Solving linear equations · Solving linear equations MQ9 Vic ch 07 Page 209 Monday, September 17,...

7John 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?

In this chapter we learn what linear equations are and how to solve them. We also discover how valuable linear equations are to computer programmers and problem solvers.

Solving linear equations

MQ9 Vic ch 07 Page 209 Monday, September 17, 2001 9:24 AM

210

M a t h s Q u e s t 9 f o r V i c t o r i a

What is a linear equation?

This chapter represents the culmination of our exploration of algebra, and enables us,finally, to answer that all-important question — what is the

value

of that pronumeral?Remember that up until now, when we used

x

,

y

or any other pronumeral in anexpression, we implied that the pronumeral was a number whose value we didn’t knowyet . . . now you will know! In essence, all algebra leads to the problem of solving equa-tions and finding that value. Without it, algebra has no real purpose.

Identifying linear equations

Recall that an equation is a statement of equality of two algebraic expressions (that is,two expressions with an equals sign between them). For example:

x

2

+

4

x

=

0

xy

+

y

=

4

+

yx

n

+

y

n

=

z

n

are all equations. However, in this chapter we will concern ourselves only with the simplest forms,

which are called linear equations. How do we know whether the given equation islinear? The answer relies on a number of ‘rules’, which are summarised in the tablebelow, showing examples of linear and non-linear equations.

Note

: Equations involving fractions may need to be examined very carefully to deter-mine if they are indeed linear. There is really no hard and fast rule here.

Rule Linear equationsEquations which

are not linear

There must be only 1

distinct

pronumeral, although it may appear more than once.

4(

x

+

1)

=

−

53

x

+

5

=

2

x

−

43

x

+

y

=

2

x

−

y

The pronumeral must not have any index or power other than 1. (Remember

x

1

=

x

.)

4

x

+

7

=

07

−

x

=

6

+

8

x

4

x

6

+

7

=

0

x

2

+

2

x

+

1

=

0

If there are fractions, the pronumeral must not appear in both numerator and denominator.

−

5

=

0

+

=

6

There must be

exactly

one equals sign. 7

−

3(2

−

x

)

=

9 2

x

=

5

=

6

−

x

3x 1–2

--------------- 1x--- x

2---

State whether each of the following equations is linear.

a 2x + 6 = x + 1 b x2 − 5 = x c − = d 3x + 6 = x(y − 1)

THINK WRITEa There is only one pronumeral (x), which has

a power of 1. There are no fractions. Finally, there is only one equals sign. So all rules are observed. State your conclusion.

a The equation 2x + 6 = x + 1 is linear.

b There is a pronumeral with a power of 2 (x2). b The equation x2 − 5 = x is not linear.c The pronumeral appears in both numerator

and denominator.c The equation − = is not linear.

d There are 2 pronumerals, x and y. d The equation 3x + 6 = x(y − 1) is not linear.

12--- 1

x--- x 2+

3------------

12--- 1

x--- x 2+

3------------

1WORKEDExample

MQ9 Vic ch 07 10 Monday, September 17, 2001 9:24 AM

C h a p t e r 7 S o l v i n g l i n e a r e q u a t i o n s 211

Converting worded equations to algebraic equationsAs we know, algebra is a form of ‘language’ and there are times when a ‘translation’ isrequired from words to a suitable algebraic form. For example, a number 5 larger thana certain number in words, can be written as x + 5 in algebra. When performing atranslation, look for keywords such as:

less than or smaller thanmore than or larger thanmultiplied by or timesdividinga certain number (which will represent our pronumeral).

As with most languages, there is no simple rule for translation; you need to applyexperience as well as expertise. This is particularly important when translating Englishinto what will become an algebraic equation.

With practice, the translation process will become easier; just remember to use thosekeywords to guide you as to the type of arithmetic required.

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 Break up the sentence into parts which indicate the kind of arithmetic or algebra required.When 6 is subtracted from: − 6 A certain number: x The result is 15: = 15.

a

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

x − 6 = 15

b Break up the sentence into parts which indicate the kind of arithmetic or algebra required.Three more than: + 3 Seven times a certain number: 7x Is zero: = 0.

b

Form the algebraic equation from the individual parts.

7x + 3 = 0

c Break up the sentence into parts which indicate the kind of arithmetic or algebra required.When dividing a certain number by 2: The answer is: = 4 more than a certain number: x + 4.

c

Form the algebraic equation from the individual parts.

= x + 4

1

2

1

2

1

x2---

2x2---

2WORKEDExample


212 M a t h s Q u e s t 9 f o r V i c t o r i a

What is a linear equation?

