Equations - web2.hunterspt-h.schools.nsw.edu.auweb2.hunterspt-h.schools.nsw.edu.au/studentshared... · TERMINOLOGY 3 Equations Absolute value: the distance of a number from zero on

TERMINOLOGY

3 Equations

Absolute value: the distance of a number from zero on a number line.

Equation: A mathematical statement that has a pronumeral or unknown number and an equal sign. An equation can be solved to fi nd the value of the unknown number e.g. 2 3 5x - =

Exponential equation: Equation where the unknown pronumeral is the power or index e.g. 2 8x

=

Inequation: A mathematical statement involving an inequality sign, , , or1 2 # $ that has an unknown

pronumeral that is solved to fi nd values that make the statement true e.g. 2 3 4x 2-

Quadratic equation: An equation involving x2 as the highest power of x that may have two, one or no solutions

Simultaneous equations: Two or more independent equations that can be solved together to produce a solution that makes each equation true at the same time. The number of equations required is the same as the number of unknowns

ch3.indd 94 7/24/09 10:18:05 PM

95Chapter 3 Equations

INTRODUCTION

EQUATIONS ARE FOUND IN most branches of mathematics. They are also important in many other fi elds, such as science, economics, statistics and engineering. In this chapter you will revise basic equations and inequations. Equations involving absolute values, exponential equations, quadratic equations and simultaneous equations are also covered here.

DID YOU KNOW?

Algebra was known in ancient civilisations. Many equations were known in Babylonia, although general solutions were diffi cult because symbols were not used in those times.

Diophantus , around 250 AD, fi rst used algebraic notation and symbols (e.g. the minus sign). He wrote a treatise on algebra in his Arithmetica , comprising 13 books. Only six of these books survived. About 400 AD, Hypatia of Alexandria wrote a commentary on them.

Hypatia was the daughter of Theon, a mathematician who ensured that she had the best education. She was the fi rst female mathematician on record, and was a philosopher and teacher. She was murdered for her philosophical views by a fanatical Christian sect.

In 1799 Carl Friedrich Gauss proved the Fundamental Theorem of Algebra: that every algebraic equation has a solution.

PROBLEM

The age of Diophantus at his death can be calculated from this epitaph:

Diophantus passed one-sixth of his life in childhood, one-twelfth in youth, and one-seventh more as a bachelor; fi ve years after his marriage a son was born who died four years before his father at half his father’s fi nal age. How old was Diophantus?

Simple Equations

Here are the four rules for changing numbers or pronumerals from one side of an equation to the other.

If a number is • added, subtract it from both sides If a number is • subtracted, add it to both sides If a number is • multiplied, divide both sides by the number If a number is • divided, multiply both sides by the number

Do the opposite operation to take a number to the

other side of an equation.

ch3.indd 95ch3.indd 95 8/11/09 10:59:40 AM8/11/09 10:59:40 AM

96 Maths In Focus Mathematics Extension 1 Preliminary Course

EXAMPLES

Solve 1. x3 5 17+ =

Solution

x

x

x

x

x

3 5 17

3 5 17

3 12

3 12

4

5 5

3 3

+ =

+ =

=

=

=

- -

You can check the solution by substituting the value into the equation.

3 5

( )

x

3 4 5

12 5

17

LHS

RHS

= +

= +

= +

=

=

Since , 4xLHS RHS= = is the correct solution.

2. 4 3 8 21y y- = +

Solution

y y

y y

y

y

y

y

y

y

y y

4 3 8 21

4 3 8 21

3 4 21

3 4 21

24 4

24 4

6

6

4 4

21 21

4 4

`

- = +

- = +

- = +

- = +

- =

-=

- =

= -

- -

- -

3. x x2 3 7 6 1+ = - -] ]g g Solution

( ) ( )x x

x x

x

x x

x

x x

2 3 7 6 1

6 14 6 1

7

6 14 7

7 14 7

+ = - -

+ = - +

= -

+ = -

+ =

+ +

Check these solutions by substituting them into the equation.

ch3.indd 96 5/20/09 11:54:44 AM


x

x

x

x

7 14 7

7 7

7 7

1

14 14

7 7

+ =

= -

=-

= -

- -

1. t 4 1+ = -

2. . .z 1 7 3 9+ = -

3. y 3 2- = -

4. . .w 2 6 4 1- =

5. x5 7= -

6. 1.5 6x =

7. 35y = 1

8. 7

5b=

9. n28

- =

10. 6 3

2r=

11. y2 1 19+ =

12. 33 4 9k= +

13. d7 2 12- =

14. x2 5 27- = -

15. y

34 9+ =

16. x2

3 7- =

17. m5

7 11+ =

18. x3 5 17+ =

19. a4 7 21+ = -

20. y7 1 20- =

21. b8 4 36- = -

22. 3( 2) 15x + =

23. ( )a2 3 1 8- + =

24. t t7 4 3 12+ = -

25. 3 6 9x x- = -

26. 2( 2) 4 3a a- = -

27. 5 2 3( 1)b b+ = - -

28. 3( 7) 2(2 9)t t+ = -

29. ( ) ( )p p p2 5 1 5 2+ - = - -

30. . . . .x x3 7 1 2 5 4 6 3+ = -

3.1 Exercises

Solve

A STARTLING FACT!

Half full half empty

full empty`

=

=

ch3.indd 97 5/20/09 11:54:46 AM


Equations involving fractions

There are different ways to solve this type of equation. One way is to multiply both sides of the equation by the common denominator of the fractions.

EXAMPLES

Solve

1. m3

421

- =

Solution

2

2

( )

m

m

m

m

m

m

m

34

21

34

2 24 3

2 24 3

2 27

2 27

227

13

6 6 6

24 24

2 2

- =

- =

- =

- =

=

=

=

=

+ +

1

1

c cm m

2. x x3

14

5++ =

Solution

(5)

( )

12 12 12

x x

x x

x x

x x

x

x

x

x

x

31

45

31

44 1 3 60

4 4 3 60

7 4 60

7 4 60

7 56

7 56

8

4 4

7 7

++ =

++ =

+ + =

+ + =

+ =

+ =

=

=

=

- -

c cm m

Multiply by the common denominator, 6.

The common denominator of 3 and 4 is 12.

ch3.indd 98 5/20/09 11:54:46 AM


3. y y

51

3

2

65+

--

=

Solution

4

( ) ( )

y y

y y

y y

y y

y

y

y

y

y

51

3

2

65

51

3

2

65

6 1 10 2 25

6 6 10 20 25

4 26 25

4 26 25

4 1

4 1

30 30 30

26 26

4 4

+-

-=

+-

-=

+ - - =

+ - + =

- + =

- + =

- = -

-=

-

=

- -

- -

1

e e co o m

When there is a fraction on either side of the equation, multiplying by the common denominator is the same as cross multiplying.

The common denominator of 5, 3

and 6 is 30.

EXAMPLES

1. Solve ( )x x538 0!=

Solution

=

xx

x

x

538

8 15

8 15

187

8 8

=

=

=

2. Solve 0n

n58

23

!= ^ h

Solution

nn

n

n

58

23

16 15

16 15

1615

16 16

=

=

=

=

ch3.indd 99 5/20/09 11:54:47 AM


1. b5 3

2=

2. ( )x x751 0!=

3. ( )y y4109 0!=

4. 45

711x

=

5. ( )k

k54

29 0!=

6. x3

4 8- =

7. 45

43t

=

8. 7

572x+

=

9. y

2 53

= -

10. x9 3

2 7- =

11. 2

3 5w -=

12. t t52

32- =

13. x4 2

1 4+ =

14. x x5 2 10

3- =

15. 3

42

1x x++ =

16. p p

23

32

2-

+ =

17. t t7

33

1 4++

-=

18. x x9

55

2 1+-

+=

19. q q

31

42

2-

--

=

20. x x5

3 22

7++ =

+

21. b b43

51

2- =

22. a3 4

385

+ =

23. x x2

5 3+

= ,x 0 2! -^ h

24. y y1

13 1

1+

=-

,y 131

! -c m

25. t t3

24

1 0-

++

= ,t 3 4! -^ h

3.2 Exercises

Solve

Substitution

Sometimes substituting values into a formula involves solving an equation.

