simultaneous equations sample · ... then x+y =4+3=7, and 2x+3y = 2(4)+3(3) = 17. ... 7x+3y = ¡5 f...

15Chapter 15Simultaneous

equations

Contents:

A Graphical solution

B Solution by substitution

C Solution by elimination

D Problem solving

AUS_10magentacyan yellow black

0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\AUS_10\AUS10_15\307AUS10_15.cdr Tuesday, 4 December 2012 9:15:49 AM EMMA

SAMPLE

308

Opening problem

SIMULTANEOUS EQUATIONS (Chapter 15)

Ewen wants to buy a pie, but only has $3:50 . He sees

his friend Andre leaving the canteen with a pie, and

asks him how much it cost.

“I’m not sure,” said Andre, “I bought a pie and a

sandwich. They cost me $7 altogether.”

Ewen saw another friend Samuel with a pie, and asked

him how much it cost.

“Well, I bought 2 pies and 3 sandwiches, and they cost

me $17 altogether.”

Things to think about:

a Can Ewen determine the cost of a pie using:

i Andre’s information only ii Samuel’s information only?

b Suppose a pie costs $x, and a sandwich costs $y.

Can you explain why: i x + y = 7 ii 2x + 3y = 17?

c Can you find values for x and y which make both equations true?

d Does Ewen have enough money to buy a pie?

In the Opening Problem, we found values for x and y which satisfied both x + y = 7 and

2x + 3y = 17 at the same time.

We say that

½x + y = 7

2x + 3y = 17is a set of simultaneous equations.

Notice that if x = 4 and y = 3, then x + y = 4 + 3 = 7, and

2x + 3y = 2(4) + 3(3) = 17.

So, x = 4, y = 3 is a solution to the set of simultaneous equations

½x + y = 7

2x + 3y = 17.

In this chapter we will investigate some techniques for solving simultaneous equations.

One way to solve simultaneous equations is to draw the graph of both equations on the same set

of axes. The point of intersection corresponds to a solution to the simultaneous equations.

A set of linear simultaneous equations can have either one, zero, or infinitely many solutions.

one solution

intersecting linesno solutions

parallel lines

infinitely many solutions

coincident lines

GRAPHICAL SOLUTIONA


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


SAMPLE

SIMULTANEOUS EQUATIONS (Chapter 15) 309

Self Tutor

Solve the following simultaneous equations graphically:

½y = 2x ¡ 4

2x + 3y = 12

We draw the graphs of y = 2x ¡ 4 and

2x + 3y = 12 on the same set of axes.

The graphs meet at the point (3, 2).

) the solution is x = 3, y = 2.

Check:

Substituting these values into:

² y = 2x ¡ 4 gives 2 = 2(3) ¡ 4 X

² 2x+3y = 12 gives 2(3)+3(2) = 12 X

EXERCISE 15A.1

1 Solve the following simultaneous equations graphically:

a

½y = 4x ¡ 1

y = 2x ¡ 3b

½y = 3x

y = ¡2x + 5c

½y = x + 2

y = ¡3x ¡ 6


a

½y = 2x ¡ 8

2x + 5y = 20b

½4x + y = 8

2x ¡ 3y = 18c

½3x ¡ y = ¡6

3x + 4y = ¡36

3 Try to solve the following simultaneous equations graphically. State the number of solutions

in each case.

a

½y = 4x + 1

y = 4x ¡ 3b

½4x ¡ 2y = ¡12

y = 2x + 6c

½5x + 3y = 2

5x + 3y = ¡7

USING TECHNOLOGY

If the solutions to simultaneous equations are not integers, it may be difficult to find the solution

on a graph drawn by hand. In these situations, we can use technology to graph the equations and

find the solution.

Example 1

y

x

y = 2x - 4y = 2x - 4

(3 2),(3 2),

2x + 3y = 122x + 3y = 12

6

4

-4-4

GRAPHICS

CALCULATOR

INSTRUCTIONS

GRAPHING

PACKAGEYou can use your graphics calculator or the graphing

package to graph the equations. When using your graphics

calculator, any equations in the general form Ax+By = C

will need to be rearranged to make y the subject.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\AUS_10\AUS10_15\309AUS10_15.cdr Wednesday, 5 December 2012 3:37:52 PM BRIAN

SAMPLE

310 SIMULTANEOUS EQUATIONS (Chapter 15)

Self Tutor

Use technology to solve the following simultaneous equations:

½y = 4x + 13

x + y = 7

We rearrange the second equation, so the

system is now:

½y = 4x + 13

y = 7 ¡ x

So, the solution is x = ¡1:2, y = 8:2 .