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

e = 4 f 5 + y = 9 g x3 + 3 = 5 h = 8

i y2 − x2 = 9 j x2 + 2x + 3 = 0 k 5 = 2m − 11 l −9 = m2 + 3

m 3b + 7 = 16 n y2 = x2 + 1 o −4a + 7b = 0 p = 18

2 Write linear equations for each of the following statements, using x to represent theunknown. (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.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 Five times a certain number gives a result of −7.g When a certain number is subtracted from 14, the answer is −3.h When 5 is added to three times a number, the answer is 8.i When 12 is subtracted from two times a number, the result is 15.j Dividing 3 times a certain number by 2 equals 5.

3Which equation matches the following statement? a A certain number, when divided by 2, gives a result of −12.

A x = B 2x = −12 C = −12 D = −2 E = −12

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

A B C D E

c Subtracting twice a certain number from 8 gives 12.

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

A 15 + 4x = 10 B 10 = C = 10 D = 10 E 10 =

A 2x − 8 = 12 B 8 − 2x = 12 C 2 − 8x = 12D 8 − (x + 2) = 12 E 12 = 8x − 2

rememberTo decide whether the equation is linear, check that each of the following rules are observed.1. There must be only one distinct pronumeral.2. The only power of the pronumeral allowed is 1.3. The pronumeral should not appear in both the numerator and the denominator

of fractional terms.4. There must be exactly one equals sign.

remember

7AWWORKEDORKEDEExample

12a 1+

3--------------- y

2---

3m 2–5

----------------

WWORKEDORKEDEExample

2

mmultiple choiceultiple choice

12–2

--------- x2--- x

12------ 2

x---

x4–

------ 9= 4x–7

--------- 9= 7 x+4–

------------ 9= 7x4–

------ 9= 4–7x------ 9=

x4--- 15+ x 15+

4--------------- 15

4x---+ 15

4------ x+



Career profileA L E X A N D E R I A M P O L S K Y —

C o m p u t e r p r o g r a m m e r

Qualifications: Bachelor of Computer ScienceEmployer: IBMPosition: VB (Visual Basic) developer

The information technology industry appeals to me because it is very dynamic. It gives me an opportunity to continually learn new skills; as new languages are being developed all the time. It also allows me to participate in a variety of projects, with each project being a new and exciting challenge. To keep up-to-date I read a lot of professional magazines and attend seminars and short courses, provided by the company or other organisations.

In my work I use the computer language Microsoft Visual Basic to develop front-end GUI (graphic use interface) functionality for large communications clients.

A typical day involves work at one of the stages of the software development life cycle. The stages of the cycle include: identifying customer’s requirements, analysing the problem and creating the initial design.Once the design is approved by the customer, the stages that follow are: creating the algorithm, coding (translating from pseudo code of algorithm to a computer language), testing the product and the final implementation.

If, as a result of testing, some problems (known as bugs) are identified, these need to be fixed prior to presenting the final product to

the customer. Support and customer training may also be included in some projects. My current project is a billing and customer information system for a large telecommunications company.

In my work I use different areas of mathematics. In particular, I have to deal a lot with variables and equations (rules). While writing codes I use a lot of inequations. For example, when billing certain calls, different charges apply during peak and off-peak hours. When the call is made, the only information received is the time and duration of a call. I write a program for the computer to compare the input variable (the time of the call) with a certain number that defines the end-point of peak time; that is, the computer checks whether the inequation ‘time of call’ ≤ ‘end point of peak time’ is true. If the input number (time of call) is indeed less than or equal to the specified time, then the call was made during peak time and peak time charges will be applied. Otherwise an off-peak charge will be applied. I then program the computer to calculate the total charge of the call by using the equation ‘total charge of call’ = ‘length of call’ × ‘charge’.

I also use my programming skills at home. When I have time, I write computer games for my children, Rachel and Nathan. Their favourites include jigsaw puzzles of various levels of difficulty and Rubik’s cube. Writing these types of games also involves a lot of mathematics, particularly an extensive knowledge of geometry and trigonometry.

Questions1. With which computer language does

Alexander work?2. List three of the stages of the software

development cycle.3. What is a computer algorithm?4. Find out which tertiary institutions offer

computer science courses to become a computer programmer.



Solving one-step equationsFor an equation, whatever is on the left-hand side of the equals sign has the same valueas what is on the right-hand side. Even the simplest equation, such as 5 + 4 = 9 followsthis definition.

Consider changing our 5 + 4 = 9 equation, by adding 3 to both sides.5 + 4 + 3 = 9 + 3

The left-hand side (5 + 4 + 3), now 12, is still equal to the right-hand side (9 + 3).

