-making tool. Although it is
practical and popular for many businesses, mainly start-ups, it has limitations.
• Can you think at some advantages and disadvantages of this analysis?




x 0 100
y 0 50,000
Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 75

2.5.2 Problem solving using simultaneous equations

Simultaneous equations can be used to solve many real-life problems.
Depending on the problem:
• Highlight the important information in the problem;
• Identify and define the variables (unknowns);
• Write the equations (relationships between the variables);
• Use one of the methods for solving the set of simultaneous equations;
• Ensure the results found make sense for the real-life problem;
• Check your answers by substituting the results in the original equations;
• Answer the questions in complete sentences.

E.g. 1. Today, the sum of the ages of Isabel and her mum is 52. Two years ago, Isabel’s age was a third of
her mum’s age. (a) How old are Isabel and her mum today? (b) After how many years will the mum’s age
be double the age of Isabel?
Solution:
• let x = Isabel’s age today and y = the mum’s age today
• + = 52 + = 52
− 2 = −2
3
or 3 − = 4
• Using the method of elimination, we get: = 14; = 38
• Check if it makes sense: > 0; > 0, < 
• Test the solution: 14 + 38 = 52 
3 × 14 − 38 = 4
• (a) Answer: Today, Isabel is 14 years old and her mum is 38 years old.
• (b) Answer: 38 + = 2(14 + ) (where n = number of years) ⇒ = 10
10 years until the mother’s age is double the daughter’s age

E.g. 2. In a math test there are 40 questions. For each correct answer, a student receives 2 marks, and for
each incorrect answer, there is a half mark penalty.
(a) Alex had 28 correct answers and 12 wrong answers. What is his mark?
(b) How many wrong and correct answers did he have if his mark was 55?

Solution: let x = number of correct answers; y = number of wrong answers
(a) 2 × 28 − 0.5 × 12 = 50 marks
(b) + = 40
2 − 0.5 = 55 A: = 30; = 10 (both positive )

Test the solution , then give the answer: ‘Alex had 30 answers correct and 10 wrong’.
Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 76



E.g. 3. On a farm there are 130 pigs, chickens and ducks. There are 268 legs altogether.
(a) How many pigs and birds are there on the farm? (b) If there are 5 times more chickens than ducks,
how many of each type are there?

Solution: let x = number of pigs; y = number of chickens and ducks;
c = number of chickens; d = number of ducks
(a) + = 130
4 + 2 = 268 A: = 4; = 126 (both positive )

Test the solution , then give the answer: ‘There are 4 pigs and 126 birds’.

(b) = 5
+ = 126 A: = 105; = 21 (both positive )

Test the solution , then give the answer: ‘There are 105 chickens and 21 ducks’.


E.g. 4. The telephone company offers two plans. With plan A, you can make an unlimited number of local
calls per month for $18.50. With plan B, you pay $6.50 monthly, plus 10 cents for each minute of calls
after the first 40 min.
(a) How many minutes would you have to use the telephone for the two plans to cost the same? (b) At
least how many minutes would you have to use the phone each month to make plan A cheaper?

Solution: let x = number of minutes used; y = total bill;
Plan A: = 18.50
Plan B: = 6.50 + 0.10( − 40)
(a) Same cost: 6.50 + 0.10( − 40) = 18.50 ⇒ = 160minutes (substitution method).
A: ‘The two plans cost the same if 160 minutes are used per month’.
(b) Draw the cost functions for both plans and see which line is above (when x>160)
OR try 161 minutes for plan B and see which plan is cheaper
A: ‘If more than 160 minutes are used every month, plan A is cheaper than plan B’.







Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 77

Exercises – Lesson 2.5

1. The sum of two numbers is 80 and the difference between double the first and triple the second is 35. Find
the numbers.

2. A truck full weighs 3500kg and half full, it weighs 2950kg. What is the weight of the empty truck?

3. Find two whole numbers if their sum is 102 and by dividing one to the other, the quotient is 2 and the
remainder 12.

4. Ravi has $200 in notes of $10 and $5. If there are 28 notes altogether, find how many of each Ravi has.

5. In a building, there are 2 bedrooms and 3-bedroom apartments. In total there are 38 apartments and 92
bedrooms. How many 2 bedrooms and 3-bedroom apartments are there?

6. A taxi fare consists of a fixed charge which every passenger must pay and also an extra cost depending on
the distance travelled. A person travelling 10km pays $28. Another person travelling 15 km pays $39.50.
Find the fixed charge and the extra cost per km.

7. The age of the father is 3 years more than 3 times the son’s age. In 3 years’ time, the age of the father will
be 10 years more than twice the age of the son. Find their present age.

8. How many litres of 30% alcohol solution and how many of 60% alcohol solution must be mixed to produce
72 litres of 50% solution? (30% alcohol solution means 30% of the solution is alcohol)

9. It takes 3 hours for a boat to travel 43 km upstream. The same boat can travel 48 km downstream in 2
hours. Find the speeds of the boat and the current, rounded to the nearest unit.

10. Mr. Ates invests his savings in two term deposits A and B. The total money invested is $15,500. He gets 3%
interest income on deposit A and 5% interest income on deposit B. If the total income earned is $649, find the
amount invested in A and B separately.
