EXERCISE 15A.2

1 Use technology to solve the following simultaneous equations:

a

½y = 5x ¡ 2

y = x + 1b

½y = 3x ¡ 6

y = 1 ¡ 2xc

½y = 3x ¡ 4

x + y = ¡5

d

½x + y = 6

6x + y = 15e

½x + 2y = ¡8

7x + 3y = ¡5f

½3x ¡ 4y = ¡2

2x + 3y = ¡10

2 Try to solve the following simultaneous equations using technology. State the number of

solutions in each case.

a

½y = 1:5x + 2

3x ¡ 2y = ¡4b

½4x + 5y = ¡3

y = ¡0:8x + 1c

½x ¡ 4y = 7

2x ¡ 8y = ¡1

Example 2

Rearranging an

equation does not

change its solutions.

TI-84 PlusCasio fx-9860G Plus TI- spiren


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


SAMPLE


We will now consider some algebraic methods for solving simultaneous equations.

The method of solution by substitution is used when at least one equation is given with either

x or y as the subject of the formula. We substitute an expression for this variable into the other

equation.

Self Tutor

Solve simultaneously by substitution:

½y = x + 5

3x ¡ y = 1

y = x + 5 .... (1)

3x ¡ y = 1 .... (2)

Substituting (1) into (2) gives 3x ¡ (x + 5) = 1

) 3x ¡ x ¡ 5 = 1

) 2x = 6

) x = 3

Substituting x = 3 into (1) gives y = 3 + 5

) y = 8

The solution is x = 3, y = 8.

Check: (1) 8 = 3 + 5 X

(2) 3(3) ¡ 8 = 9 ¡ 8 = 1 X

EXERCISE 15B

1 Solve simultaneously:

a

½y = x + 1

4x ¡ 3y = 1b

½y = x ¡ 3

2x + 3y = 26c

½y = 3x ¡ 2

5x ¡ 2y = 5

d

½7x + 4y = ¡7

y = 8 ¡ 5xe

½y = 2x ¡ 12

y = 13 ¡ 3xf

½y = 3x + 4

5x + 3y = 5


a

½x = y ¡ 4

3x + 2y = 3b

½x = y + 9

5x + 2y = 10c

½7x ¡ 5y = 1

x = 2y + 4

d

½x = 2y ¡ 7

x = 13 ¡ 6ye

½y = 2x ¡ 2

x = 5y + 7f

½2x + 4y = 1

x = 10y ¡ 7


a

½y = 1

2x ¡ 2

3x ¡ 8y = 11b

½x = ¡2

3y

6x + 7y = 6c

½5x ¡ 12y = 18

y = 1

4x ¡ 1

SOLUTION BY SUBSTITUTIONB

Example 3

(x + 5) is substituted

for y in equation (2).


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


SAMPLE


d

½y = ¡1

2x ¡ 1

4x + 3y = ¡2e

½2x ¡ y = 3

x = 1

4y + 2

f

½14x + 15y = 25

y = 2

5x ¡ 3

4 a Try to solve by substitution:

½y = 3x + 1

9x ¡ 3y = 5

b What is the simultaneous solution for the equations in a?

5 a Try to solve by substitution:

½y = 3x + 1

6x ¡ 2y = ¡2

b How many simultaneous solutions do the equations in a have?

6 Consider the simultaneous equations:

½4x + 5y = ¡4 .... (1)

3x + 2y = ¡5 .... (2)

a Rearrange equation (1) so that y is the subject.

b Solve the simultaneous equations by substitution.

If both equations are presented in the general form Ax+By = C, then solution by substitution

is tedious. We instead use the method of elimination.

In this method, we make the coefficients of x (or y) the same size but opposite in sign, and then

add the equations. This has the effect of eliminating one of the variables.