Investigation

Body mass index (BMI) is a formula that is used to measure body fatness and is used by health professionals to screen for weight categories that may lead to health problems.

ch3.indd 100 5/20/09 11:54:47 AM


This is not the only measure that is used when looking for health problems, however. For example, there are other factors in cardiac (heart) disease. Research these to fi nd out what other things doctors look for.

The BMI is used in a different way with children and teens, and is taken in relation to the child’s age.

The formula for BMI is hwBMI

2= where w is weight in kg and h is height

in metres.

For adults over 20, a BMI under 18.5 means that the person is underweight and over 25 is overweight. Over 30 is obese.

The BMI may not always be reliable in measuring body fat. Can you think of some reasons?

Is it important where the body fat is stored? Does it make a difference if it is on the hips or the stomach?

Research these questions and fi nd out more about BMI generally.

EXAMPLES

1. The formula for the surface area of a rectangular prism is given by ( ) .S lb bh lh2= + + Find the value of b when 180, 9 6.S l hand= = =

Solution

( )

( )

( )

.

S lb bh lh

b b

b

b

b

b

b

b

2

180 2 9 6 9 6

2 15 54

30 108

30 108

72 30

72 30

2 4

108 108

30 30

180

#

= + +

= + +

= +

= +

= +

=

=

=

- -

CONTINUED

Another way of doing this would be to change

the subject of the formula fi rst.

ch3.indd 101 5/20/09 11:54:48 AM


2. The volume of a cylinder is given by V r h2r= . Evaluate the radius r, correct to 2 decimal places, when 350V = and . .h 6 5=

Solution

( . )

( . )

.

.

..

. .

V r h

r

r

r

r

r

r

350 6 5

350 6 5

6 5350

6 5350

6 5350

4 14

6 5 6 5

2

2

2

2

2

r

r

r

r

r

r

r r

=

=

=

=

=

=

=

1. Given that u atv = + is the formula for the velocity of a particle at time t, fi nd the value of t when . ,u 17 3= . . .v a100 6 9 8and= =

2. The sum of an arithmetic series is

given by ( ) .S n a l2

= + Find l if

3, 26 1625.a n Sand= = =

3. The formula for fi nding the area

of a triangle is 2 .A bh= 1 Find b

when 36 9.A hand= =

4. The area of a trapezium is given

by 2 ( ) .A h a b= +1 Find

the value of a when 120,A =

.h b5 7and= =

5. Find the value of y when 3,x = given the straight line equation 5 2 7 0.x y- - =

6. The area of a circle is given by .A r2r= Find r correct to 3 signifi cant fi gures if 140.A =

7. The area of a rhombus is given by

the formula 2A xy= 1 where x and

y are its diagonals. Find the value of x correct to 2 decimal places when 7.8 25.1.y Aand= =

8. The simple interest formula is

.PrnI100

= Find n if 14.5,r =

150 326.25.P Iand= =

9. The gradient of a straight

line is given by .x xy y

m2 1

2 1

-

-=

Find y1 when 6,m = - 5

, .y x x7 3 1and2 2 1= = - =

10. The surface area of a cylinder is given by the formula .S r r h2r= +] g Evaluate h correct to 1 decimal place if 232 4.5.S rand= =

3.3 Exercises

ch3.indd 102 5/20/09 11:54:50 AM


11. The formula for body mass index

is hwBMI

2= . Evaluate

the (a) BMI when 65w = and 1.6h =

(b) w when 21.5BMI = and 1.8h =

(c) h when 19.7BMI = and . .w 73 8=

12. A formula for depreciation is .D P r1 n= -] g Find r if ,D P12 000 15 000= = and 3.n =

13. The x -value of the midpoint is

given by x x

x2

1 2+= . Find x 1

when 2x = - and .x 52 =

14. Given the height of a particle at time t is ,h t5 2= evaluate t when .h 23=

15. If ,y x 12= + evaluate x when .y 5=

16. If the surface area of a sphere is ,S r4 2r= evaluate r to 3 signifi cant fi gures when . .S 56 3=

17. The area of a sector of a circle

is .A r2

21

i= Evaluate r when

24.6A = and . .0 45i=

18. If x

y1

23 -

= , fi nd the value of x

when .y 3=

19. Given 2 5,y x= + evaluate x when .y 4=

20. The volume of a sphere is

V r3

34r= . Evaluate r to 1 decimal

place when .V 150=

There are two solutions to this question.

Inequations

• 2 means greater than • 1 means less than • $ means greater than or equal to • # means less than or equal to

In order to solve inequations, we need to see what effect one operation applied to both sides has on the inequality sign.

If a b2 then a c b c2+ + for all c

For example, 3 22 and 3 1 2 12+ + are both true.

If a b2 then a c b c2- - for all c

For example, 3 22 and 3 1 2 12- - are both true.

ch3.indd 103 5/20/09 11:54:50 AM


If a b2 then ac bc2 for all c 02

For example, 3 22 and 3 2 2 2# #2 are both true.

If a b2 then ac bc1 for all c 01

On the number plane, we graph inequalities using arrows and circles (open for greater than and less than and closed in for greater than or equal to and less than or equal to)

1

2

#

$

If a b2 then a c b c' '2 for all 0c 2

If a b2 then a c b c' '1 for all c 01

If a b2 then a b1 11 for all positive numbers a and b

The inequality sign reverses when: multiplying by a negative • dividing by a negative • taking the reciprocal of both sides •

For example, 3 22 but .3 2 2 2# #1- -

For example, 6 42 and 6 2 4 2' '2 are both true.

For example, 6 42 but .6 2 4 2' '1- -

For example, 3 22 but .31

21

1

ch3.indd 104 5/20/09 11:54:51 AM


EXAMPLES

Solve and show the solutions on a number line 1. x5 7 17$+

Solution

x

x

x

x

x

5 7 17

5 7 17

5 10

5 10

2

7 7

5 5

$

$

$

$

$

+

+ - -

-4 -3 -2 -1 0 1 2 3 4

2. t t3 2 5 42- +

Solution

t t

t t

t

t

t

t

t

t t

t t

t

t

t

t

t

t t

t t

3 2 5 4

3 2 5 4

2 2 4

2 2 4

6 2

6 2

3

3 2 5 4

3 2 5 4

2 2 4

2 2 4

2 6

2 6

3

3 3

4 4

2 2

5 5

2 2

2 2

or

2

2

2

2

2

2

2

2

2

2

2

2

2

1

- +

- +

- +

- +

-

-

-

- +

- +

- -

- -

-

-

-

- -

- -

- -

+ +

- -

-4 -3 -2 -1 0 1 2 3 4

CONTINUED

Remember to change the inequality sign when

dividing by -2.

ch3.indd 105 5/20/09 12:28:46 PM


3. Solve .z1 2 7 111 #+

Solution

Method 1: Separate into two separate questions. (i) z

z

z

z

z

1 2 7

1 2 7

6 2

6 2

3

7 7

2 2

1

1

1

1

1

+

+

-

-

-

- -

(ii) z

z

z

z

z

2 7 11

2 7 11

2 4

2 4

2

7 7

2 2

#

#

#

#

#

+

+ - -

Putting these together gives the solution .z3 21 #-

Method 2: Do as a single question.

z

z

z

z

z

1 2 7 11

1 2 7 11

6 2 4

6 2 4

3 2

7 7 7

2 2 2

1

1

1

1

1

#

#

#

#

#

+

+

-

-

-

- - -

Solving this inequation as a single question is quicker than splitting it into two parts.

