4A Solving linear equations 4B Contents 4C 4D 4E linear equationslgs8maths.weebly.com/.../22594084/ch_4_linear_equations.pdf · 2019-10-24 · number AnD AlgebrA • lineAr AnD non-lineAr

4

opening Question

John supervises building sites where he regularly organises concrete pours. There are two companies that he can use. Angelico’s Concrete charges $700 plus $20 per cubic metre of concrete; Baux Cementing charges $1200 plus $15 per cubic metre of concrete. For what volumes of concrete should John use each of the two companies?

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

4A Solving linear equations 4B Solving linear equations with brackets 4C Solving linear equations with

pronumerals on both sides 4D Solving problems with linear equations 4E Rearranging formulas

WhAt Do You KnoW?

1 List what you know about linear equations. Create a concept map to show your list.

2 Share what you know with a partner and then with a small group.

3 As a class, create a large concept map that shows your class’s knowledge of linear equations.

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linear equations

ContentsLinear equationsAre you ready?Solving linear equationsSolving linear equationsSolving linear equations with bracketsSolving linear equations with bracketsSolving linear equations with pronumerals on both sidesSolving linear equations with pronumerals on both sidesSolving problems with linear equationsSolving problems with linear equationsRearranging formulasRearranging formulasSummaryChapter reviewActivities


102 maths Quest 9 for the Australian Curriculum

Are you ready?Try the questions below. If you have diffi culty with any of them, extra help can be obtained by completing the matching SkillSHEET on your eBookPLUS.

Solving one-step equations 1 Solve each of the following equations (that is, fi nd the value for the pronumeral that makes

each statement true).a x + 2 = 8 b y - 9 = 4

c 5k = 75 d x3

= -17

Checking solutions to equations 2 State whether the value given in brackets is the solution to the equation for each of the

following.a x + 23 = 51 (x = 28) b 4x = 108 (x = 25)c 2x - 7 = -3 (x = -2) d -5 - x = 2 (x = -7)

Solving equations 3 Solve each of the following equations.

a 2x + 1 = 7 b 3a - 5 = 13c 7m + 3 = 3 d 5y - 3 = -8

Expanding brackets 4 Expand each of the following.

a 6(x - 2) b -2(y + 1)c 4(2a + 3) d -5(2p - 7)

Writing equations from worded statements 5 Write an equation for each of the following statements, using x to represent the unknown

number.a When 2 is added to a certain number, the result is 9.b Eight times a certain number is 40.c When 11 is subtracted from a certain number, the result is 3.d Dividing a certain number by 6 gives a result of 2.

Transposing and substituting into a formula 6 a If P = 2l + 2w and P = 22, l = 7, fi nd the value of w.

b If A = lw and A = 78, w = 6, fi nd the value of l.

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x = 6 y = 13

k = 15 x = -51

Yes No

No Yes

x = 3 a = 6m = 0 y = -1

6x - 12 -2y - 28a + 12 -10p + 35

x + 2 = 98x = 40

x - 11 = 3x6

= 2

w = 4l = 13


103Chapter 4 linear equations

solving linear equationsWhat is a linear equation?

■ An equation is a mathematical statement of two equal algebraic expressions that contain one or more variables.

■ Solving linear equations involves fi nding a numerical value for the unknown pronumeral(s) or variable.

■ In a linear equation, the variable can only have an index of 1.

State whether each of the following equations is linear.

a 2x + 6 = x + 1 b x2 - 5 = x c12

- 1x

= x + 2

3

thinK Write

a Check that all characteristics have been satisfi ed for a linear equation and answer the question.

a The equation 2x + 6 = x + 1 is linear because it is an equation and the variable (x) has a power of 1.

b Check that all characteristics have been satisfi ed for a linear equation and answer the question.

b The equation x2 - 5 = x is not linear, because there is a variable with a power of 2; that is, x2.

c Check that all characteristics have been satisfi ed for a linear equation and answer the question.

c The equation 12

- 1x

= x + 23

is not linear, because

the variable that appears in the denominator on

the left hand side has a power of -1 1 1

xx=

− .

solving one-step equations ■ For an equation, the expression on the left-hand side of the equals sign has the same value as the expression on the right-hand side.

■ If the same arithmetic operation is performed to both sides of an equation, the equation remains valid. That is, the equation remains a true statement.

■ To solve an equation means to fi nd the value of the variable or pronumeral that, when substituted, will make the equation a true statement.

■ To solve a linear equation, perform the same arithmetic operations to both sides of the equation until the pronumeral is left by itself.

■ To solve a one-step equation, decide which operation was performed with the pronumeral in the process of forming the equation and then perform the inverse operation, as shown in the following table. This inverse operation has the effect of undoing the original operation.

Operation Inverse operation

+ -- +ì óó ì

4A

WorKeD eXAmple 1



WorKeD eXAmple 2

Solve each of the following linear equations.a x - 79 = 153 b x + 46 = 82

c 6x = 100 dx7

= 19

thinK Write

a 1 Write the equation. 79 is taken from the unknown number x to give 153. Therefore the inverse of subtraction must be applied.

a x - 79 = 153

2 Add 79 to both sides of the equation. (This is the inverse operation.)

x - 79 + 79 = 153 + 79

3 Simplify both sides to obtain the value of x.

x = 232

b 1 Write the equation. 46 has been added to x to give 82. Therefore the inverse of addition must be applied.

b x + 46 = 82

2 Subtract 46 from both sides of the equation.

x + 46 - 46 = 82 - 46

3 Simplify both sides to obtain the value of x. x = 36

c 1 Write the equation. The unknown number x has been multiplied by 6 to give 100. Therefore the inverse of multiplication must be applied.

c 6x = 100

2 Divide both sides of the equation by 6.66x

= 1006

3 Simplify both sides to obtain the value of x. x = 162

3

d 1 Write the equation. The unknown x has been divided by 7 to result in 19. Therefore the inverse of division must be applied.

d x7

= 19

2 Multiply both sides of the equation by 7.x7

ì 7 = 19 ì 7

3 Simplify both sides of the equation to obtain the value of x.

x = 133

Note: In each case the result can be checked by substituting the value obtained for x back into the original equation and confi rming that it will make the equation a true statement.

solving two-step equations ■ If two operations have been performed on the pronumeral it is known as a two step equation. ■ To solve two-step equations, establish the order in which the operations were performed.



■ Perform inverse operations in the reverse order to both sides of the equation. ■ Each inverse operation must be performed one step at a time. ■ This principle will apply to any equation with two or more steps, as shown in the examples that follow.

Solve the following linear equations.a 2y + 4 = 12b -6 - 2x = 12

thinK Write

a 1 Write the equation. Identify the order of operations. y has been multiplied by 2 and then 4 has been added to give the result 12 (ì 2 then + 4). The inverse operations needed in reverse order are subtract 4 from both sides of the equation and then divide by 2 (- 4 then ó 2).

a 2y + 4 = 12

2 Subtract 4 from both sides of the equation fi rst, then simplify.

2y + 4 - 4 = 12 - 4 2y = 8

3 Divide both sides of the equation by 2 to obtain the value of y.

22y

= 82

y = 4

b 1 Write the equation and to identify the order of operations. x has been multiplied by -2 and then 6 is subtracted from the result to give 12 (ì -2 then - 6). The inverse operations needed in reverse order are add 6 to both sides and then divide by -2 (+ 6 then ó -2).

b -6 - 2x = 12

2 Add 6 to both sides of the equation fi rst, then simplify.

-6 - 2x + 6 = 12 + 6-2x = 18

3 Divide both sides of the equation by -2 to obtain the value of x.

−−2−2−2−2−x

= 182−2−2

x = -9

■ Sometimes there may be just a negative sign in front of the pronumeral in the equation. This type of equation is still solved like the previous two-step equations because -x is the same as -1x.