Self Tutor

Solve simultaneously by elimination:

½4x + 3y = 2

x ¡ 3y = 8

The coefficients of y are the same size but opposite in sign.

We add the LHSs and the RHSs to get an equation which contains x only.

4x + 3y = 2 .... (1)

x ¡ 3y = 8 .... (2)

Adding, 5x = 10

) x = 2

Substituting x = 2 into (1) gives 4(2) + 3(y) = 2

) 8 + 3y = 2

) 3y = ¡6

) y = ¡2

The solution is x = 2, y = ¡2.

Check: In (2): (2) ¡ 3(¡2) = 2 + 6 = 8 X

SOLUTION BY ELIMINATIONC

Example 4

By adding the equations,

we eliminate the y variable.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


SAMPLE


In problems where the coefficients of x (or y) are not the same size or opposite in sign, we must

first multiply one or both equations by a constant.

Self Tutor

Solve simultaneously by elimination:

½3x + 2y = 7

2x ¡ 5y = 11

3x + 2y = 7 .... (1) 2x ¡ 5y = 11 .... (2)

To make the coefficients of y the same size but opposite in sign, we multiply

(1) by 5 and (2) by 2.

) 15x + 10y = 35 f(1) £ 5g

4x ¡ 10y = 22 f(2) £ 2g

Adding, 19x = 57

) x = 3

Substituting x = 3 into (1) gives 3(3) + 2y = 7

) 9 + 2y = 7

) 2y = ¡2

) y = ¡1

The solution is x = 3, y = ¡1.

Check: In (2): 2(3) ¡ 5(¡1) = 6 + 5 = 11 X

We can choose to eliminate either x or y. In Example 5 we could have eliminated x by multiplying

(1) by 2 and (2) by ¡3. We choose the variable which is easiest to eliminate.

EXERCISE 15C

1 What equation results when the following are added vertically?

a 3x + 4y = 6

8x ¡ 4y = 5

b 2x ¡ y = 7

¡2x + 5y = 5

c 7x ¡ 3y = 2

2x + 3y = 7

d 6x ¡ 11y = 12

3x + 11y = ¡6

e ¡7x + 2y = 5

7x ¡ 3y = 6

f 2x ¡ 3y = ¡7

¡2x ¡ 8y = ¡4

2 Solve using the method of elimination:

a

½5x ¡ y = 4

2x + y = 10b

½3x ¡ 2y = 7

3x + 2y = ¡1c

½¡5x ¡ 3y = 14

5x + 8y = ¡29

d

½4x + 3y = ¡11

¡4x ¡ 2y = 6e

½2x ¡ 5y = 14

4x + 5y = ¡2f

½¡6x ¡ y = 23

6x + 5y = ¡13

3 Give the equation that results when both sides of the equation:

a 2x + 5y = 1 are multiplied by 5 b 3x ¡ y = 4 are multiplied by ¡1

c x ¡ 7y = 8 are multiplied by 3 d 5x + 4y = 9 are multiplied by ¡2

e ¡3x ¡ 2y = 2 are multiplied by 6 f 4x ¡ 2y = 3 are multiplied by ¡4.

Example 5


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


SAMPLE



a

½2x + y = 8

x ¡ 3y = 11b

½3x + 2y = 7

x + 3y = 7c

½5x ¡ 2y = 17

3x ¡ y = 9

d

½2x + 5y = ¡14

¡6x + 2y = ¡9e

½7x + 3y = 5

5x ¡ 6y = 9f

½4x + 9y = 24

12x ¡ 7y = ¡30


a

½2x + 3y = 13

3x + 2y = 17b

½4x ¡ 3y = 1

2x + 5y = 7c

½2x + 5y = 14

5x ¡ 3y + 27 = 0

d

½7x + 2y = 20

13x + 3y = 34e

½3x + 7y = 5

5x + 11y = 10f

½5x ¡ 7y ¡ 9 = 0

4x ¡ 5y ¡ 5 = 0

6 Use the method of elimination to attempt to solve:

a

½2x ¡ y = 3

4x ¡ 2y = 6b

½3x + 4y = 6

6x + 8y = 7Comment on your results.