Notice that the circle is not fi lled in for 1 and fi lled in for #.

1. Solve and plot the solution on a number line

(a) x 4 72+ (b) y 3 1#-

2. Solve (a) t5 352 (b) x3 7 2$- (c) 2( 5) 8p 2+ (d) 4 ( 1) 7x #- - (e) y y3 5 2 42+ - (f) a a2 6 5 3#- - (g) 3 4 2(1 )y y$+ - -

(h) 2 9 1 4 ( 1)x x1+ - +

(i) a2

3# -

(j) y

83

22

(k) b2

5 41+ -

(l) x3

4 62-

(m) x41

51#+

(n) m4

332

2-

3.4 Exercises

-4 -3 -2 -1 0 1 2 3 4

ch3.indd 106 7/8/09 5:31:43 PM


(o) b52

21 6$-

(p) r2

3 6#-

-

(q) z9

1 2 32+

+

(r) w w6 3

2 5 41++

(s) x x2

13

2 7$+

--

(t) t t7

22

3 2#+

-+

(u) q q

32

243

1-

+

(v) x x32

21

92

2--

(w) b b8

2 5 312

6#

-+

+

3. Solve and plot the solutions on a number line

(a) 3 2 9x 11 + (b) p4 2 101#- (c) 2 3 1 11x1 1- (d) y6 5 9 34# #- + (e) 2 3 (2 1) 7y1 1- -

PROBLEM

Find a solution for this sum. Is it a unique solution?

D A N G E R

C R O S S

R O A D S+

Equations and Inequations Involving Absolute Values

On a number line, x means the distance of x from zero in either direction.

EXAMPLES

Plot on a number line and evaluate x 1. x 2=

Solution

x 2= means the distance of x from zero is 2 (in either direction).

x 2!=

CONTINUED

2 2

-4 -3 -2 -1 0 1 2 3 4

ch3.indd 107 7/8/09 5:32:03 PM


Class Discussion

What does a b- mean as a distance along the number line? Select different values of a and b to help with this discussion.

The solution of | x | 21 would be

2 x 2.1 1-

The solution of | x | 2$ would be x 2, x 2.# $-

x a= means x a!=

x a1 means a x a1 1-

x a2 means ,x a x a2 1 -

We use absolute value as a distance on a number line to solve equations and inequations involving absolute values.

2. x 2#

Solution

x 2# means the distance of x from zero is less than or equal to 2 (in either direction).

2 2

-4 -3 -2 -1 0 1 2 3 4

Notice that there is one region on the number line. We can write this as the single statement .x2 2# #-

3. 2x 2

Solution

2x 2 means the distance of x from zero is greater than 2 (in either direction).

2 2

-4 -3 -2 -1 0 1 2 3 4

There are two regions on the number line, so we write two separate inequalities , .x x2 21 2-

ch3.indd 108 5/20/09 11:54:54 AM


EXAMPLES

Solve

1. 4x 7+ =

Solution

This means that the distance from 4x + to zero is 7 in either direction. So .x 4 7!+ = 4x 7+ =

7-

7 4

=

- -=

x

x

x

4

4

11

4

+

+

= -

-

x

x

x

4 7

4 7

3

4 4

or+ =

+ =

=

- -

2. 2 1y 51-

Solution

This means that the distance from 2 1y - to zero is less than 5 in either direction. So it means .y5 2 1 51 1- -

y

y

y

y

5 2 1 5

5 2 1 5

4 2 6

2 3

1 1 1

2 2 2

1 1

1 1

1 1

1 1

- -

- -

-

-

+ + +

3. 5 7b 3$-

Solution

b5 7 3$- means that the distance from 5 7b - to zero is greater than or equal to 3 in either direction.

.

b b

b b

b b

b b

5 7 3 5 7 3

5 4 5 10

5 4 5 10

54 2

54

5 5 5 5

# $

# $

# $

# $

- - -

+ + + +

,

b b

b b

5 7 3 5 7 3

2

7 7 7 7

So

# $

# $

- - -

You could solve these as two separate inequations.

These must be solved and written as two

separate inequations.

ch3.indd 109 7/9/09 3:05:15 AM


While it is always a good habit to check solutions to equations and inequations by substituting in values, in these next examples it is essential to check, as some of the solutions are impossible!

EXAMPLES

Solve

1. x x2 1 3 2+ = -

Solution

x x2 1 3 2+ = - means that x2 1+ is at a distance of x3 2- from zero.

x x2 1 3 2!+ = -] g This question is impossible if x3 2- is negative. Can you see why? If x2 1+ is equal to a negative number, this is impossible as the absolute value is always positive. Case (i)

x x

x

x

x xx x

x

2 1 3 2

1 2

3

2 2

2 2

2 1 3 2

1 2

+ = -

= -

=

- -

+ +

+ = -

= -

Check solution is possible: Substitute x 3= into .x x2 1 3 2+ = -

3 3 2

2 3 1

77

9 2

7

LHS

RHS

#

#

= +

=

= -

= -

=

=

Since , 3xLHS RHS= = is a solution. Case (ii)

( )x x

x

x x

x

x

x

x

x

x x

2 1 3 2

3 2

2 1 3 2

5 1 2

5 1 2

5 1

5 1

51

3 3

1 1

5 5

+ = - -

= - +

+ = - +

+ =

+ =

=

=

=

+ +

- -

ch3.indd 110 7/8/09 5:33:01 PM


Check: Substitute x

51

= into .x x2 1 3 2+ = -

3 2

251 1

152

152

51

53 2

52

LHS

RHS

#

#

= +

=

= -

= -

1

=

= -

Since , x51LHS RHS! = is not a solution.

So the only solution is .x 3=

2. x x2 3 1 9- + + =

Solution

In this question it is diffi cult to use distances on the number line, so we use the defi nition of absolute value.

( )

( )

x x xx x

x x xx x

2 3 2 3 2 3 02 3 2 3 0

1 1 1 01 1 0

whenwhenwhenwhen

1

1

$

$

- =- -

- - -

++ +

- + +=

''

This gives 4 cases: (i) ( ) ( )x x2 3 1 9- + + = (ii) ( ) ( )x x2 3 1 9- - + = (iii) ( ) ( )x x2 3 1 9- - + + = (iv) ( ) ( )x x2 3 1 9- - - + =

Case (i)

( ) ( )x x

x x

x

x

x

x

x

2 3 1 9

2 3 1 9

3 2 9

3 2 9

3 11

3 11

332

2 2

3 3

- + + =

- + + =

- =

- =

=

=

=

+ +

Check by substituting x 332

= into .x x2 3 1 9- + + =

CONTINUED

It is often easier to solve these harder equations graphically. You will do

this in Chapter 5.

ch3.indd 111 7/8/09 5:33:22 PM


2 332 3 3

32 1

431 4

32

431 4

32

9

LHS #= - + +

= +

=

RHS

= +

=

So x 332

= is a solution.

Case (ii)

( ) ( )x x

x x

x

x

x

2 3 1 9

2 3 1 9

4 9

4 9

13

4 4

- - + =

- - - =

- =

- =

=

+ +

Check by substituting x 13= into .x x2 3 1 9- + + =

2 13 3 13 1

23 1423 14

37

LHS

RHS

#

!

= - + +

= +

= +

=

So x 13= is not a solution. Case (iii)

( ) ( )x x

x x

x

x

x

x

x

2 3 1 9

2 3 1 9

4 9

4 9

5

5

5

4 4

1 1

- - + + =

- + + + =

- + =

- + =

- =

-=

= -

- -

- -

Check by substituting x 5= - into .x x2 3 1 9- + + =

2 5 3 5 1

13 413 4

17

LHS #= - - + - +

- -

= +

=

RHS!