Algebraic fractions — pronumeral in the numerator ■ Situations where there is a pronumeral in a numerator of an algebraic fraction will require an extra step. Again it is important to observe the order in which the steps occur, as we will see in these multi-step equations.

WorKeD eXAmple 3



Solve the following linear equations.

ax +++++

=====1

211 b

75−−−−− x

= -6.3

thinK Write

a 1 Write the equation. Identify the order of operations. 1 has been added to x and 2 has been divided into this result to give 11 (+ 1 then ó 2). The inverse operations, in reverse order, are to multiply both sides of the equation by 2 and then subtract 1 from both sides of the equation (ì 2 then - 1).

a x +12

= 11

2 Multiply both sides of the equation by 2.x +1

2 ì 2 = 11 ì 2

x + 1 = 22

3 Subtract 1 from both sides of the equation to obtain the value of x.

x + 1 - 1 = 22 - 1x = 21

b 1 Write the equation. Identify the order of operations. x has been multiplied by -1 and then 7 has been added. 5 has then been divided into this result to give -6.3 (ì -1, + 7 then ó 5). The inverse operations, in reverse order, are to multiply both sides by 5, subtract 7 from both sides and then divide by -1 (ì 5, - 7 then ó -1).

b 75− x

= -6.3

2 Multiply both sides of the equation by 5.7

5− x

ì 5 = -6.3 ì 5

7 - x = -31.5

3 Subtract 7 from both sides of the equation. 7 - x - 7 = -31.5 - 7-x = -38.5

4 Divide both sides by -1 to obtain the value of x. x = 38.5

Algebraic fractions — pronumeral in the denominator ■ If a pronumeral is in the denominator, there is an extra step involved in fi nding the solution.Consider the following example:

4 32x

=

In order to solve this equation manipulate the equation so that the x is removed from the denominator and placed into the numerator.Step 1: Multiply both sides of the equation by x.

4 32

4 32

432

x

xx xx x

x

=

× =x x× =x x×x x×x x

=

WorKeD eXAmple 4



Step 2: The equation can now be solved by multiplying both sides by 2 and then dividing both sides of the equation by 3.

432

4 2 3

4 23

33

83

=

× =4 2× =4 2

4 2×4 2 =

=

x

x

x

x

Therefore x = 83

.

■ This result could also be obtained directly by one of two methods:Method 1: Inverting each side of the equation

4 32

4232 4

383

xx

x

=

=

= 2 4×2 4 =

Method 2: Cross-multiplication4 3

24 2 3

8 3

83

xx

x

x

=

× =4 2× =4 2 ×8 3=8 3

=

Note: These last two methods can be applied directly to equations of the type 2 57x

= . It cannot

be applied to equations of the form 2

625x

+ =6+ =6 .

Solve each of the following linear equations by using either the method of inverting the equation or cross-multiplication.

a 3a

= 45

b5b

= −2−2−7

thinK Write

a 1 Write the equation. a3a

= 45

2 To use the method of inverting the equation, invert (interchange) the numerator and the denominator on both sides of the equation.

a3

= 54

3 Multiply both sides of the equation by 3 to obtain the value of a.

a3

ì 3 = 54

ì 3

a = 154

or a = 334

WorKeD eXAmple 5



b 1 Write the equation. b5b

= -27

2 To use the cross-multiplication method, multiply the numerator of the fi rst fraction by the denominator of the second fraction and vice versa. (This is called cross-multiplication.)

5 ì 7 = -2 ì b35 = -2b

3 Divide both sides of the equation by -2. 352−2−2

= −−2−2−2−2−b

-1712 = b

or b = -17.5

remember

1. A linear equation in terms of one variable, x, is an algebraic equation in which the pronumeral must not have any index or power other than 1. (Remember x1 = x.)

2. To solve an equation means to fi nd the value of the pronumeral that, when substituted, will make an equation a true statement.

3. If you perform the same arithmetic operation on both sides of an equation, the equation remains valid.

4. To fi nd the solution to the equation, perform identical arithmetic operations to both sides of the equation until the pronumeral is left by itself.

5. The solution can be verifi ed by substituting it into the original equation and checking whether it makes a true statement.

6. The inverse operations, which will allow us to obtain the pronumeral on its own, must be performed in the reverse order.

7. A CAS calculator can be used to solve linear equations.

solving linear equationsfluenCY

1 We 1 Which of the following equations are linear?a x + 3 = 7 b 2a - 5 = 9 c x2 - 2 = 9

d1x

x+ = 7 e2 1

3a2 1a2 12 1+2 1

= 4 f 5 + y = 9

g x3 + 3 = 5 hy2

= 8 i y2 - x2 = 9

j x2 + 2x + 3 = 0 k 5 = 2m - 11 l -9 = m2 + 3m 3b + 7 = 16 n y2 = x2 + 1 o -4a3 + 7b = 0

p3 2

53 2m3 23 2−3 2

= 18

2 We2 Solve each of the following linear equations. Check your answers by substitution.a x - 43 = 167 b x - 17 = 35 c x + 286 = 516d 58 + x = 81 e x - 78 = 64 f 209 - x = 305g 5x = 185 h 60x = 1200 i 5x = 250

eXerCise

4A

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Activity 4-A-2Riddle B

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Activity 4-A-3Riddle C

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inDiViDuAl pAthWAYs a, b, e, f, h, k, m, p

x = 210 x = 52 x = 230

x = 23 x = 142 x = -96

x = 37 x = 20 x = 50



jx

23 = 6 k

x17

= 26 l x9

= 27

m y - 16 = -31 n 5.5 + y = 7.3 o y - 7.3 = 5.5p 6y = 14 q 0.2y = 4.8 r 0.9y = -0.05

sy5

= 4.3 t y

7 57 5.7 5 = 23 u

y8

= -1.04

3 We3a Solve each of the following linear equations.a 2y - 3 = 7 b 2y + 7 = 3 c 5y - 1 = 0d 6y + 2 = 8 e 7 + 3y = 10 f 8 + 2y = 12g 15 = 3y - 1 h -6 = 3y - 1 i 6y - 7 = 140j 4.5y + 2.3 = 7.7 k 0.4y - 2.7 = 6.2 l 600y - 240 = 143

4 We3b Solve each of the following linear equations.a 3 - 2x = 1 b -3x - 1 = 5 c -4x - 7 = -19d 1 - 3x = 19 e -5 - 7x = 2 f -8 - 2x = -9g 9 - 6x = -1 h -5x - 4.2 = 7.4 i 2 = 11 - 3xj -3 = -6x - 8 k -1 = 4 - 4x l 35 - 13x = -5

5 Solve each of the following linear equations.a 7 - x = 8 b 8 - x = 7 c 5 - x = 5d 5 - x = 0 e 15.3 = 6.7 - x f 5.1 = 4.2 - xg 9 - x = 0.1 h 140 - x = 121 i -30 - x = -4j -5 = -6 - x k -x + 1 = 2 l -2x - 1 = 0

6 Solve each of the following linear equations.

ax4

= 3 bx3

= -1 cx8

= 12

d - x3

= 5

e - x2

= -8 f - x6

= 11 g23x

= 6 h52x

= -3

i - 34x = - 7 j -8

3x = 6 k

27x

= -2 l -310

x = - 15

7 We4 Solve each of the following linear equations.

az −1

3 = 5 b

z +14

= 8 cz − 4

2 = -4 d

67− z

= 0

e3

2− z

= 6 f− −z− −z− − 50

22 = -2 g

z − 4 42 1

.4 4.4 4.2 1.2 1

= -3 hz + 27 4.7 4.7 4

= 1.2

i140

150− z

= 0 j− −z− −z− − 0 4

20 4.0 4

= -0.5 kz − 6

9 = -4.6 l

z + 6573

= 1


a5 1

35 1x5 15 1+5 1

= 2 b2 5

72 5x2 52 5−2 5

= 3 c3 4

23 4x3 43 4+3 4

= -1 d4 13

9x4 1x4 14 1−4 1

= -5

e4 3

24 3−4 3x

= 8 f1 2

61 2−1 2x

= -10 g− −5 3− −5 3− −

95 3x5 3− −5 3− −x− −5 3− −

= 3 h− −10− −10− − 4

3x− −x− −

= 1

i4 2 6

5x4 2x4 24 2+4 2.