Many problems can be described mathematically by a pair of linear equations. The Opening

Problem is one example.

Once the equations are formed, they can then be solved simultaneously, and we can then answer

the original problem. The following method is recommended:

Step 1: Decide on two unknowns such as x and y. Do not forget the units.

Step 2: Write down two equations connecting x and y.

Step 3: Solve the equations simultaneously.

Step 4: Check your solutions with the original data given.

Step 5: Give your answer in sentence form.

The form of the original equations will help you decide whether to use substitution or elimination.

Self Tutor

Two numbers have a sum of 45 and a difference of 13. Find the numbers.

Let x and y be the unknown numbers, where x > y.

Then x + y = 45 .... (1) f‘sum’ means addg

and x ¡ y = 13 .... (2) f‘difference’ means subtractg

Adding, 2x = 58

) x = 29

Substituting x = 29 into (1) gives 29 + y = 45

) y = 16

The numbers are 29 and 16.

Check: In (1): 26 + 16 = 45 X In (2): 29 ¡ 16 = 13 X

PROBLEM SOLVINGD

Example 6

We need to find two

equations containing

two unknowns.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


SAMPLE


EXERCISE 15D

1 The sum of two numbers is 72, and their difference is 40. Find the numbers.

2 Find two numbers whose sum is 30 and half their difference is 7.

3 The larger of two numbers is three times the smallest number, and their difference is 34. Find

the two numbers.

Self Tutor

When shopping in Jamaica, 5 coconuts and 14 bananas cost me $8:70, and 8 coconuts and

9 bananas cost $9:90 .

Find the cost of each coconut and each banana.

Let each coconut cost x cents, and each banana cost y cents.

) 5x + 14y = 870 .... (1)

and 8x + 9y = 990 .... (2)

To eliminate x, we multiply (1) by 8 and (2) by ¡5.

) 40x + 112y = 6960 f(1) £ 8g

¡40x ¡ 45y = ¡4950 f(2) £ ¡5g

Adding, 67y = 2010

) y = 30

Substituting y = 30 in (2) gives 8x + 9(30) = 990

) 8x + 270 = 990

) 8x = 720

) x = 90

Check: In (1): 5 £ 90 + 14 £ 30 = 450 + 420 = 870 X

In (2): 8 £ 90 + 9 £ 30 = 720 + 270 = 990 X

The coconuts cost 90 cents each, and bananas cost 30 cents each.

4 At a plant shop, 2 geraniums and 3 marigolds cost $25, whereas 5 geraniums and 6 marigolds

cost $58. Find the cost of each type of plant.

5 Three pieces of fish and two serves of chips cost a

total of $8:10. Five pieces of fish and three serves

of chips cost a total of $13:25. Find the cost of each

piece of fish and each serve of chips.

6 Seven cups of coffee and four muffins cost a total of

$25:30, whereas two cups of coffee and three muffins

cost a total of $9:55. Find the cost of each item.

Example 7

The units are cents

on both sides of

each equation.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


SAMPLE


Self Tutor

I have only five cent and ten cent coins in my pocket. I have 24 coins altogether,

and their total value is $1:55. How many of each type of coin do I have?

Let x be the number of five cent coins, and y be the number of ten cent coins.

) x + y = 24 .... (1) fthe total number of coinsg

and 5x + 10y = 155 .... (2) fthe total value of the coins in centsg

To eliminate x, we multiply (1) by ¡5.

) ¡5x ¡ 5y = ¡120 f(1) £ ¡5g

5x + 10y = 155 f(2)g

Adding, 5y = 35

) y = 7

Substituting y = 7 into (1) gives x + 7 = 24

) x = 17

Check: In (1): 17 + 7 = 24 X

In (2): 5 £ 17 + 10 £ 7 = 85 + 70 = 155 X

I have 17 five cent coins and 7 ten cent coins in my pocket.

7 Margaret saves 50-cent and 10-cent coins. She has 56 of these coins, and their total value is

$17:60. How many of each coin type does she have?

8 Milk is sold in either 600 mL or 1 L cartons. A supermarket manager ordered 79:8 litres of

milk and received 93 cartons.

How many of each size carton did the manager receive?

9 The triangle alongside is equilateral.

Find a and b.