= +

So x 5= - is not a solution. Case (iv)

( ) ( )x x

x x

x

x

x

2 3 1 9

2 3 1 9

3 2 9

3 2 9

3 7

2 2

- - - + =

- + - - =

- + =

- + =

- =

- -

ch3.indd 112 7/8/09 5:33:39 PM


x

x

3 7

231

3 3-

=

= -

- -

Check by substituting x 231

= - into .x x2 3 1 9- + + =

2 231 3 2

31 1

732 1

31

732 1

31

9

LHS #= - - + - +

- -

= +

=

RHS

= +

=

So x 231

= - is a solution.

So solutions are , .x 332 2

31

= -

While you should always check solutions, you can see that there are some cases where this is really important.

You will learn how to solve equations involving

absolute values graphically in Chapter 5. With

graphical solutions it is easy to see how many

solutions there are.

1. Solve (a) x 5=

(b) y 8=

(c) a 41

(d) k 1$

(e) x 62

(f) p 10#

(g) x 0=

(h) a 142

(i) y 121

(j) b 20$

2. Solve (a) x 2 7+ =

(b) n 1 3- =

(c) 2a 42

(d) 5x 1#-

(e) x9 2 3= +

(f) x7 1 34- =

(g) y4 3 111+

(h) x2 3 15$-

(i) x3

4=

(j) a2

3 2#-

3. Solve (a) x x2 5 3+ = - (b) a a2 1 2- = + (c) b b3 2 4- = - (d) k k3 2 4- = - (e) y y6 23 7+ = - (f) x x4 3 5 4+ = - (g) m m2 5- = (h) d d3 1 6+ = +

(i) y y5 4 1- = + (j) t t2 7 3- = -

4. Solve (a) x x3 3 1+ = -

(b) y y2 5 2- = -

(c) a a3 1 2 9+ = -

(d) x x2 5 17+ + =

(e) d d3 2 4 18- + + =

5. (a) Solve .t t4 3 1 11- + - = By plotting the solutions on (b)

a number line and looking at values in between the solutions, solve .t t4 3 1 111- + -

3.5 Exercises Remember to check solutions

in questions 3, 4 and 5.

ch3.indd 113 7/8/09 5:34:00 PM


To solve equations, use inverse operations: For squares, take the square root For cubes, take the cube root For square roots, take the square For cube roots, take the cube

You have previously used these rules when substituting into formulae involving squares and cubes.

EXAMPLES

Solve

1. x 92 =

Solution

x

xx

9

93

2

2 !

` !

=

=

=

2. n5 403 =

Solution

n

n

n

nn

5 40

5 40

8

82

5 5

3

3

3

33 3

=

=

=

=

=

There are two possible solutions for x – one positive and one negative since 3 92

= and ( 3) 9.2

- =

There is only one answer for this question since 2 83

= but ( 2) 8.3

- = -

Exponential Equations

An exponential equation involves an unknown index or power e.g. .2 8x = We can also solve other equations involving indices. In order to solve

these, you need to understand their relationship. For example, squares and square roots are the reverse of each other (we call them inverse operations). Similarly cubes and cube roots are inverses, and this extends to all indices.

ch3.indd 114 7/24/09 10:18:11 PM


Investigation

Investigate equations of the type x kn = where k is a constant, for example, .x 9n =

Look at these questions:

What is the solution when 1. ?n 0= What is the solution when 2. ?n 1= How many solutions are there when 3. ?n 2= How many solutions are there when 4. ?n 3= How many solutions are there when 5. n is even? How many solutions are there when 6. n is odd?

3. 3a 4=2

Solution

We use the fact that 32

23

.a a a= =2

33

2

` `j j

3

3

2

22

a

a

a

a

4

4

4

428

3

3

`

=

=

=

=

=

=

2

2

3

33

^

`

h

j

In other types of equations, the pronumeral (or unknown variable) is in the index. We call these exponential equations , and we use the fact that if the base numbers are equal, then the powers (or indices or exponents) must be equal.

EXAMPLES

Solve 1. 3 81x =

Solution

3 81x = Equating indices:

x

3 34

x 4

`

=

=

CONTINUED

ch3.indd 115 7/8/09 5:34:34 PM


2. 5 25k2 1 =-

Solution

5 5

k

k

k

k

5 25

2

2 1 2

2 3

2 3

121

1 1

2 2

k

k

2 1

2 1 2

=

=

=

- =

=

=

=

+ +

-

-

2 1k` -

3. 8 4n =

Solution

It is hard to write 8 as a power of 4 or 4 as a power of 8, but both can be written as powers of 2.

( )

n

n

n

8 4

2 2

2 23 2

3 2

32

3 3

n

n

n

3 2

3 2

`

=

=

=

=

=

=

We can check this solution

by substituting k 12

1= into

the equation 5 25.2k 1=

-

1. Solve (a) x 273 = (b) y 642 = (c) n 164 = (d) ( )x 20 give the exact answer2 = (e) p 10003 = (f) x2 502 = (g) y6 4864 = (h) w 7 153 + =

(i) n6 4 922 - = (j) q3 20 43 + = -

2. Solve and give the answer correct to 2 decimal places.

(a) p 452 = (b) x 1003 = (c) n 2405 = (d) x2 702 = (e) y4 7 343 + =

(f) d3

144

=

(g) k2

3 72

- =

(h) x5

1 23 -

=

(i) y2 9 202 - = (j) y7 9 2003 + =

3.6 Exercises

ch3.indd 116 8/1/09 11:46:31 PM


3. Solve

(a) 3n 9=2

(b) 4t 8=3

(c) 5x 4=2

(d) 3t 16=4

(e) 5p 27=3

(f) 4m2 250=3

(g) 3b 3 39+ =2

(h) 3y5 405=4

(i) 7a3 2 10- =2

(j) 4t

39=

3

4. Solve (all pronumerals ! 0) (a) x 51 =- (b) a 83 =- (c) y 325 =- (d) x 1 502 + =- (e) n2 31 =-

(f) a813 =-

(g) x412 =-

(h) b911 =-

(i) x 2412 =-

(j) b81164 =-

5. Solve (all pronumerals ! 0)

(a) 3x 8=-

1

(b) -

2x125

8=

3

(c) 4a 3=-

1

(d) -

4k 125=3

(e) 3x3 12=-

2

(f) 2x81

=-

3

(g) -

3y41

=2

(h) 5n94

=-

2

(i) 3b321

=-

5

(j) -

3m4936

=2

6. Solve (a) 2 16n = (b) 3 243y = (c) 2 512m = (d) 10 100 000x = (e) 6 1m = (f) 4 64x = (g) 4 3 19x + = (h) )(5 3 45x =

(i) 4 4x =

(j) 26 18

k

=

7. Solve (a) 3 81x2 = (b) 2 16x5 1 =- (c) 4 4x 3 =+ (d) 3 1n 2 =- (e) 7 7x2 1 =+ (f) 3 27x 3 =- (g) 5 125y3 2 =+ (h) 7 49x3 4 =-

(i) 2 256x4 = (j) 9 9a3 1 =+

8. Solve (a) 4 2m = (b) 27 3x = (c) 125 5x =

(d) 491 7

k

=c m

(e) 1000

1 100k

=c m

(f) 16 8n = (g) 25 125x = (h) 64 16n =

(i) 41 2

k3

=c m

(j) 8 4x 1 =-

9. Solve (a) 2 8x x14 =+ (b) 3 9x x5 2= - (c) 7 7k k2 3 1=+ - (d) 4 8n n3 3= + (e) 6 216x x5 =- (f) 16 4x x2 1 4=- - (g) 27 3x x3=+

(h) 21

641x x2 3

=+c cm m

ch3.indd 117 5/20/09 11:54:58 AM


PUZZLE

Test your logical thinking and that of your friends. How many months have 28 days? 1. If I have 128 sheep and take away all but 10, how many 2. do I have left? A bottle and its cork cost $1.10 to make. If the bottle costs $1 more 3. than the cork, how much does each cost? What do you get if you add 1 to 15 four times? 4. On what day of the week does Good Friday fall in 2016? 5.