= 8.8 j5 0 7

0 33 1

5 0x5 05 0−5 0−

= −.0 3.0 3

3 1.3 1 k1 0 5

41 0−1 0. x

= -2.5 l− −3 8− −3 8− −

143 8x3 8− −3 8− −x− −3 8− −

= 12

9 We5 Solve each of the following linear equations by using either the method of inverting the equation or cross-multiplication.

a2x

= 12

b3x

= 7 c−4x

= 72

d5x

= −34

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x = 138 x = 442 x = 243

y = -15 y = 1.8 y = 12.8y = 21

3y = 24 y = - 1

18

y = 21.5 y = 172.5 y = -8.32

y = 5 y = -2 y = 0.2y = 1 y = 1 y = 2

y = 51

3y = -12

3y = 24.5

y = 1.2 y = 22.25 y = 383

600

x = 1 x = -2 x = 3x = -6 x = -1 x = 1

2x = 12

3x = -2.32 x = 3

x = - 56

x = 114

x = 3 1

13

x = -1 x = 1 x = 0

x = 5 x = -8.6 x = -0.9

x = 8.9 x = 19 x = -26x = -1 x = -1 x = −1

2

x = 12 x = -3 x = 4 x = -15

x = 16 x = -66 x = 9 x = -115

x = 913

x = -214

x = -7 x = 23

x = 4 x = 37

x = -117 x = -62

3

7 a z = 16 b z = 31 c z = -4

d z = 6 e z = -9 f z = -6

g z = -1.9 h z = 6.88 i z = 140

j z = 0.6 k z = -35.4 l z = 8

8 a x = 1 b x = 13 c x = -2

d x = -8 e x = -4

f x = 301

2

g x = -6 h x = - 7

10 i x = 10.35

j x = 0.326 k x = 22 l x = -5



e0 40 4.0 4x

= 92

f8x

= 1 g−4x

= 23

h−6x

= −45

i1 7.1 7.1 7x

= 13

j6x

= -1 k4x

= −1522

l50x

= −35

4310 mC a The solution to the equation 82 - x = 44 is:

A x = 126 B x = -126 C x = 122 D x = -38 E x = 38b What is the solution to the equation 5x - 12 = -62?

A x = -14.8 B x = 14.8 C x = 10 D x = -10 E x = -50

c What is the solution to the equation x −1

2 = 5.3?

A x = 9.6 B x = 10.6 C x = 11.6 D x = 2 E x = 12.611 Solve each of the following linear equations.

a 3a + 7 = 4 b 5 - b = -5 c 4c - 4.4 = 44 dd − 467

= 0

e 5 - 3e = -10 f23f

= 8 g 100 = 6g + 4.2 hh + 2

6 = 5.5

i 452i - 124 = -98 j6 1

17j6 1j6 16 1−6 1

= 0 k12

5− k

= 4 ll − 5 2

3 45 2.5 2

3 4.3 4 = 1.5

unDerstAnDing

12 Write the following worded statements as a mathematical sentence and then solve for the unknown.a Seven is added to the product of x and 3, which gives the result of 4.b Four is divided by x and this result is equivalent to 2

3.

c Three is subtracted from x and this result is divided by 12 to give 25. 13 Driving lessons are usually quite expensive but a discount of $15 per lesson is given if a

member of the family is a member of the automobile club. If 10 lessons cost $760 (after the discount), fi nd the cost of each lesson before the discount.

14 Anton lives in Australia and his pen pal, Utan, lives in USA. Anton put a post on his Facebook page that his home town of Horsham experienced one of the hottest days on record with a temperature of 46.7 èC. Utan commented on Anton’s status saying that his home town had experienced a day hotter than that, with the temperature reaching 113 èF. Which hometown experienced the hotter day?

reAsoning

15 Santo solved the following linear equation 9 = 5 - 4x. His second inverse step was to divide by 4. Trudy, his mathematics buddy, said that he was incorrect. a Explain why Trudy thinks Santo’s method is

incorrect. b Using Santo’s second step, add a third step that

would result in the correct answer of x = -1.

solving linear equations with brackets

■ Equations can be expressed in factorised form (containing brackets). ■ To solve equations containing brackets, either expand the bracket fi rst or divide both sides by the coeffi cient of the bracket. If the coeffi cient is not a factor of the other side, then a fraction will be created.

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How are linear equations defined?

4b

x = 445

x = 8 x = -6 x = 7.5

x = 5.1 x = -6 x = -51315

x = -6137

✔

✔

✔

a = -1 b = 10 d = 4

e = 5 f = 12g = 1529

30

h = 31

j = 16

k = -8 l = 10.3i = 13226

-1

6

303

$91

Anton (Utan’s temperature was 45 èC.)

Answers will vary.

c = 12.1



Solve each of the following linear equations.a 7(x - 5) = 28 b 6(x + 3) = 7

thinK Write

a 1 Write the equation. a 7(x - 5) = 28

2 Observe that 7 is a factor of 28, so divide both sides of the equation by 7.

7 57

( )7 5( )7 57 5( )7 5x7 5( )7 57 5( )7 5−7 5( )7 5 =

287

3 Add 5 to both sides of the equation to solve for x.

x - 5 = 4x - 5 + 5 = 4 + 5

x = 9

b 1 Write the equation. b 6(x + 3) = 7

2 Observe that 6 is not a factor of 7, so it will be easier to expand the brackets. (The other method will not give you an incorrect answer but may make the calculations more tedious.)

6x + 18 = 7

3 Subtract 18 from both sides of the equation.

6x + 18 - 18 = 7 - 186x = -11

4 Divide both sides of the equation by 6 to solve for x.

66x

= −116

x = -11

6 (or -15

6)

Note: The equation 7(x - 5) = 28 in part a could also be solved by expanding the brackets; however, it would take an unnecessary extra step.

remember

If an equation contains brackets, expand the brackets fi rst, unless dividing both sides by the coeffi cient in front of the brackets does not create a fraction.

solving linear equations with bracketsfluenCY

1 We 6a Solve each of the following linear equations.a 5(x - 2) = 20 b 4(x + 5) = 8c 6(x + 3) = 18 d 5(x - 41) = 75e 8(x + 2) = 24 f 3(x + 5) = 15g 5(x + 4) = 15 h 3(x - 2) = -12i 7(x - 6) = 0 j -6(x - 2) = 12k 4(x + 2) = 4.8 l 16(x - 3) = 48

WorKeD eXAmple 6

eXerCise

4b

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Activity 4-B-1Crossword A

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inDiViDuAl pAthWAYs

x = 6 x = -3x = 0 x = 56x = 1 x = 0x = -1 x = -2

x = 6 x = 0x = -0.8 x = 6



2 We6b Solve each of the following equations.a 6(b - 1) = 1 b 2(m - 3) = 3c 2(a + 5) = 7 d 3(m + 2) = 2e 5(p - 2) = -7 f 6(m - 4) = -8g -10(a + 1) = 5 h -12(p - 2) = 6i -9(a - 3) = -3 j -2(m + 3) = -1k 3(2a + 1) = 2 l 4(3m + 2) = 5

3 Solve each of the following equations.a 9(x - 7) = 82 b 2(x + 5) = 14c 7(a - 1) = 28 d 4(b - 6) = 4e 3(y - 7) = 0 f -3(x + 1) = 7g -6(m + 1) = -30 h -4(y + 2) = -12i -3(a - 6) = 3 j -2(p + 9) = -14k 3(2m - 7) = -3 l 2(4p + 5) = 18

4 Solve the following linear equations. Round the answers correct to 3 decimal places where appropriate.a 2(y + 4) = -7 b 0.3(y + 8) = 1c 4(y + 19) = -29 d 7(y - 5) = 25e 6(y + 3.4) = 3 f 7(y - 2) = 8.7g 1.5(y + 3) = 10 h 2.4(y - 2) = 1.8i 1.7(y + 2.2) = 7.1 j -7(y + 2) = 0k -6(y + 5) = -11 l -5(y - 2.3) = 1.6

5 mC a The best fi rst step in solving the equation 7(x - 6) = 23 would be to:A add 6 to both sidesB subtract 7 from both sidesC divide both sides by 23D expand the bracketsE multiply both sides by 0.7

b The solution to the equation 84(x - 21) = 782 is closest to:A x = 9.31 B x = 9.56C x = 30.31 D x = -11.69E x = 21

unDerstAnDing

6 In 1974 a mother is 6 times as old as her daughter. The mother turned 50 in the year 2000. In what year was the mother double her daughter’s age?