If you perform the same arithmetic operation on both sides of an equation, the equation remains valid. That is, the equation remains a true statement.

Let us now apply this principle to a simple linear equation:x − 3 = 10.

Add 3 to both sides and simplify.x − 3 + 3 = 10 + 3

x + 0 = 13 x = 13

To solve an equation means to find the value of the pronumeral which, when substituted,will make the equation a true statement. By applying a simple arithmetic operation (add3), we have ‘discovered’ the value of x, namely 13.

Let’s check whether x = 13 is the solution by substituting 13 for x into our originalequation.

LHS (left-hand side) = x − 3= 13 − 3= 10= RHS (right-hand side)

Since the LHS is equal to the RHS, x = 13 is the solution to the equation.

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

The equation discussed above (x − 3 = 10) is an example of a one-step equation, sincewe needed to perform only one arithmetic operation (adding 3) to obtain the value of x.The technique for all one-step equations is to find the one arithmetic operation that leavesthe pronumeral by itself.

To find such an operation we need to ask ourselves what operation was performed withthe pronumeral in the process of forming the equation. The operation that needs to beperformed in order to leave the pronumeral on its own will then be the inverse (that is,the opposite) operation.

In the example with the equation x − 3 = 10, 3 was subtracted from x to give the resultof 10. So to find x, an inverse operation was performed; that is, 3 was added.

To solve a one-step equation, decide which operation was performed with the pronumeral in the process of forming the equation and then perform an inverse operation, as shown in the table at right.

Operation Inverse operation

+ −

− +

× ÷

÷ ×



Note: In each case the result can be checked by substituting your found value for x backinto the original equation and confirming that it will make the equation a truestatement.

For example, to verify that 232 is the solution of the equation x − 79 = 153, substi-tute 232 for x:

LHS = x − 79 = 232 − 79

= 153 = RHS

Since the left-hand side is equal to the right-hand side, 232 is the correct solution.

In the previous worked example, the solution to each equation was a positive wholenumber. However, there is no reason why your answer has to be a positive wholenumber. The pronumeral can take any value, as shown in the next worked example.

Solve each of the following linear equations.

a x − 79 = 153 b x + 46 = 82 c 6x = 102 d = 19

THINK WRITE

a Write the equation. a x − 79 = 153Add 79 to both sides of the equation. (This is the inverse operation.)

x − 79 + 79 = 153 + 79

Simplify both sides to find the value of x.

x = 232

b Write the equation. b x + 46 = 82Subtract 46 from both sides of the equation.

x + 46 − 46 = 82 − 46


x = 36

c Write the equation. c 6x = 102

Divide both sides of the equation by 6.

=


x = 17

d Write the equation. d = 19

Multiply both sides of the equation by 7.

× 7 = 19 × 7

Simplify both sides of the equation to get the solution.

x = 133

x7---

12

3

12

3

1

26x6

------ 1026

---------

3

1x7---

2x7---

3

3WORKEDExample



Solve each of the following linear equations.a 4 + y = 2 b 5y = 11

THINK WRITE

a Write the equation. a 4 + y = 2Subtract 4 from both sides of the equation.

4 + y − 4 = 2 − 4

Simplify the equation. This gives the answer.(In this case the result is a negative whole number.)

y = −2

b Write the equation. b 5y = 11

Divide both sides of the equation by 5.

=

Simplify to find the value of y.(In this case the result is a mixed number.)

y = 2

12

3

1

25y5

------ 115

------

315---

4WORKEDExample

remember1. To solve an equation means to find the value of the pronumeral which, when

substituted, will make an equation a true statement.

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

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

4. To solve a one-step equation, decide which operation was performed with the pronumeral in the process of forming the equation and then perform an inverse operation.

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

6. Solutions to the equations do not have to be whole positive numbers.

Operation Inverse operation

+ −

− +

× ÷

÷ ×

remember



Solving one-step equations

1 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

j = 6 k = 26 l = 27

2 Solve each of the following linear equations.a y + 5 = 3 b y − 5 = 3 c y + 45 = 12d y − 16 = −31 e 5.5 + y = 7.3 f y − 7.3 = 5.5g 6y = 14 h 0.2y = 4.8 i 0.9y = −0.05

j = 4.3 k = 23 l = −1.04

3a The solution to the equation m + 9 = −5 is:

b What is the solution to the equation x − 6.3 = 5.7?

c The solution to the equation 4k = 3.4 is:

d Which is the solution to the equation = −15?

e Which is the equation with solution a = −2?