Quadratic Equations

A quadratic equation is an equation involving a square. For example, .x 4 02 - =

Solving by factorisation

When solving quadratic equations by factorising, we use a property of zero.

For any real numbers a and b , if ab 0= then a 0= or b 0=

EXAMPLES

Solve 1. x x 6 02 + - =

Solution

( ) ( )

x xx x

6 03 2 0

2 + - =

+ - =

(i) 43

6427x x2 3

=-c cm m

(j) 5251x

x 9

=--] cg m

10. Solve (a) 4 2m =

(b) 259

53k 3

=+c m

(c) 2

1 4 x2 5= -

(d) 3 3 3k =

(e) 271

813n3 1

=+c m

(f) 52

25n n3 1

=+ -c cm m

(g) 32161x =-

(h) 9 3 3b b2 5 =+

(i) 81 3x x1 =+

(j) 2551m

m3 5

=--c m

ch3.indd 118 5/20/09 11:54:59 AM


0

0

2

=

=

=

x x

x x

x

3 0 2

3 0 2

3

3 3 2 2

or

or

` + = -

+ = -

= -

- - + +

x

So the solution is .x 3 2or= -

2. y y7 02 - =

Solution

( )y y

y y

y y

7 07 0

0 7 0or

2

`

- =

- =

= - =

y

y

7 0

7

7 7- =

=

+ +

So the solution is y 0= or 7.

3. a a3 14 82 - = -

Solution

( ) ( )

a a

a a

a aa a

a a

a a

a a

a

a

3 14 8

3 14 8

3 14 8 03 2 4 0

3 2 0 4 0

3 2 0 4 0

3 2 4

3 2

32

8 8

2 4 4

3 3

or

or

2

2

2

`

- = -

- = -

- + =

- - =

- = - =

- = - =

= =

=

=

+ +

+ + +

So the solution is .a 4or32

=

Solve

3.7 Exercises

1. y y 02 + =

2. b b 2 02 - - =

3. p p2 15 02 + - =

4. t t5 02 - =

5. x x9 14 02 + + =

6. q 9 02 - =

ch3.indd 119 7/9/09 3:05:20 AM


7. x 1 02 - =

8. a a3 02 + =

9. x x2 8 02 + =

10. x4 1 02 - =

11. x x3 7 4 02 + + =

12. y y2 3 02 + - =

13. b b8 10 3 02 - + =

14. x x3 102 - =

15. x x3 22 =

16. x x2 7 52 = -

17. x x5 02- =

18. y y 22 = +

19. n n8 152= +

20. x x12 7 2= -

21. m m6 52 = -

22. ( ) ( )x x x1 2 0+ + =

23. ( ) ( ) ( )y y y1 5 2 0- + + =

24. ( ) ( )x x3 1 32+ - =

25. ( ) ( )m m3 4 20- - =

Application

A formula for displacement s at time t is given by 21

s ut at2= + where u is the

initial velocity and a is the acceleration. Find the time when the displacement will be zero, given 12u = - and 10.a =

2

20 12 (10)

12 5

( 12 5 )

0 or 12 5 0

s ut at

t t

t t

t t

t t

2

2

2

`

= +

= - +

= - +

= - +

= - + =

1

1

12 5 0

5 12

5 12

2.4

12 12

5 5

t

t

t

t

+ =+ +-

=

=

=

So displacement will be zero when 0t = or 2.4.

Solving by completing the square

Not all trinomials will factorise, so other methods need to be used to solve quadratic equations.

ch3.indd 120 5/20/09 5:22:45 PM


EXAMPLES

Solve 1. x 72 =

Solution

.

x

x

7

72 6

2

!

!

=

=

=

2. x 3 112+ =] g

Solution

. , .

x

x

x

x

3 11

3 11

3 11

11 30 3 6 3

3 3

2

!

!

!

+ =

+ =

+ =

= -

= -

- -

] g

3. y 2 72- =^ h

Solution

. , .

y

y

y

y

2 7

2 7

2 7

7 24 6 0 6

2 2

2

!

!

!

- =

- =

- =

= +

= -

+ +

^ h

Take the square root of both sides.

To solve a quadratic equation like ,x x6 3 02 - + = which will not factorise, we can use the method of completing the square.

You learnt how to complete the square in

Chapter 2.

EXAMPLES

Solve by completing the square 1. x x6 3 02 - + = (give exact answer)

Solution

2

x x

x x

6 3 0

6 3 3 9

2

22

2

- + =

- = - = =6c m Halve 6, square it and

add to both sides of the equation.

CONTINUED

ch3.indd 121 7/9/09 3:05:21 AM


x x

x

x

x

x

6 3

3 6

3 6

3 6

6 3

9 9

3 3

2

2

` !

!

!

- = -

- =

- =

- =

= +

+ +

+ +

] g

2. y y2 7 02 + - = (correct to 3 signifi cant fi gures)

Solution

2

. .

y y

y y

y y

y

y

y

y

y

2 7 0

2 7 1 1

2 7

1 8

1 8

1 8

8 1

2 2 11 83 3 83

1 1

1 1

or

2

22

2

2

2

` !

!

!

!

+ - =

+ = = =

+ =

+ =

+ =

+ =

= -

= -

= -

+ +

- -

2c

^

m

h

1. Solve by completing the square, giving exact answers in simplest surd form

(a) x x4 1 02 + - = (b) a a6 2 02 - + = (c) y y8 7 02 - - = (d) x x2 12 02 + - = (e) p p14 5 02 + + = (f) x x10 3 02 - - = (g) y y20 12 02 + + = (h) x x2 1 02 - - =

(i) n n24 7 02 + + = (j) y y3 1 02 - + =

2. Solve by completing the square and write your answers correct to 3 signifi cant fi gures

(a) x x2 5 02 - - = (b) x x12 34 02 + + = (c) q q18 1 02 + - = (d) x x4 2 02 - - = (e) b b16 50 02 + + = (f) x x24 112 02 - + = (g) r r22 7 02 - - = (h) x x8 5 02 + + =

(i) a a6 012 + - = (j) y y40 3 02 - - =

3.8 Exercises

Solving by formula

Completing the square is diffi cult with harder quadratic equations, for example .x x2 5 02 - - = Completing the square on a general quadratic equation gives the following formula.

ch3.indd 122 7/8/09 5:36:02 PM


For the equation ax bx c 02 + + =

xa

b b ac2

42!=

- -

Proof

Solve ax b c 02 + + = by completing the square.

ax bx c

ax bx c

x abx

ac

x abx

ac

x abx

ac

ab

ab

ab

x abx

ac

xa

bac

ab

aac b

xab

aac b

ab ac

xa

ba

b ac

xab

ab ac

ab b ac

a a a a

ac

ac

ab

ab

ab

ab

0

0

0

0

22 4

2 4

44

2 44

24

2 24

2 24

24

4 4

2 2

2

2

2

2

22 2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

'

!

!

!

!

!

+ + =

+ + =

+ + =

+ + =

+ = - = =

+ = -

+ = - +

=- +

+ =- +

=-

+ =-

=- -

=- -

- -

+ +

- -

b c

c

l m

m

EXAMPLES

1. Solve x x 2 02 - - = by using the quadratic formula.

Solution

, ,

( )( ) ( ) ( ) ( )

a b c

xa

b b ac

1 1 2

24

2 11 1 4 1 2

21 1 8

2

2

!

!

!

= = - = -

=- -

=- - - - -

=+

CONTINUED

ch3.indd 123 5/20/09 11:55:02 AM


21 9

21 3

2 1or

!

!

=

=

= -

2. Solve y y2 9 3 02 - + = by formula and give your answer correct to 2 decimal places.

Solution

, ,

. .

a b c

xa

b b ac

y

2 9 3

24

2 29 9 4 2 3

49 81 24

49 57

4 14 0 36or

2

2

!

!

!

!