7 New edging is to be placed around the perimeter of a rectangular children’s playground. The width of the playground is x m and the length is 7 metres longer than the width. a Write down an expression that could be used to determine the perimeter of the

playground. Write your answer in factorised form. b If the minimum amount, in metres, of edging required is 54. Determine the dimensions,

in metres, of the playground.

reAsoning

8 Juanita is solving the following equation: 2(8 - x) = 10. She performs the following operations to both sides of the equation in order, +8, ó2. Explain why Juanita will not fi nd the correct value of x using her order of inverse operations.

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Activity 4-B-2Crossword B

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Activity 4-B-3Crossword C

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inDiViDuAl pAthWAYs

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Digital docHistory of Maths:

Grace Murray Hopper

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refleCtion

Explain why there are two possible methods for solving equations in factorised form.

x = 1619

x = 2a = 5 b = 7

y = 7 x = -313

m = 4 y = 1a = 5 p = -2

m = 3 p = 1

y = -7.5 y = -4.667y = -26.25 y = 8.571

y = -2.9 y = 3.243y = 3.667 y = 2.75

y = 1.976 y = -2y = -3.167 y = 1.98

✔

✔

1990

2(2x + 7)

Width 10 m, length 17 m

Answers will vary.

2 a b = 116

b m = 412

c a = -112

d m = -113

e p = 35

f m = 223

g a = -112

h p = 112

i a = 313

j m = -212

k a = -16

l m = - 14



solving linear equations with pronumerals on both sides

■ Equations can be expressed with pronumerals on both sides of the equation. ■ To solve these types of equations, perform an operation to both sides of the equation so that the pronumeral is removed from one of the sides.

Solve each of the following linear equations.a 5y = 3y + 3 b 7x + 5 = 2 - 4xc 3(x + 1) = 14 - 2x d 2(x + 3) = 4(x + 7)

thinK Write

a 1 Write the equation. We want to move y to the left-hand side of the equation.

a 5y = 3y + 3

2 Create a single pronumeral term by subtracting 3y from both sides of the equation.

5y - 3y = 3y + 3 - 3y2y = 3

3 Divide both sides of the equation by 2 to fi nd the value of y.

22y

= 32

y = 32

(or 112)

b 1 Write the equation. We want to move x to the left hand side of the equation.

b 7x + 5 = 2 - 4x

2 Create a single pronumeral term by adding 4x to both sides of the equation.

7x + 5 + 4x = 2 - 4x + 4x11x + 5 = 2

3 Subtract 5 from both sides of the equation. 11x + 5 - 5 = 2 - 511x = -3

4 Divide both sides of the equation by 11 to fi nd the value of x.

1111

x = −3

11

x = −311

c 1 Write the equation. c 3(x + 1) = 14 - 2x

2 Before we can isolate the pronumeral to one side of the equation, we have to expand the bracket.

3x + 3 = 14 - 2x

3 We want to move the x to the left hand side of the equation. Create a single pronumeral term by adding 2x to both sides of the equation.

3x + 3 + 2x = 14 - 2x + 2x5x + 3 = 14

4 Subtract 3 from both sides of the equation. 5x + 3 - 3 = 14 - 35x = 11

5 Divide both sides of the equation by 5. 5x = 11

x = 115

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InteractivitySolving

equationsint-2764

WorKeD eXAmple 7



d 1 Write the equation. d 2(x + 3) = 4(x + 7)

2 Before we can isolate the pronumeral on one side of the equation, we will have to either (i) expand the brackets fi rst or(ii) divide both sides of the equation by 2.Either method is correct, but the latter will give a more direct answer.

2(x + 3) = 4(x + 7)

3 Divide both sides of the equation by 2. 2 32

( )2 3( )2 32 3( )2 3x2 3( )2 32 3( )2 3+2 3( )2 3 =

4 72

( )4 7( )4 7( )x( )4 7( )4 7x4 7( )4 74 7( )4 7+4 7( )4 7

x + 3 = 2(x + 7)

4 Expand the bracket. x + 3 = 2x + 14

5 Isolate x on one side of the equation. For ease take x from both sides.

x + 3 = 2x + 14x + 3 - x = 2x + 14 - x

3 = x + 14

6 Finally subtract 14 from both sides of the equation.

3 - 14 = x + 14 - 14-11 = x

\ x = -11

remember

If there are pronumerals on both sides of an equation, combine the pronumeral terms into one by either addition or subtraction.

solving linear equations with pronumerals on both sides

fluenCY

1 We7a Solve each of the following linear equations.a 5y = 3y - 2 b 6y = -y + 7c 10y = 5y - 15 d 25 + 2y = -3ye 8y = 7y - 45 f 15y - 8 = -12yg 7y = -3y - 20 h 23y = 13y + 200i 5y - 3 = 2y j 6 - 2y = -7yk 24 - y = 5y l 6y = 5y - 2

2 mC a To solve the equation 3x + 5 = -4 - 2x, the fi rst step is to:A add 3x to both sidesB add 5 to both sidesC add 2x to both sidesD subtract 2x from both sidesE subtract 4 from both sides

b To solve the equation 6x - 4 = 4x + 5, the fi rst step is to:A subtract 4x from both sidesB add 4x to both sidesC subtract 4 from both sidesD add 5 to both sidesE add 6x to both sides

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Activity 4-C-1What is the word? A

doc-4015

Activity 4-C-2What is the word? B

doc-4016

Activity 4-C-3What is the word? C

doc-4017

inDiViDuAl pAthWAYs

eXerCise

4C

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doc-6158

y = -1 y = 1y = -3 y = -5

y = -45 y = 827

y = -2 y = 20y = 1 y = -11

5y = 4 y = -2

✔

✔



3 We7b Solve each of the following linear equations.a 2x + 3 = 8 - 3x b 4x + 11 = 1 - xc x - 3 = 6 - 2x d 4x - 5 = 2x + 3e 3x - 2 = 2x + 7 f 7x + 1 = 4x + 10g 5x + 3 = x - 5 h 6x + 2 = 3x + 14i 2x - 5 = x - 9 j 10x - 1 = -2x + 5k 7x + 2 = -5x + 2 l 15x + 3 = 7x - 3

4 Solve each of the following linear equations.a x - 4 = 3x + 8 b 3x + 12 = 4x + 5c 2x + 9 = 7x - 1 d -2x + 7 = 4x + 19e -3x + 2 = -2x - 11 f 11 - 6x = 18 - 5xg 6 - 9x = 4 + 3x h x - 3 = 18x - 1i 5x + 13 = 15x + 3

5 mC a The solution to 5x + 2 = 2x + 23 is:A x = 3 B x = -3 C x = 5 D x = 7 E x = -7

b The solution to 3x - 4 = 11 - 2x is:A x = 15 B x = 7 C x = 3 D x = 5 E x = 11

6 We7c, d Solve each of the following.a 5(x - 2) = 2x + 5 b 7(x + 1) = x - 11c 2(x - 8) = 4x d 3(x + 5) = xe 6(x - 3) = 14 - 2x f 9x - 4 = 2(3 - x)g 4(x + 3) = 3(x - 2) h 5(x - 1) = 2(x + 3)i 8(x - 4) = 5(x - 6) j 3(x + 6) = 4(2 - x)k 2(x - 12) = 3(x - 8) l 4(x + 11) = 2(x + 7)

unDerstAnDing

7 Aamir’s teacher gave him an algebra problem and told him to solve it. Can you help him? 3x + 7 = x2 + k = 7x + 15 What is the value of k?