10 A rectangle has perimeter 56 cm. If 4 cm is taken

from the length and added to the width, the rectangle

becomes a square. Find the dimensions of the original

rectangle.

11 A car park charges a certain rate per hour before 3 pm,

and a higher rate per hour after 3 pm.

Parking from 11 am to 5 pm costs $18, and parking

from 9 am to 7 pm costs $31. Find the rates per hour

charged before and after 3 pm.

Example 8

(b + 6)� cm� (2a + b)� � cm

(4a - 1)� � cm


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


SAMPLE


12 An art store sells small and large painting sets. The small

sets contain 2 brushes and 3 tubes of paint, and the large

sets contain 5 brushes and 8 tubes of paint. Leanne buys

some of each of these sets, and receives a total of 37brushes and 58 tubes of paint. How many of each set

did she buy?

Simultaneous equations like

½2x + 3y = 11

3x ¡ 2y = 10are called 2 by 2 systems of equations,

because there are 2 equations involving 2 unknowns.

In this Investigation we will examine a method for solving 3 by 3 systems of equations.

Consider the 3 by 3 system of equations:

8<:

2x + 3y ¡ z = 5

¡x ¡ 3y + 4z = 2

x + 6y ¡ 5z = 7.

To solve this system of equations, we write the first two equations with the z variable on the

RHS, and find x and y in terms of z:

) 2x + 3y = 5 + z .... (1)

¡x ¡ 3y = 2 ¡ 4z .... (2)

Adding, x = 7 ¡ 3z .... (3)

Substituting x = 7 ¡ 3z into (1) gives 2(7 ¡ 3z) + 3y = 5 + z

) 14 ¡ 6z + 3y = 5 + z

) 3y = 7z ¡ 9

) y =7z ¡ 9

3.... (4)

We then substitute these values into the third equation to solve for z:

(7 ¡ 3z) + 6

µ7z ¡ 9

3

¶¡ 5z = 7

) 7 ¡ 3z + 14z ¡ 18 ¡ 5z = 7) 6z = 18) z = 3

So, using (3) and (4), x = 7 ¡ 3(3) = ¡2, and y =7(3)¡ 9

3= 4.

The solution is x = ¡2, y = 4, z = 3.

What to do:

1 Use the method above to solve these systems of linear equations:

a

8<:

x ¡ y + z = 9

3x + y ¡ z = 7

4x ¡ 2y + 5z = 35

b

8<:

2x ¡ y ¡ z = 3

3x + 2y ¡ 2z = 13

5x + y ¡ 3z = 18

Investigation Solving 3 by 3 systems of equations


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


SAMPLE

Review set 15

(x + y) cm

(x + 6) cm (2y + 1) cm


2 At a greengrocer:

² 3 pineapples, 2 cabbages, and 6 bananas cost a total of $27

² 2 pineapples, 3 cabbages, and 4 bananas cost a total of $23

² 4 pineapples, 1 cabbage, and 8 bananas cost a total of $31.

Find the cost of each item.


a

½y = 5x + 2

y = 2x ¡ 7b

½2x ¡ 3y = 18

x + y = 4

2 Use technology to solve simultaneously:

a

½y = 3x ¡ 4

y = 9 ¡ xb

½3x ¡ 4y = 3

7x ¡ 11y = 8


a

½x = 8 ¡ 3y

2x ¡ 7y = ¡10b

½y = 5x + 3

4x + y = 1

4 a Try to solve

½4x ¡ 2y = 3

y = 2x ¡ 5by substitution.

b How many solutions do the simultaneous equations in a have?

5 What equation results when the following are added vertically?

a

½4x ¡ 3y = 8

2x + 3y = ¡14b

½5x ¡ y = 7

¡5x ¡ 3y = 33

6 Solve simultaneously using elimination:

a

½¡2x + 7y = 13

2x ¡ 3y = ¡9b

½4x ¡ 6y = 21

3x + 5y = ¡8


a

½3x ¡ 2y = 18

y = x ¡ 7b

½4x + 5y = 2

5x ¡ 11y = 14

8 6 adult tickets and 5 student tickets for the theatre cost $168. 2 adult tickets and 3 student

tickets cost $72. Find the cost of each type of ticket.