Z

= = - =

=- -

=- - - -

=-

=

]] ] ] ]

gg g g g

x2

1 3!= gives two

separate solutions, 2

1 3+

and .2

1 3-

These solutions are irrational.

1. Solve by formula, correct to 3 signifi cant fi gures where necessary

(a) y y6 2 02 + + = (b) x x2 5 3 02 - + = (c) b b 9 02 - - = (d) x x2 1 02 - - = (e) x x8 3 02- + + = (f) n n8 2 02 + - = (g) m m7 10 02 + + = (h) x x7 02 - =

(i) x x5 62 + = (j) y y3 12 = -

2. Solve by formula, leaving the answer in simplest surd form

(a) x x 4 02 + - = (b) x x3 5 1 02 - + = (c) q q4 3 02 - - = (d) h h4 12 1 02 + + = (e) s s3 8 2 02 - + = (f) x x11 3 02 + - = (g) d d6 5 2 02 + - = (h) x x2 72 - =

(i) t t 12 = + (j) x x2 1 72 + =

3.9 Exercises

Class Investigation

Here is a proof that .1 2= Can you see the fault in the proof?

( ) ( ) ( )x x x xx x x x x x x

x x x

x x2

1 2

2 2 2 2

`

- = -

- = + -

= +

=

=

ch3.indd 124 7/8/09 5:36:25 PM


EXAMPLES

1. Solve .x1 31

Solution

x 0!

.x

xx

x

x

x x

1 3

1 3

1 3

1 3

313 3

Solve

# #

=

=

=

=

=

.x x31 1 3is not a solution of the inequation 1=

Place 0x = and x31

= on a number plane and test x values on either side

of these values in the inequation.

Test for 0,x 1 say x 1= - Substitute into the inequation:

( )

x1 3

11 3

1 3 true

1

1

1-

-

So 0x 1 is part of the solution.

Test for ,x310 1 1 say x

101

=

1011 31

( )10 3 false1

So x310 1 1 is not part of the solution.

Test for ,x31

2 say 1x =

Substitute into the inequation:

0

1 is undefi ned.

Circle these values as they are not included in the

solution.

CONTINUED

Further Inequations

Inequations involving pronumerals in the denominator can be solved in several ways. Here is one method. You will use a different method in Chapter 10.

-3 -2 -1 0 1 2 3 4 513

ch3.indd 125 7/8/09 5:36:45 PM


( )

11 3

1 3 true

1

1

So x31

2 is part of the solution.

Solution is .,x x3101 2

2. Solve .x 3

6 1$+

Solution

x 3! -

.

( ) ( )

x

xx

x

x

x x

36 1

36 1

6 3

6 3

3

3 3

3 3

Solve

# #

+=

+=

= +

= +

=

+ +

- -

3x = is a solution of the inequation .x 3

6 1$+

Place 3x = - and 3x = on a number plane and test values on either side in the inequation.

-3 -2 -1 0 1 2 3 4 5

Test for ,x 31 - say 4x = - Substitute into the inequation:

( )

x 36 1

4 36 1

6 3 false

$

$

$

+

- +

-

So x 31 - is not part of the solution.

Test for ,x3 31 #- say x 0=

( )

0 36 1

2 1 true

$

$+

So x3 31 #- is part of the solution.

Test for ,x 3$ say x 4= Substitute into the inequation:

( )

4 36 1

76 1 false

$

$

+

So x 3$ is not part of the solution.

0

6 is undefi ned.

Circle x 3-= and fi ll in x 3= since it is a part of the solution.

-3 -2 -1 0 1 2 3 4 513

ch3.indd 126 5/20/09 11:55:03 AM


Solution is 3 3x1 #-

3. Solve .yy 6

12

#-

Solution

0y!

.yy

yy

y y

y y

y yy y

y y

y y

61

61

6

6

6 03 2 0

Solve2

2

2

2

2

# #

-=

-=

- =

- =

- - =

- + =

- -

^ ^h h

,

,

,

y y

y y

y y

3 0 2 0

3 0 2 0

3 2

3 3 2 2

- = + =

- = + =

= = -

+ + - -

Sketch these on a number line and test values on either side.

Test for ,y 2# - say y 3= - Substitute into the inequation:

( )

yy 6

1

33 6

1

1 1 true

2

2

#

#

#

-

-

- -

-

] g

So y 2# - is part of the solution.

Test for ,y2 01#- say y 1= -

( )

11 6

1

5 1 false

2

#

#-

- -] g

So y2 01#- is not part of the solution.

Test ,y0 31 # say y 1=

( )

11 6 1

5 1 true

2

#

#

-

-

So y0 31 # is part of the solution.

Test 3,y $ say y 4=

CONTINUED

-3 -2 -1 0 1 2 3 4 5

-3 -2 -1 0 1 2 3 4 5

ch3.indd 127 7/8/09 5:37:10 PM


( )

44 6 1

221 1 false

2

#

#

-

So y 3$ is not part of the solution. The solution is ,y y2 0 31# #-

1. y1 11

2. x1 22

3. x3 21

4. m2 7$

5. x3 52 -

6. b2 1# -

7. x 1

1 42-

8. z 3

1 51+

-

9. x 2

3 4$-

10. x2

1 61-

-

11. x 4

5 9#+

-

12. x3 4

2 52-

13. a2 5

3 21+

-

14. xx

2 152

-

15. y

y

121

+

16. xx

43 1

31

$-

+

17. p

p

2 98 7

52-

+

18. xx5 1

243

#+

-

19. tt

3 87 4 1$

-

+-

20. m

m2

5 441

1+

21. xx 5 4

2

1-

-

22. nn 8 6

2

$+

23. xx 15 2

2

2-

24. mm

18 4

2

#+

-

25. x

x3

4$

-

26. xx

3 22 1

2

#-

-

27. x

x2

3#

-

28. nn n

352

-

+

29. xx

7 43 2

2

1+

-

30. ( )xx x

12 4

7#-

-

3.10 Exercises

Solve

-3 -2 -1 0 1 2 3 4 5

ch3.indd 128 7/8/09 5:37:31 PM


In Chapter 10 you will look at how to use the number plane to solve these quadratic inequations. Here are some examples of solving quadratic inequations using the number line.

Quadratic Inequations

Solving quadratic inequations is similar to solving quadratic equations, but you need to do this in two stages. The fi rst is to solve the equation and then the second step is to look at either the number line or the number plane for the inequality.

To solve a quadratic inequation: Factorise and solve the quadratic equation 1. Test values in the inequality 2.

EXAMPLES

Solve 1. x x 6 02 2+ -

Solution

( ) ( )x x

x x

x

6 02 3 0

2 3

First solve

or

2

`

+ - =

- + =

= -

Now look at the number line.

-4 -3 -2 -1 0 1 2 3 4

Choose a number between 3- and 2, say .x 0= Substitute x 0= into the inequation.

x x 6 0

0 0 6 06 0 (false)

2

2

2

2

2

+ -

+ -

-

So the solution is not between 3- and 2. the solution lies either side of 3- and 2. Check by choosing a number on either side of the two numbers. Choose a number on the RHS of 2, say .x 3=

Be careful: x x 6 02 2+ - does not mean x 2 02-

and x 3 0.2+

CONTINUED

ch3.indd 129 5/20/09 11:59:45 AM


Substitute x 3= into the inequation.

6 0

3 3 6 0(true)

2 2

2

+ -

So the solution is on the RHS of 2. Choose a number on the LHS of ,3- say x 4= - Substitute x 4= - into the inequation

( ) ( )4 4 6 0

6 0 (true)

2 2

2

- + - -

So the solution is on the LHS of .3-

-4 -3 -2 -1 0 1 2 3 4

This gives the solution , .x x3 21 2-

2. x9 02 $-

Solution

( ) ( )x

x x

x

9 03 3 0

3

First solve 2

` !