A m = 4 B m = −4 C m = −14 D m = 14 E m = −45

A x = 0.6 B x = −0.6 C x = 12 D x = −12 E x = 13

A k = −1.6 B k = 0.85 C k = −0.6 D k = 13.6 E k = 7.4

A x = 6 B x = −6 C x = 37.5 D x = −37.5 E None of these

A 5 + a = −7 B 12 − a = 14 C a + 8 = 10D 7a = 14 E −4 − a = −6

7BWWORKEDORKEDEExample

3

Mathcad

Solvingone-stepequations

x23------ x

17------ x

9---

GC program

Solvinglinear

equations

SkillSH

EET 7.3


4 SkillSHEET 7.2

SkillSH

EET 7.1

y5--- y

7.5------- y

8---


x2.5-------

MA

TH

SQUEST

C H A L L

EN

GE

MA

TH

SQUEST

C H A L L

EN

GE

1 A family on holiday rented a cabin for 14 days on the condition that theywould pay $40 a day every day that it did not rain and $10 a day forevery day that it did rain. At the end of the two weeks, the family paid$410. How many days did it rain?

2 Buffy, the spaniel, ate a total of 34 dog treats in 4 days,each day eating 3 more treats than the day before.How many dog treats did Buffy eat on the first day?

3 In the triangle shown, angle A is 50∞ less than angleB. Angle B is 10∞ more than angle C. What is the sizeof angle A?

A

B

C



History of mathematicsG R AC E M U R R AY H O P P E R ( 1 9 0 6 – 9 2 )

During her life ...Two world wars are fought.Neil Armstrong walks on the moon.The HIV/AIDS disease appears.

Grace Hopper was an American mathematician who developed computer languages and invented the term ‘computer bug’. She had an innovative mind and was fond of saying that, ‘we’ve always done it that way’, is not a good reason to continue to do so. Many people thought of her as ‘Amazing Grace’.

Hopper was born in New York and was the oldest of three children. As a child she was interested in science and was especially interested in gadgets such as alarm clocks. She enjoyed taking them apart to see how they worked. She went on to study mathematics and physics at Vassar College. Her mother also liked mathematics and had studied geometry even though it was considered unusual for women at the time.

Hopper became a teacher and then a professor of mathematics at Vassar College. She continued her studies at Yale University and in 1934 was awarded a PhD. The topic of

her thesis was New Types of Irreducibility Criteria. In 1936, she published a paper on The Ungenerated Seven as an index to Pythagorean Number Theory.

In 1930, Grace married, but her husband was killed in World War II. She never had children. In 1943, she joined the Naval Reserve where she used computers and mathematics to aid the war effort. One day while programming a computer she could not understand why a particular program was not working. She checked the mechanical switches and found that one was stuck on open because a moth was trapped in it. She put the moth in her note book and described it as a computer ‘bug’.

After the war, Hopper developed computer languages while working at Harvard University and for private enterprises. In the late 1950s she became interested in making computer languages easier to use. The development of the computer language COBOL was largely due to her efforts. During the 1960s and 1970s, the Navy sought her help several times when others were unable to solve problems related to computers. In 1969, the Data Processing Management Association presented her with the ‘Man of the Year’ award. In 1985 she was promoted to Rear Admiral and became the first woman to reach this rank.

Questions1. What computer term did Hopper

invent?2. Which computer language did Hopper

develop?3. What rank did Hopper reach in the

Navy?

ResearchFind out about the Year 2000 computer bug. How did this problem come about?

mq9.07 hist



Solving multi-step equationsSolving two-step equationsOnce you have mastered the one-step problems, the next level is solving two-step ones.There are several different types of two-step problems, but the principle and techniquesyou have learned in the previous section still apply.

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

2. To solve an equation, perform identical arithmetic operations on both sides of an equation, until the pronumeral is left by itself.

A common two-step problem can arise when there is a coefficient in front of thepronumeral.

Consider the equation 3x + 5 = 12. In the process of forming this equation, x was firstmultiplied by 3 and then 5 was added to give the result of 12. That is, the order ofoperations was:

1. × 32. + 5.

The inverse operations, which will allow us to get x on its own, must be performedin the reverse order. That is, we need to subtract 5 from both sides first and then divideboth sides by 3:

1. − 5 reduces the equation to 3x = 72. ÷ 3 leaves the answer x = 2 .

This principle will apply to any two- (or more) step equation, as shown in theexamples that follow.

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 as −x is thesame as −1x.