9 Carmel has some 50-cent coins and some 20-cent coins in her purse. She has 21 coins in

total, and their total value is $8:40. How many of each type of coin does Carmel have?

10 The triangle alongside has perimeter 25 cm.

Find x and y.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\AUS_10\AUS10_15\318AUS10_15.cdr Tuesday, 4 December 2012 2:28:41 PM EMMA

SAMPLE

SIMUL

Practice test 15B Short response

Practice test 15A Multiple Choice

TANEOUS EQUATIONS (Chapter 15) 319

Click on the icon to obtain this printable test.

1 Solve graphically:

½y = 3x ¡ 8

y = ¡2x + 7.

2 Solve by substitution:

½y = 4x ¡ 7

x = 9 ¡ 2y.

3 Try to solve

½y = 3

5x ¡ 1

3x ¡ 5y = 5using technology.

State how many solutions this set of linear equations has.

4 State the equation that results when both sides of 2x¡ 3y = ¡4 are multiplied by ¡3.

5 Solve by elimination:

½13x + 6y = 1

7x + 4y ¡ 5 = 0.


a

½6x ¡ 5y = 2

y = 3x + 2b

½5x ¡ 7y = 2

¡3x + 4y = 5

7 a Try to solve

½y = 4x ¡ 2

8x ¡ 2y = 4by substitution.

b How many solutions do the simultaneous equations in a have?

8 The sum of two numbers is 42, and their difference is 12. Find the two numbers.

9 In an egg and spoon race, children must transport

eggs from one end of a course to the other. Points

are awarded for each egg successfully transported,

and points are deducted for each egg which drops

and breaks.

Jenny transported 5 eggs and dropped 2 eggs, for

a score of 11 points. Heath transported 8 eggs

and dropped 5 eggs, for a score of 5 points. How

many points are deducted for dropping an egg?

10 A sports store sells packs of 3 tennis balls for $5, and packs of 8 tennis balls for $12. In

one day the store sells $151 worth of tennis balls. Given that they sold 97 tennis balls in

total, how many packs of 3 balls did they sell?

PRINTABLE

TEST


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


SAMPLE

320

Practice test 15C Extended response

SIMULTANEOUS EQUATIONS (Chapter 15)

1 Consider the simultaneous equations

½y = 3x + 4 .... (1)

x = 2y ¡ 3 .... (2)

a Solve the simultaneous equations by substituting (1) into (2).

b State the equation that results when 2 £ (1) is added to (2). Hence check your

answer to a.

2 a On the same set of axes, draw the graphs of:

i y = 2x ¡ 5 ii x ¡ 4y = ¡8 iii y = ¡x ¡ 5

b Hence, solve the following simultaneous equations:

i

½y = 2x ¡ 5

x ¡ 4y = ¡8ii

½y = 2x ¡ 5

y = ¡x ¡ 5iii

½x ¡ 4y = ¡8

y = ¡x ¡ 5

3 a At a school fete, Nicole buys 5 books and 2 CDs for $50, and Sharyn buys 2 books

and 3 CDs for $31. Find the price of:

i a book ii a CD.

b Later in the day, the prices at the fete are dropped. Nicole’s and Sharyn’s purchases

would now cost $29 and $16 respectively. Find the new price of:

i a book ii a CD.

4 a Rearrange 2x + 5y = 7 so that y is the subject.

b Hence, solve

½2x + 5y = 7

7x ¡ 4y = 2by substitution.

c Check your answer by solving

½2x + 5y = 7

7x ¡ 4y = 2by elimination.

5 Jason paints signs on buildings. He can paint

letters which contain all straight edges quite easily,

but letters which contain a curved edge take

longer.

It takes Jason 25 minutes to paint the word

CHEMIST, and 43 minutes to paint the word

SUPERMARKET.

a On average, how long does Jason take to paint

each:

i straight edged letter

ii curved edged letter?

b How long would it take Jason to paint the

word ACCOUNTANT?


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


SAMPLE

simultaneous equations sample · ... then x+y =4+3=7, and 2x+3y = 2(4)+3(3) = 17. ... 7x+3y = ¡5 f...

Documents