- =

- + =

=

-4 -3 -2 -1 0 1 2 3 4

Choose a number between 3- and 3, say .x 0= Substitute x 0= into the inequation.

x9 0

9 0 09 0 (true)

2

2

$

$

$

-

-

So the solution is between 3- and 3, that is x3 3# #- . On the number line:

-4 -3 -2 -1 0 1 2 3 4

Check numbers on the RHS and LHS to verify this.

Earlier in the chapter you learned how to solve inequations with the unknown in the denominator. Some people like to solve these using quadratic inequations. Here are some examples of how to do this.

ch3.indd 130 5/20/09 11:59:45 AM


EXAMPLES

Solve

1. x1 31

Solution

x 0! First, multiply both sides by .x2

xx x

x x

1 3

3

0 3

2

2

1

1

1 -

( )x x

x x

x

3 03 1 0

031

Now, solve

or

2 - =

- =

=

By checking on each side of 0 and 3,1 for ,x x0 3 21 - the solution is

30, .x x1 2 1

2. x 5

3 2$+

Solution

x 5! -

First, multiply both sides by ( ) .x 5 2+

( ) ( )

( ) ( )( ) [ ( ) ]

( ) ( )

xx x

x xx x

x x

53 2

3 5 2 5

0 2 5 3 50 5 2 5 3

0 5 2 7

2

2

$

$

$

$

$

+

+ +

+ - +

+ + -

+ +

( ) ( )x x

x x

x x

5 2 7 0

5 0 2 7 0

5 3

Now, solve

or

2

`

+ + =

+ = + =

= - = - 1

Check by choosing a number on each side of 5- and 321

- for

( ) ( )x x0 5 2 7$ + + that the solution is .x5 321

1 #- -

( )x 52

+ is positive, so the inequality sign does not

change.

Check this factorisation carefully.

x cannot be -5 as this would give 0 in the denominator.

x2 is positive, so the inequality sign does not change.

-2 -1 1 20 13

-6 -5 -4 -3 -212-3

ch3.indd 131 7/24/09 10:18:11 PM


Solve

1. x x3 02 1+

2. y y4 02 1-

3. n n 02 $-

4. x 4 02 $-

5. n1 02 1-

6. n n2 15 02 #+ -

7. c c 2 02 2- -

8. x x6 8 02 #+ +

9. x x9 20 02 1- +

10. b b4 10 4 02 $+ +

11. a a1 2 3 02 1- -

12. y y2 6 02 2- -

13. x x3 5 2 02 $- +

14. b b6 13 5 02 1- -

15. x x6 11 3 02 #+ +

16. y y 122 #+

17. x 162 2

18. a 12 #

19. x x 62 1 +

20. x x2 32 $ +

21. x x22 1

22. a a2 5 3 02 #- +

23. y y5 6 82 $+

24. m m6 152 2 -

25. x x3 7 42 # -

26. x1 22

27. x3 6#

28. y 1

1 51+

29. n 3

1 2$-

30. x 5

3 1$+

-

31. x5 2

1 71-

32. x 5

4 5$-

-

33. x

x1

5#+

34. xx

22 1 12

-

+

35. xx

5 32 3 6$

+

-

3.11 Exercises

Simultaneous Equations

Two equations, each with two unknown pronumerals, can be solved together to fi nd one solution that satisfi es both equations.

There are different ways of solving simultaneous equations. The elimination method adds or subtracts the equations. The substitution method substitutes one equation into the other.

ch3.indd 132 5/20/09 11:59:47 AM


EXAMPLES

Solve simultaneously 1. a b3 2 5+ = and a b2 6- = -

Solution

( )

( )

: ( )

( ): ( )

a b

a b

a b

a ba

a

3 2 5 1

2 6 2

2 2 4 2 12 3

1 3 3 2 5 17 7

1

#

+ =

- = -

- = -

+ + =

= -

= -

]]

gg

( )a 1 1Substitute in= -

( ) b

b

b

b

3 1 2 5

3 2 5

2 8

4

- + =

- + =

=

=

,a b1 4solution is` = - =

2. x y5 3 19- = and x y2 4 16- =

Solution

( )

( )

( ) : ( )

( ) : ( )

( ) ( ):

x y

x y

x y

x y

x

x

5 3 19 1

2 4 16 2

1 4 20 12 76 3

2 3 6 12 48 4

3 4 14 28

2

#

#

- =

- =

- =

- =

- =

=

( )x 2 2Substitute in=

( ) y

y

y

y

2 2 4 16

4 4 16

4 12

3

- =

- =

- =

= -

Linear equations

These equations can be solved by either method. Many students prefer the elimination method.

ch3.indd 133 5/20/09 11:59:47 AM


Solve simultaneously

1. a b a b2 4and- = - + =

2. x y x y5 2 12 3 2 4and+ = - =

3. p q p q4 3 11 5 3 7and- = + =

4. y x y x3 1 2 5and= - = +

5. x y x y2 3 14 3 4and+ = - + = -

6. t v t v7 22 4 13and+ = + =

7. x y4 5 2 0 and+ + = x y4 10 0+ + =

8. x y x y2 4 28 2 3 11and- = - = -

9. x y x y5 19 2 5 14and- = + = -

10. m n m n5 4 22 5 13and+ = - = -

11. w w w w4 3 11 3 2and1 2 1 2+ = + =

12. a b a b3 4 16 2 3 12and- = - + =

13. p q5 2 18 0 and+ + =

p q2 3 11 0- + =

14. x x x x7 3 4 3 5 2and1 2 1 2+ = + = -

15. x y x y9 2 1 7 4 9and- = - - =

16. s t5 3 13 0 and- - = s t3 7 13 0- - =

17. a b a b3 2 6 3 2and- = - - = -

18. k h3 2 14 and- = - k h2 5 13- = -

19. v v2 5 16 0 and1 2+ - = v v7 2 6 01 2+ + =

20. . . .x y1 5 3 4 7 8 and+ = . . .x y2 1 1 7 1 8- =

3.12 Exercises

PROBLEM

A group of 39 people went to see a play. There were both adults and children in the group. The total cost of the tickets was $939, with children paying $17 each and adults paying $29 each. How many in the group were adults and how many were children? (Hint: let x be the number of adults and y the number of children.)

Non-linear equations

In questions involving non-linear equations there may be more than one set of solutions. In some of these, the elimination method cannot be used. Here are some examples using the substitution method.

ch3.indd 134 5/20/09 11:59:48 AM


EXAMPLES

Solve simultaneously 1. xy 6= and x y 5+ =

Solution

( )

( )

( ): ( )

xy

x y

y x

6 1

5 2

2 5 3From

=

+ =

= -

Substitute (3) in (1)

( )

( ) ( )

x x

x x

x xx x

x x

x x

5 6

5 6

0 5 60 2 3

2 0 3 0

2 3

or

or

2

2

`

- =

- =

= - +

= - -

- = - =

= =

Substitute x 2= in (3) y 5 2 3= - = Substitute x 3= in (3) y 5 3 2= - = solutions are ,x y2 3= = and ,x y3 2= =

2. x y 162 2+ = and x y3 4 20 0- - =

Solution

( )( )

:

( )

x yx y

x yx y

16 13 4 20 0 2

2 3 20 4

43 20 3

From

2 2+ =

- - =

- =

-=

] g

Substitute (3) into (1)

( )

.

x x

x x x

x x x

x x

xx

x

x

43 20 16

169 120 400 16

16 9 120 400 256

25 120 144 0

5 12 05 12 0

5 12

2 4

22

22

2 2

2

2

`

+-

=

+- +

=

+ - + =

- + =

- =

- =

=

=

cd

mn

.( . )

.

. , . .

x

y

x y

2 4 3

43 2 4 20

3 2

2 4 3 2

Substitute into

So the solution is

=

=-

= -

= = -

] g

ch3.indd 135 5/20/09 11:59:48 AM


Equations with 3 unknown variables

Three equations can be solved simultaneously to fi nd 3 unknown pronumerals.