Macquarie University College Foundation Program Numeracy Skills for Everyday Life (WFNS0001) 209
Lesson 2.1.1:
1. (a) 11y (b) 5t (c) 2h+6 (d) 8C-2d (e) 14+6xy
(f) -13x+16y
2. (a) -30a3 (b) a7b7 (c) 2a2b3 (d) m6 (e) 1 (f) a7x
(g) -48x5 (h) x6 (i) xy (j) ahmst
3. (a) 10a+23 (b) -2x+12 (c) 6x+2y-5 (d) 6x+2
(e) 11a2+2a+6 (f) 2x2+11x-21 (g) 2x-4y-xy-y2+12
(h) x2-6x-9
4. (a) 8a(a+3b-2) (b) 7n(m-7p) (c) 3x(4y+5x) (d)
xy(xy-7) (e) (a+2)(a+b) (f) 2(3b-5d)(2a+1)
(g) (2x-3)(x+y) (h) (a+3b)(5+2a)
5. (a)( + )( − ) (b)( + )( − )
(c)( + )( − ) (d)( + )(− )

Lesson 2.1.2:
1. (a) -6 (b) 4 (c) -8 (d) 6 (e) -1 (f) -1 (g) 1
(h) all real numbers (i) 7
2. 4x+8; x(x+4) 3. 4h+6=86; 20, 26, 40 yrs old 4.
40 5. 133 6. 32 7. $120
8. 125 9. 18 cents


Lesson 2.1.3:
1. (a) -3,-3 (d) 9,9 (b) -3/2,-3/2 (c) 9/2,9/2 2.
(a) -11,11 (c) -3,3 (e) 2,8 (g) -30,30 3. (a) -2, -
4 (b) -2,-6 (c) 2,-4 (d) 7,-5 (e) 5,1/2 (f) 3,2 (g)
7/2,3/2 (h) 8,-7
4. (a) -1,-2 (b) -2,-6 (c) -7,-4 (d) -7,-2 (e) -3,1 (f) 2,3
(g) 6,-2 (h) 8,-7
5. (a) -3/2,-5 (b) -1,-5/2 (c) -2,3/7 (d) -2/3,1/4 (e) -
3,-1/3 (f) 1,-1/3 (g) -1,1/5 (h) 1,3/2
6. (33,34),(-34,-33) 7. (11,13), (-13,-11) 8.
(6,2); (-2,-6) 9. 1.6m(2sf) 10. (a) -22,2 (c) k<-2 or k>2 11. (a) m>25/4 (b) 25/4 (c)
m<25/4 12. 1.5m


Lesson 2.2.3:
(a) y=2x+7 (b) b=-4a+20 (c) y=2x+5
2. (a) Y, (b) Y, (c) N, (d) Y 3. (a) Q4, Q1, Q2, Q3
(b) √26,√29,√73,√74
(c) �5
2
, 3
2
� , �1
2
, 3� , �− 7
2
,−2� (d) −3
4
, 5
4
, 5
7
4.
(a) (3,7), (b) (6,10), (c) (1,-2) 5. (3,-5)
6. (a) y=2/3x+3 (b) y=-3x-6 (c) y=7/3x+10/3
(d) y=-12/7x-2 (e) y=3/4x (f) y=-9
7. x + 3y = 11



Lesson 2.2.3:
8.
grad(A,B) grad(P,Q)
(a) 1/4 -4 Perpendicular
(b) 3 3 Parallel
(c) -2 -2 Parallel
(d) 2/3 -3/2 Perpendicular
9. 3x - y - 5 = 0


Lesson 2.4:
1. (a) no (b) yes (c) no (d) yes 2. (a) (4,7) (b) (3,-1)
3. (a) (-4,3) (b) (3,4) 4. (a) (0,2) (b) (5,1) 5. (a)
(4,0) (b) (1,3) 6. (-3, -5) ;(7,15)


Lesson 2.5:
1. 55,25 2. 2400kg 3. 72,30 4. 12 of $10; 16 of
$5 5. 22(2br) 16(3br) 6. $5, $2.30/km 7. 33,10
8. 24L; 48L 9. 5km/h; 19km/h 10. $6300;
$9200


Lesson 3.1:
1. R; [5,+ꚙ) 2. [1,+ꚙ); [0,+ꚙ) 3. R\{6}; R\{0} 4. (a) 5
(b) 33 (c) 5 (d) − +
(e) ( + ) − ( + ) + 5. (a) N (b) Y (c) Y 6.
(a) Y (b) N (c) Y (d) N (e) Y (f) N
7. (a) {1,2,3,4}; {2,4,6,8,10,12,14}; {2,4,14}
(b) {a,b,c,d}; {0,2,7,9,11}; {0,2,7,9}
8. (a) (-2,0);(0,4) (b) (-2,0);(0,2) (c) (1,0);(0,1.5)
9. (a) N (b) Y, {-3,-2,-1,0,1,2,3}; {0,1,4,9} (c) N
10. 1; 2a2


Lesson 3.2:
1. (a) ¾, ascending (b) -1/2, descending (c) -3/4,
descending (d) 0, constant
2. (a) (0,1), (1/3,0) (b) () = − +
(c) () =

+ 3. (i) (a) (3/4,1/8)
(b) (1/6,-25/12) (c) (2,1) (d) (3/2,5/4) (ii) (a)
down (b) up (c) down (d) down (iii) (a) (0,-1),
(1/2,0), (1,0) (b) (0,-2), (-2/3,0), (1,0) (c) (0,-3),
(1,0), (3,0) (d) (0,-10), (1,0), (2,0) (iv) (a) x=3/4
(b) up (c) down (d) down 4. (a) 9 (b) -49/8 (c) 6
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