8 A classroom contained an equal number of boys and girls. Six girls left to play hockey, leaving twice as many boys as girls in the classroom. What was the original number of students present?

reAsoning

9 Express the following information as an equation, then use it to show that n = 29.

150 − 31n20n + 50

n − 36

−98

n − 36

n − 36

solving problems with linear equationsConverting worded sentences to algebraic equations

■ As previously stated, algebra is a form of mathematical language. ■ Translations are required to write statements or words in algebraic form. ■ Looking for keywords such as ‘less than’, ‘larger than’, ‘dividing’ enable translations to be made.

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Digital docWorkSHEET 4.2

doc-6159

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InteractivityBad egg

int-0741

refleCtion

Draw a diagram that could represent 2x + 4 = 3x + 1.

4D

x = 1 x = -2x = 3 x = 4

x = 9 x = 3x = -2 x = 4x = -4 x = 1

2x = 0 x =

−34

x = -6 x = 7x = 2 x = -2

x = 13 x = -7x = 1

6x = - 2

17x = 1

✔

✔

x = 5x = -8

x = 4x = -18x = 2

3x = 0

-3

24

3(n - 36) - 98 = -11n + 200

x = -3 x = -71

2 x = 10

11 x = 32

3 x = -13

7 x = -15



Write linear equations for each of the following statements, using x to represent the unknown. (Do not attempt to solve the equations.)a When 6 is subtracted from a certain number, the result is 15.b Three more than seven times a certain number is zero.c When dividing a certain number by 2, the answer is 4 more than that certain number.

thinK Write

a 1 Let x be the certain number. Break up the sentence into parts which indicate the kind of arithmetic or algebra required. ‘When 6 is subtracted from’: - 6‘A certain number’: xThe result is 15: = 15.

a

2 Form the algebraic equation from the individual parts. Note the change in order of the parts.

x - 6 = 15

b 1 Let x be the certain number. Break up the sentence into parts that indicate the kind of arithmetic or algebra required. ‘Three more than’: + 3 ‘Seven times a certain number’: 7xIs zero: = 0.

b

2 Form the algebraic equation from the individual parts.

7x + 3 = 0

c 1 Let x be the certain number. Break up the sentence into parts that indicate the kind of arithmetic or algebra required.

‘When dividing a certain number by 2’: x2

‘The answer is’: =‘4 more than that certain number’: x + 4.

c

2 Form the algebraic equation from the individual parts.x2

= x + 4

■ The ability to solve linear equations is important in the real-world.Many different occupations require the skill of solving linear equations, such as economics, engineering and computer programming.

■ Converting worded problems into algebraic form requires looking for keywords in the statement of the problem.

WorKeD eXAmple 8



If 3 is added to a certain number and the result is multiplied by 12, the answer is 108. Determine the original number.

thinK Write

1 Defi ne a pronumeral representing the certain number.

Let x = a certain number.

2 Follow the instructions, to build up the equation.‘3 is added to a number’: x + 3.‘The result is multiplied by 12’: 12(x + 3).The answer is 108: 12(x + 3) = 108.

12(x + 3) = 108

3 Solve the equation. Since 12 is a factor of 108, divide both sides by 12 fi rst. (Then you can omit the brackets.)

1212

( )3( )3( )x( )( )+( ) =

10812

x + 3 = 9

4 Subtract 3 from both sides. x + 3 - 3 = 9 - 3x = 6

5 Give a worded answer. The original number was 6.

Taxi charges are $3.60 plus $1.38 per kilometre for any trip in Melbourne. If Elena’s taxi fare was $38.10, how far did she travel?

WorKeD eXAmple 9

WorKeD eXAmple 10



thinK Write

1 The question being asked is how far Elena travelled. Let the pronumeral stand for the distance travelled.

Let x = distance travelled.

2 If it costs $1.38 to travel 1 kilometre, write the distance-based cost; that is, the cost to travel x kilometres.

Distance-based cost = $1.38 ì x= $1.38x

3 Write the constant charge; that is, the cost for booking and getting into the taxi, for 0 km travelled.

Constant charge = $3.60

4 Write the total cost by adding together the constant charge and the distance-based cost.

Total cost = $3.60 + $1.38x

5 Write the total cost of Elena’s trip, as given in the question.

Total cost = $38.10

6 Form an equation by equating the two expressions for the total cost of the trip. (Omit the dollar signs.)

3.60 + 1.38x = 38.10

7 Solve the equation: fi rst subtract 3.60 from both sides.

3.60 + 1.38x - 3.60 = 38.10 - 3.601.38x = 34.50

8 Divide both sides of the equation by 1.38 to fi nd the value of x.

x = 34 501 38

.1 3.1 3

= 25

9 State the solution in words. Elena’s journey was 25 kilometres.

The SIVA car rental company charges $50 per day plus $1.20 per kilometre for a car rental.The HURTS company charges $40 per day plus $1.35 per kilometre.Nathan wishes to rent a car for 3 days.How far can he travel so that the cost from either company is the same?

WorKeD eXAmple 11



thinK Write

1 We need to fi nd the cost of renting a car from each company. This cost is dependent on the distance travelled. Defi ne a pronumeral representing the distance travelled.

Let x = distance travelled in km.

2 Write an equation for the total cost of renting the car.

Total cost = fl at fee ì 3 + cost per km ì x

3 Write the amount of the fl at fee and cost per km charged by SIVA.

SIVA: fl at fee = $50; cost per km = $1.20

4 Determine the total cost of renting from SIVA. Total cost = 50 ì 3 + 1.20 ì x= 150 + 1.2x

5 Write the amount of the fl at fee and the cost per km charged by HURTS.

HURTS: fl at fee = $40; cost per km = $1.35

6 Determine the total cost of renting from HURTS.

Total cost = 40 ì 3 + 1.35 ì x= 120 + 1.35x

7 Equate the two equations. Total cost (SIVA) = Total cost (HURTS)150 + 1.2x = 120 + 1.35x

8 Subtract 1.2x from both sides of the equation. 150 + 1.2x - 1.2x = 120 + 1.35x - 1.2x150 = 120 + 0.15x

9 Subtract 120 from both sides of the equation. 150 - 120 = 120 + 0.15x - 12030 = 0.15x

10 Divide both sides of the equation by 0.15.30

0 150 1.0 1 =

0 150 15

0 1.0 10 1.0 1

x

200 = x

11 Give a worded answer to the question. If Nathan travels 200 km over 3 days the cost will be the same.

remember

How to solve worded problems.1. Identify the unknown quantity and use a pronumeral to represent it.2. Search for keywords that indicate the steps needed for the solution.3. Create a linear equation from the information provided in the question.4. Solve the equation.5. Interpret the result and write the worded answer.

solving problems with linear equationsfluenCY

1 We8 Write linear equations for each of the following statements, using x to represent the unknown. (Do not attempt to solve the equations.)a When 3 is added to a certain number, the answer is 5.b Subtracting 9 from a certain number gives a result of 7.

eXerCise

4D

x + 3 = 5x - 9 = 7



c Seven times a certain number is 24.d A certain number divided by 5 gives a result of 11.e Dividing a certain number by 2 equals -9.f Three subtracted from fi ve times a certain number gives a

result of -7.g When a certain number is subtracted from 14 and this result is then

multiplied by 2, the result is -3.h When 5 is added to three times a certain number, the answer is 8.i When 12 is subtracted from two times a certain number, the result

is 15.j The sum of 3 times a certain number and 4 is divided by 2, which gives a result of 5.