Solve the simultaneous equations.

1. y x2= and y x=

2. y x2= and x y2 0+ =

3. x y 92 2+ = and x y 3+ =

4. x y 7- = and xy 12= -

5. y x x42= + and x y2 1 0- - =

6. y x2= and x y6 9 0- - =

7. x t2= and x t 2 0+ - =

8. m n 162 2+ = and m n 4 0+ + =

9. xy 2= and y x2=

10. y x3= and y x2=

11. y x 1= - and y x 32= -

12. y x 12= + and y x1 2= -

13. y x x3 72= - + and y x2 3= +

14. xy 1= and x y4 3 0- + =

15. h t2= and h t 1 2= +] g

16. x y 2+ = and x xy y2 82 2+ - =

17. y x3= and y x x62= +

18. | |y x= and y x2=

19. y x x7 62= - + and x y24 4 23 0+ - =

20. x y 12 2+ = and x y5 12 13 0+ + =

3.13 Exercises

Four unknowns need 4 equations, and so on.

EXAMPLE

Solve simultaneously ,a b c a b c7 2 4- + = + - = - and .a b c3 3- - =

Solution

b2

b

b

b

b

b

=

( )

( )

( )

( ) ( ):

( )

( ) ( ):

( )

( ) ( ):

4 8

a c

a b c

a c

a

a b c

a c

a

or a

a

a

7 1

2 4 2

3 3 3

1 2 7

2 3 4

1 3 7

3

4 10

2 5 5

4 5 2

2

- + =

+ - = -

- - =

+ =

+

+ - + =

- -

- =

- =

+ +

=

=

b

a

ba c

3

- +

=

=

a b c2

3

+ - 4= -

ch3.indd 136 7/25/09 3:34:44 PM


Substitute a 2= in (4)

( ) b

b

b

2 2 3

4 3

1

+ =

+ =

= -

Substitute a 2= and b 1= - in (1)

( ) c

c

c

c

2 1 7

2 1 7

3 7

4

- - + =

+ + =

+ =

=

solution is , ,a b c2 1 4= = - =

Solve the simultaneous equations.

1. ,x x y2 2 4= - - = and x y z6 0- + =

2. ,a a b2 2 3 1= - - = - and a b c5 9- + =

3. ,a b c a b2 1 2+ + = + = - and c 7=

4. ,a b c a b c0 4+ + = - + = - and a b c2 3 1- - = -

5. ,x y z x y z7 2 1+ - = + + = and x y z3 2 19+ - =

6. ,x y z x y z1 2 9- - = + - = - and x y z2 3 2 7- - =

7. ,p q r2 5 25+ - = p q r2 2 24- - = - and

p q r3 5 4- + =

8. ,x y z2 3 9- + = x y z3 2 2+ - = - and x y z3 5 14- + =

9. ,h j k3 3+ - = - h j k2 3+ + = - and h j k5 3 2 13- - = -

10. ,a b c2 7 3 7- + = a b c3 2 4+ + = - and a b c4 5 9+ - =

3.14 Exercises

You will solve 3 simultaneous

equations in later topics (for example,

in Chapter 10).

ch3.indd 137 5/20/09 11:59:49 AM


Test Yourself 3 1. Solve

(a) b8 3 22= -

(b) a a4 3

2 9-+

=

(c) ( )x x4 3 1 11 3+ = -

(d) x 3

4 3#+

-

(e) p p3 1 9#+ +

2. The compound interest formula is

.A P r1100

n

= +c m Find correct to 2

decimal places. (a) A when ,P r1000 6= = and n 4= (b) P when , .A r12 450 5 5= = and n 7=

3. Complete the square on (a) x x82 - (b) k k42 +

4. Solve these simultaneous equations. (a) x y 7 0- + = and x y3 4 26 0- + = (b) xy 4= and x y2 7 0- - =

5. Solve (a) 3 81x 2 =+ (b) 16 2y =

6. Solve (a) b3 1 5- =

(b) g g5 3 3 1- = +

(c) x2 7 1$-

7. The area of a trapezium is given by

2 ( )A h a b= +1 . Find

(a) A when ,h a6 5= = and b 7= (b) b when ,A h40 5= = and .a 4=

8. Solve x x2 3 1 02 - + = by factorisation (a) quadratic formula. (b)

9. Solve ,y2 3 1 101 #- + and plot your solution on a number line.

10. Solve correct to 3 signifi cant fi gures (a) x x7 2 02 + + = (b) y y2 9 02 - - = (c) n n3 2 4 02 + - =

11. The surface area of a sphere is given by .A r4 2r= Evaluate to 1 decimal place

(a) A when .r 7 8= (b) r when .A 102 9=

12. Solve .x7

343 92

--

13. Solve .x x11 18 02 2- +

14. Solve the simultaneous equations x y 162 2+ = and .x y3 4 20 0+ - =

15. The volume of a sphere is .V r34 3r=

Evaluate to 2 signifi cant fi gures (a) V when r 8= (b) r when V 250=

16. Which of the following equations has (i) 2 solutions (ii) 1 solution

(iii) no solutions?

(a) x x6 9 02 - + =

(b) x2 3 7- =

(c) x x2 7- = -

(d) x x 4 02 - + =

(e) x x2 1 2+ = -

17. Solve simultaneously , , .a b a b c a b c5 2 4 5+ = + + = - - =

18. Solve ,n3 5 52+ and plot the solution on a number line.

19. Solve , .x x x

13 4 0 1!+

= -^ h

ch3.indd 138 7/8/09 5:39:13 PM


20. Solve .9 27x x2 1 =+

21. Solve (a) y y2 3 5 52- +^ h (b) n n3 02 #+ (c) 3 27x2 1 =- (d) x5 1 393 - = (e) x5 4 11- = (f) t2 1 3$+ (g) x x2 8 02 #+ - (h) 8 4x x1 =+

(i) y 4 02 2- (j) x1 02 #-

(k) 27 9x2 1 =- (l) b4 3 5#-

(m) x x3 2 2 3+ = - (n) t t4 5 2- = + (o) x x2 32 1 + (p) m m 62 $+

(q) t

t2 3 51-

(r) y

y

1

122

-

+

(s) n

n2 4

3$-

(t) xx

2 13 2 1#

+

--

1. Find the value of y if .aa1y3 5

2=-

2. Solve .x a2 22

3. The solutions of x x6 3 02 - - = are in the form .a b 3+ Find the values of a and b .

4. Solve x x1

21

1 1-

-+

= correct to 3

signifi cant fi gures. ( )x 1!!

5. Solve .yy 6

12

#-

6. Factorise .x x x9 8 725 3 2- - + Hence solve .x x x9 8 72 05 3 2- - + =

7. Solve simultaneous equations y x x3 2= + and .y x 1= +

8. Find the value of b if x x b82 2- + is a perfect square. Hence solve x x8 1 02 - - = by completing the square.

9. Considering the defi nition of absolute

value, solve ,x

xx

3

3

-

-= where .x 3!

10. Solve .t t2 3 1 51+ + -

11. Solve .x x4 1 28#- -] ]g g

12. Solve 2 .x81

=3

13. The volume of a sphere is given by

.V r34 3r= Find the value of r when

.V 51 8= (correct to three signifi cant fi gures).

14. Solve .x x x3 4 2- + + = -

15. Find the solutions of x ax b2 02 - - = by completing the square.

16. Solve .y

y

3 2

63

2

#-

-

17. Given ,A P r1100

n

= +c m fi nd P

correct to 2 decimal places when . , .A r3281 69 1 27= = and .n 30=

18. Solve ( )x x3 8 2 12 = - and write the solution in the simplest surd form.

19. Solve .xx x

45 3 22

+

+

20. Solve .y y3 1 2 3 52- + +

Challenge Exercise 3

ch3.indd 139 7/25/09 3:35:05 PM

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