2 mC Which equation matches the following statement?a A certain number, when divided by 2, gives a result of -12.

A x = −122

B 2x = -12

Cx2

= -12 Dx

12 = -2

E2x

= -12

b Dividing 7 times a certain number by -4 equals 9.

Ax

−=

49 B

− =47

9x

C7

49

+−

=xD

74

9x

−=

E− =47

9x

c Subtracting twice a certain number from 8 gives 12.A 2x - 8 = 12 B 8 - 2x = 12C 2 - 8x = 12 D 8 - (x + 2) = 12E 12 = 8x - 2

d When 15 is added to a quarter of a number, the answer is 10.

A 15 + 4x = 10 B 10 = x4

15+

Cx +15

4 = 10 D 15

4+x

= 10

E 10 = 154

+ x

unDerstAnDing

3 We 9 When a certain number is added to 3 and the result is multiplied by 4, the answer is the same as when the same number is added to 4 and the result is multiplied by 3. Find the number.

4 One half my age is 10 years more than one-third my age. How old am I? 5 We 10 The Green Cab taxi company charges $3.25 plus $0.72 per kilometre. Michael has $12

to spend on the taxi fare.a How far can he go in the taxi?b If he lives 13 km away, will he make it home in the taxi? If not, how far will he have to

walk?

eBookpluseBookplus

Activity 4-D-1Linear equations and

problem solvingdoc-4018

Activity 4-D-2Use of linear

equations in worded problemsdoc-4019

Activity 4-D-3Applying linear

equations to real-life problemsdoc-4020

inDiViDuAl pAthWAYs

eBookpluseBookplus


doc-6154

7x = 24x5

= 11x2Three subtracted from fi ve times a certain number gives a 2Three subtracted from fi ve times a certain number gives a Three subtracted from fi ve times a certain number gives a

= -9

5x - 3 = -7

2(14 - x) = -33x + 5 = 8

2x - 12 = 153 4

2

3 4x3 43 4+3 4 = 5

✔

✔

✔

✔

0

60 years

12.15 km

He will have to walk 0.85 km (850 m).



6 The cost of running a small aeroplane between Geelong and Hamilton is $4000 per round trip. The plane holds 24 passengers. If the return fare is $180, how many passengers are required so that the company does not lose money?

7 A maker of an orange-juice drink can purchase her raw materials from two sources. The first source provides liquid with 6% orange juice, while the second source provides liquid with 3% orange juice. She wishes to make 1 litre of drink with 5% orange juice. Let x equal amount of liquid purchased from the first source.a Write an expression for the amount of orange juice from the first supplier, given that x is

the amount of liquid.b Write an expression for the amount of liquid from the second supplier, given that x is the

amount of liquid used from the first supplier.c Write an expression for the amount of orange juice from the second supplier.d Write an equation for the total amount of orange juice in the mixture of the 2 supplies,

given that 1 litre of drink is mixed to contain 5% orange juice.e How much of the first supplier’s liquid should she use?

8 The cost of producing computer CD-ROMs is quoted as $1200 plus $0.95 per disk. If Maya’s recording studio has a budget of $2100, how many CDs can she get made?

23 passengers

0.06x

(1 - x)0.03(1 - x)

0.06x + 0.03(1 - x) = 0.052

3 L

947



9 We 11 Joseph wishes to have some flyers delivered for his grocery business. Post Quick quotes a price of $200 plus 50 cents per flyer, while Fast Box quotes $100 plus 80 cents per flyer.a If Joseph needs to order 1000 flyers, which distributor would be the cheapest to use?b For what number of fliers will the cost be the same for either distributor?

10 Rachel, a bushwalker, goes on a 4-day journey. She travels a certain distance on the first day, half that distance on the second day, a third that distance on the third day and a fourth of that distance on the fourth day. If the total journey is 50 km, how far did she walk on the first day?

reAsoning

11 Svetlana, another bushwalker goes on a 5-day journey, using the same pattern as Rachel in the previous question (a certain amount, then half that amount, then one third, one fourth and one fifth). If her journey is also 50 km, show that she travelled 21123

137 km on the first day.

12 Nile.com, the internet bookstore, advertises its shipping cost to Australia as a flat rate of $20 for up to 10 books; Sheds & Meager, a competing bookstore, offers a rate of $12 plus $1.60 per book. For how many books (6, 7, 8, 9 or 10) is Nile.com’s cost a better deal?

13 A new internet bookstore, Mississippi.com, is starting up. It offers the following shipping costs: a flat rate of $10 plus $1.50 for each book over 3 books plus an additional $1 for each book over 6 books delivered.a Write an expression for the shipping cost of x books, if x > 6.b Show that this expression can be simplified to

2.5x - 0.5.c Compare the cost of Mississippi.com for 10

books with Nile.com’s flat rate of $20.d Compare the cost of Mississippi.com with

Sheds & Meager (from question 12) and find the break-even point.

rearranging formulas ■ Formulas are written in terms of two or more pronumerals. ■ One pronumeral is usually written on one side of the equal sign. ■ When rearranging formulas, use the same methods as for solving linear equations (use inverse operations in reverse order).

■ The difference between rearranging formulas and solving linear equations is that rearranging formulas does not require a value for the pronumeral(s) to be found.

■ The pronumeral required on the left hand side of the formula is known as the subject of the formula.

refleCtion

Why is it important to define the pronumeral used in forming a linear equation to solve a problem?

4e

Post Quick distributor, cost = $700

333 flyers, cost = $366.50 and $366.40

24 km

Answers will vary.

6, 7, 8, 9, or 10 books

10 + 1.5(x - 3) + 1(x - 6)

Check with your teacher.Mississippi.com is $4.50 more expensive.

14 books



solving for another pronumeral ■ Consider Ohm’s Law, V = IR, where I is the current, R is the resistance and V is the voltage. In this form, the voltage, V, can be found if the current, I and the resistance, R are given. If the resistance, R, is required and the voltage, V and current I are given then the formula needs to be rearranged. In this case, dividing both sides by I yields the formula

VI

= IRI

Which when simplifi ed, gives

VI

= R

or R = VI

■ In summary, use the same technique as for linear equations to obtain the desired pronumeral (the subject) by itself by performing additions, subtractions, multiplications and divisions.

Rearrange each formula to make x the subject.a y = kx + m b 6(y + 1) = 7(x - 2)

thinK Write

a 1 Write the equation. a y = kx + m

2 Attempt to get x by itself. First subtract m from both sides of the equation.

y - m = kx + m - my - m = kx

3 Divide both sides of the equation by k. y m

ky m−y m

= kxk

y m

ky m−y m

= x

4 Rewrite the equation so that x is on the left-hand side.

x = y m

ky m−y m

WorKeD eXAmple 12



b 1 Write the equation. b 6(y + 1) = 7(x - 2)

2 Expand the brackets fi rst. 6y + 6 = 7x - 14

3 Attempt to get x by itself. First add 14 to both sides of the equation.

6y + 6 + 14 = 7x - 14 + 146y + 20 = 7x

4 Divide both sides of the equation by 7.6 20

7y6 2y6 26 2+6 2 =

77x

6 207

y6 2y6 26 2+6 2 = x

5 Rewrite the equation so that x is on the left-hand side.

x = 6 20

7y6 2y6 26 2+6 2

For each of the following make the variable shown in brackets the subject of the formula.

a g = 6d - 3 (d ) b a = v u

tv u−v uv u−v uv u−v u−v u−v u

(v)

thinK Write

a 1 Write the equation. a g = 6d - 32 Attempt to make d the subject. First add

3 to both sides of the equation.g + 3 = 6d - 3 + 3g + 3 = 6d

3 Divide both sides of the equation by 6.g + 3

6 =

66d

g + 36

= d

4 Rewrite the equation so that d is on the left-hand side.

d = g + 3

6

b 1 Write the equation. b a = v u

tv u−v u

2 Attempt to make v the subject. First multiply both sides of the equation by t.

at = v u

tv u−v u

ì t

at = v - u

3 Add u to both sides of the equation. at + u = v - u + uat + u = v

4 Rewrite the equation so that v is on the left-hand side.

v = at + u

remember

1. Rearranging a formula may be required when there is more than one pronumeral in the equation.

2. The pronumeral that is required to be left by itself on the left-hand side of the equal sign is called the subject.

3. To rearrange the formula so that the required pronumeral is the subject, use the same techniques as for linear equations, treating the other pronumerals as numbers.

WorKeD eXAmple 13



rearranging formulasfluenCY

1 We 12 Rearrange each formula to make x the subject.a y = ax b y = ax + bc y = 2ax - b d y + 4 = 2x - 3e 6(y + 2) = 5(4 - x) f x(y - 2) = 1g x(y - 2) = y + 1 h 5x - 4y = 1i 6(x + 2) = 5(x - y) j 7(x - a) = 6x + 5ak 5(a - 2x) = 9(x + 1) l 8(9x - 2) + 3 = 7(2a -3x)

2 We 13 For each of the following, make the variable shown in brackets the subject of the formula.

a g = 4P - 3 (P) b f = 95c (c)

c f = 95c + 32 (c) d V = IR (I)

e v = u + at (t) f d = b2 - 4ac (c)

g m = y k

h

y k−y k (y) h m =

y a

x b

y a−y a

x b−x b (y)

i m = y a

x b

y a−y a

x b−x b (a) j m =

y a

x b

y a−y a

x b−x b (x)

k C = 2πr

(r) l f = ax + by (x)

m s = ut + 12 at2 (a) n F = GMmGMmGM

r2 (G)

unDerstAnDing

3 The cost to rent a car is given by the formula C = 50d + 0.2k, where d = the number of days rented and k = the number of km driven. Lin has $300 to spend on car rental for her 4-day holiday. How far can she travel on this holiday?

eXerCise

4e

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Activity 4-E-1Rearranging

formulas doc-4021

Activity 4-E-2Writing formulas in

different formsdoc-4022

Activity 4-E-3Using formulas in

practical situations doc-4023

inDiViDuAl pAthWAYs

eBookpluseBookplus


doc-6155

eBookpluseBookplus

eLessonFormulas in

the real worldeles-0113

P = g + 3

4c =

59f

c = 5 3

9

( )5 3( )5 32( )2( )5 3( )5 35 3f5 3f( )f( )5 3( )5 3f5 3( )5 35 3( )5 3−5 3( )5 3I =

VR

t = v u

av u−v u

y = hm + k y = m(x - b) + a

a = y - m(x - b) x = y a mb

m− +y a− +y a

r = 2πC

x = f bya

f b−f b

a = 22

( )( )s u( )( )t( )

t

( )s u( )−( )s u( )G =

FrMm

2

c = b d

a

2b d2b d

4

b d−b d

500 km

a x = ya b x =

y ba−

c x = y b

ay b+y b2

d x = y + 7

2

e x = 8 6

5

8 6−8 6y f x = 1

2y −

g x = y

y

+−

1

2

h x = 4 1

5

y4 1y4 14 1+4 1

i x = -5y - 12 j x = 12a

k x = 5 9

19

5 9a5 95 9−5 9

l x = 14 13

93a +



4 A cyclist pumps up a bike tyre that has a slow leak. The volume of air (in cm3) after t minutes is given by the formula: V = 24 000 - 300ta What is the volume of air in the tyre when it is first filled?b Write an equation and solve it to work out how long it takes the tyre to go completely flat.

reAsoning

5 The total surface area of a cylinder is given by the formula T = 2p r 2 + 2p rh, where r = radius and h = height. A car manufacturer wants the engine’s cylinders to have a radius of 4 cm and a total surface area of 400 cm2. Show that the height of the cylinder is approximately 11.92 cm, correct to 2 decimal places. (Hint: Express h in terms of T and r.)

refleCtion

How does rearranging formulas differ to solving linear equations?

24 000 cm3.t = 80 min = 1h 20 min

Answers will vary.



summarySolvinglinearequations

■ A linear equation in terms of one variable, x, is an algebraic equation in which the pronumeral must not have any index or power other than 1. (Remember x1 = x.)

■ To solve an equation means to fi nd the value of the pronumeral that, when substituted, will make an equation a true statement.

■ If you perform the same arithmetic operation on both sides of an equation, the equation remains valid.

■ To fi nd the solution to the equation, perform identical arithmetic operations to both sides of the equation until the pronumeral is left by itself.

■ The solution can be verifi ed by substituting it into the original equation and checking whether it makes a true statement.

■ The inverse operations, which will allow us to obtain the pronumeral on its own, must be performed in the reverse order.

■ A CAS calculator can be used to solve linear equations.

Solvinglinearequationswithbrackets

If an equation contains brackets, expand the brackets fi rst, unless dividing both sides by the coeffi cient in front of the brackets does not create a fraction.

Solvinglinearequationswithpronumeralsonbothsides

If there are pronumerals on both sides of an equation, combine the pronumeral terms into one by either addition or subtraction.

Solvingproblemswithlinearequations

How to solve worded problems. ■ Identify the unknown quantity and use a pronumeral to represent it. ■ Search for keywords that indicate the steps needed for the solution. ■ Create a linear equation from the information provided in the question. ■ Solve the equation. ■ Interpret the result and write the worded answer.

Rearrangingformulas ■ Rearranging a formula may be required when there is more than one pronumeral in the equation.

■ The pronumeral that is required to be left by itself on the left-hand side of the equal sign is called the subject.

■ To rearrange the formula so that the required pronumeral is the subject, use the same techniques as for linear equations, treating the other pronumerals as numbers.

MAPPINGYOURUNDERSTANDING

Use the summary above to construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare this concept map with the one that you created in What do you know? on page 101.Have you completed the two Homework sheets, the Rich task and two Code puzzles in your Maths Quest 9 Homework Book?

HomeworkBook



Chapter reviewfluenCY

1 The linear equation represented by the sentence ‘When a certain number is multiplied by 3, the result is 5 times the certain number plus 7’ is:A 3x + 7 = 5x B 5(x + 7) = 3x C 5x + 7 = 3x

D 5x = 3x + 7 E 3x = 35 + x

2 The solution to the equation x3

= 5 is:

A x = -15 B x = 15 C x = 123

D x = 3 E x = 5

3 What is the solution to the equation 7 = 21 + x?A x = 28 B x = -28 C x = -14D x = 14 E x = 7

4 What is the solution to the equation 5x + 3 = 37?A x = 8 B x = -8 C x = 6.8D x = 106 E x = -106

5 The solution to the equation 8 - 2x = 22 is:A x = 11 B x = 15 C x = -15D x = 7 E x = -7

6 The solution to the equation 4x + 3 = 7x - 33 is:

A x = -12 B x = 12 C x = 3611

D x = 3011

E x = −3611

7 The solution to the equation 7(x - 15) = 28 is:A x = 11 B x = 19 C x = 20D x = 6.14 E x = 10

8 When rearranging y = ax + b in terms of x, we obtain:

A x = y a

by a−y a

B x = y b

ay b−y b

C x = b y

ab y−b y

D x = y b

ay b+y b

E x = y - ab

9 Which of the following are linear equations?a 5x + y2 = 0 b 2x + 3 = x - 2

cx2

= 3 d x2 = 1

e 11 3

xx+ =1 3+ =1 3 f 8 = 5x - 2

g 5(x + 2) = 0 h x2 + y = -9i r = 7 - 5(4 - r)

10 Solve each of the following linear equations.a 3a = 8.4 b a + 2.3 = 1.7

cb21

= -0.12 d b - 1.45 = 1.65

e b + 3.45 = 0 f 7.53b = 5.64


a2 3

72 3x2 32 3−2 3

= 5 b5

2− x

= -4

c− −3 4− −3 4− −

53 4x3 4− −3 4− −x− −3 4− −

= 3 d6x

= 5

e4x

= 35

fx +1 7

2 3.1 7.1 7

2 3.2 3 = -4.1

12 Solve each of the following linear equations.a 5(x - 2) = 6 b 7(x + 3) = 40c 4(5 - x) = 15 d 6(2x + 3) = 1e 4(x + 5) = 2x - 5 f 3(x - 2) = 7(x + 4)

13 Liz has a packet of 45 Easter eggs. She saves 21 to eat tomorrow but rations the remainder so that she can eat 8 eggs each hour.a Write a linear equation in terms of the number

of hours, h, to represent this situation.b Work out how many hours it will take to eat

today’s share.

14 Solve each of the following linear equations.a 11x = 15x - 2 b 3x + 4 = 16 - xc 5x + 2 = 3x + 8 d 8x - 9 = 7x - 4e 2x + 5 = 8x - 7 f 3 - 4x = 6 - x

15 Translate these sentences into algebraic equations. Use x for the certain number.a Twice a certain number is equal to 3 minus that

certain number.b When 8 is added to 3 times a certain number,

the result is 19.c Multiplying a certain number by 6 equals 4.

✔

✔

✔

✔

✔

✔

✔

✔

b, c, f, g, i

a = 2.8 a = -0.6

b = -2.52 b = 3.1

b = -3.45 b = 0.749

x = 19 x = 13

x = -61

3x = 11

5

x = 62

3x = -11.13

8h + 21 = 45

3 hours

12 a x = 31

5b x = 25

7c x = 11

4

d x = -1 5

12e x = -12 1

2f x = -81

2

x = 1

2x = 3

x = 3 x = 5x = 2 x = -1

2x = 3 - x

3x + 8 = 196x = 4



d Dividing 10 by a certain number is one more than dividing that number by 6.

e Multiply a certain number by 2, then add 5. Multiply this result by 7. This expression equals 0.

f Twice the distance travelled is 100 m more than the distance travelled plus 50 metres.

16 Samuel decides to go on a holiday. He travels a certain distance on the fi rst day, twice that distance on the second day, three times that distance on the third day and four times that distance on the fourth day. If his total journey is 2000 km, how far did he travel on the third day?

17 For each of the following, make the variable shown in brackets the subject of the formula.a y = 6x - 4 (x)b y = mx + c (x)c q = 2(P - 1) + 2r (P)d P = 2l + 2w (w)e v = u + at (a)

f s = u v+u v+u v

2

t (t)

g v2 = u2 + 2as (a)h 2A = h(a + b) (b)

problem solVing

1 John is comparing two car rental companies, Golden Ace Rental Company and Silver Diamond Rental Company. Golden Ace Rental Company charges a fl at rate of $38 per day and $0.20 per kilometre. The Silver Diamond Rental Company charges a fl at rate of $30 per day plus $0.32 per kilometre. John plans to rent a car for three days.a Write an algebraic equation for the cost of

renting a car for three days from the Golden Ace Rental Company in terms of the number of kilometres travelled, k.

b Write an algebraic equation for the cost of renting a car for three days from the Silver Diamond Rental Company in terms of the number of kilometres travelled, k.

c How many kilometres would John have to travel so that the cost of hiring from each company is the same?

2 Frederika has $24 000 saved for a holiday and a new stereo. Her travel expenses are $5400 and her daily expenses are $260.a Write down an equation for the cost of her

holiday if she stays for d days. Upon her return from holidays Frederika wants

to purchase a new stereo system that will cost her $2500.

b How many days can she spend on her holiday if she wishes to purchase a new stereo upon her return?

eBookpluseBookplus

InteractivitiesTest yourself

Chapter 4int-2685

Word search Chapter 4int-0686

Crossword Chapter 4int-0700

7(2x + 5) = 0

2x - 100 = x + 50

600 kmCG = 114 + 0.20k

CS = 90 + 0.32k

200 km

5400 + 260d = CH

61 days

10x

- 1 = x6

17 a x = y + 4

6b x =

y cm

y c−y c

c P = q rq r−q r2q r2q r

2 + 1 d w =

P lP l−P lP l2P l

2

e a = v u

tv u−v u

f t = 2s

u v+u v+u v

g a = v u

s

2 2

2

v u−v u h b = 2A ah

h

A a−A a


eBookpluseBookplus ACtiVities

ChapteropenerDigital doc

• Hungry brain activity Chapter 4 (doc-6149) (page 101)Areyouready?

Digital docs (page 102)• SkillSHEET 4.1 (doc-6150): Solving one-step

equations• SkillSHEET 4.2 (doc-6151): Checking solutions to

equations• SkillSHEET 4.3 (doc-6152): Solving equations• SkillSHEET 4.4 (doc-6153): Expanding brackets• SkillSHEET 4.5 (doc-6154): Writing equations from

worded statements• SkillSHEET 4.6 (doc-6155): Transposing and

substituting into a formula4A Solvinglinearequations

Digital docs (pages 108–110)• Activity 4-A-1 (doc-4009): Riddle A• Activity 4-A-2 (doc-4010): Riddle B• Activity 4-A-3 (doc-4011): Riddle C• SkillSHEET 4.1 (doc-6150): Solving one-step

equations• SkillSHEET 4.2 (doc-6151): Checking solutions to

equations• SkillSHEET 4.3 (doc-6152): Solving equations• SkillSHEET 4.5 (doc-6154): Writing equations from

worded statements• WorkSHEET 4.1 (doc-6156): Solving linear

equations4B Solvinglinearequationswithbrackets

Digital docs (pages 111–112)• Activity 4-B-1 (doc-4012): Crossword A• Activity 4-B-2 (doc-4013): Crossword B• Activity 4-B-3 (doc-4014): Crossword C• SkillSHEET 4.4 (doc-6153): Expanding brackets• History of Maths (doc-6157): Grace Murray Hopper

4C Solvinglinearequationswithpronumeralsonbothsides

Digital docs (pages 114–115)• Activity 4-C-1 (doc-4015): What is the word? A• Activity 4-C-2 (doc-4016): What is the word? B

• Activity 4-C-3 (doc-4017): What is the word? C• SkillSHEET 4.7 (doc-6158): Simplifying like

terms• WorkSHEET 4.2 (doc-6159): Solving equations with

pronumerals on both sidesInteractivities

• Solving equations (int-2764) (page 113)• Bad egg (int-0741) (page 115)

4D SolvingproblemswithlinearequationsDigital docs (page 120)• Activity 4-D-1 (doc-4018): Linear equations and

problem solving• Activity 4-D-2 (doc-4019): Use of linear equations

in worded problems• Activity 4-D-3 (doc-4020): Applying linear

equations to real-life problems• SkillSHEET 4.5 (doc-6154): Writing equations from

worded statements

4E RearrangingformulasDigital docs (page 125)• Activity 4-E-1 (doc-4021): Rearranging formulas• Activity 4-E-2 (doc-4022): Writing formulas in

different forms• Activity 4-E-3 (doc-4023): Using formulas in

practical situations• SkillSHEET 4.6 (doc-6155): Transposing and

substituting into a formulaeLesson

• Formulas in the real world (eles-0113) (page 125)

ChapterreviewInteractivities (page 129)• Test yourself Chapter 4 (int-2685): Take the end-of-

chapter test to test your progress.• Word search Chapter 4 (int-0686)• Crossword Chapter 4 (int-0700)

To access eBookPLUS activities, log on to

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Documents

4A Solving linear equations 4B Contents 4C 4D 4E linear equationslgs8maths.weebly.com/.../22594084/ch_4_linear_equations.pdf · 2019-10-24 · number AnD AlgebrA • lineAr AnD non-lineAr