13---

Solve the following linear equations.a 2y + 4 = 12 b −6 − 2x = 12THINK WRITEa Write the equation. Identify the order of operations

(× 2 then + 4) and the inverse operations needed in reverse order (− 4 then ÷ 2).

a 2y + 4 = 12

Subtract 4 from both sides of the equation first. 2y + 4 − 4 = 12 − 4 2y = 8

Divide both sides of the equation by 2. =

y = 4b Write the equation. Identify the order of operations

(× −2 then −6) and the inverse operations needed in reverse order (+6 then ÷ −2).

b −6 − 2x = 12

Add 6 to both sides of the equation first. −6 − 2x + 6 = 12 + 6−2x = 18

Divide both sides of the equation by −2 to find the value of x.

=

x = −9

1

2

32y2

------ 82---

1

2

32x–2–

--------- 182–

------

5WORKEDExample



Note: You can also achieve the same result by multiplying by −1 instead of dividingby −1. For example, in part a of the worked example above, multiplying both sides of−x = 6 by −1 also produces x = −6.

A short cut can be used.

If −x is equal to a certain number (say, a), then x equals that number with the opposite sign. That is, if −x = a then x = −a.

Algebraic fractions — pronumeral in the numeratorSituations where there is a pronumeral in a numerator of an algebraic fractionwill require an extra step. Again it is important to observe the order in which thesteps occur.

Solve the following linear equations.

a 4 − x = 10 b − = 11

THINK WRITE

a Write the equation. Identify the order of operations (× −1 then + 4) and the inverse operations needed in reverse order (− 4 then ÷ −1).

a 4 − x = 10

Subtract 4 from both sides of the equation.

4 − x − 4 = 10 − 4 −x = 6

Divide both sides by −1. =

x = −6

b Write the equation. We can consider

this equation to be = 11 or

= 11. Identify the order of

operations (× −1 then ÷ 4 or ÷ −4 only). Identify the inverse operations needed.

b − = 11

= 11

Multiply both sides by 4. × 4 = 11 × 4−x = 44

Divide both sides by −1. (Alternatively, multiply both sides of

= 11 by −4 to obtain x = −44.)

x = −44

x4---

1

2

3x–1–

------ 61–

------

1

x–4

------

x4–

------

x4---

x–4

------

2x–

4------

3

x4–

------

6WORKEDExample



Algebraic fractions — pronumeral in the denominatorIf a pronumeral is in the denominator, there is an extra step involved in finding the solution.

There are two methods you can use:Method 1 Inverting the equationMethod 2 Cross-multiplication.Both methods are shown in the worked example which follows.

Solve the following linear equations.

a b

THINK WRITE

a Write the equation. Identify the order of operations (+ 1 then ÷ 2) and the inverse operations needed in reverse order (× 2 then − 1).

a = 11

Multiply both sides of the equation by 2. × 2 = 11 × 2 x + 1 = 22

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

x + 1 − 1 = 22 − 1x = 21

b Write the equation. Identify the order of operations (× −1, + 7 then ÷ 5) and the inverse operations needed in the reverse order (× 5, − 7 then ÷ −1).

b = −6.3

Multiply both sides of the equation by 5. × 5 = −6.3 × 5

7 − x = −31.5Subtract 7 from both sides of the equation. 7 − x − 7 = −31.5 − 7

−x = −38.5Divide both sides by −1 or use the short cut of opposite signs. x = 38.5

x 1+2

------------ 11= 7 x–5

------------ 6.3–=

1x 1+

2------------

2x 1+

2------------

3

17 x–

5-----------

27 x–

5-----------

3

4

7WORKEDExample

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

a = b =

THINK WRITE

a Write the equation. a =

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

=

Multiply both sides of the equation by 3. × 3 = × 3

a =

or a = 3

3a--- 4

5--- 5

b--- 2–

7------

13a--- 4

5---

2a3--- 5

4---

3a3--- 5

4---

154

------

34---

8WORKEDExample

Continued over page



It doesn’t really matter which method you use, you’ll get the answer either way. Futureexperience will tell you which method is most appropriate for any given equation.Furthermore, the requirements of the problem will indicate whether you should leaveyour answer as a fraction or a decimal.

Solving multi-step equations

1 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

2 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

THINK WRITE

b Write the equation. b =

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

5 × 7 = −2 × b35 = −2b

Divide both sides of the equation by −2.

=

−17 = b

or b = −17.5

15b--- 2–

7------

2

335

2–------ 2b–

2–---------

12---

remember1. Perform the same operation on both sides of an equation to keep the equation

valid.

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

3. If there is a pronumeral in the denominator of a fraction, either ‘invert’ or ‘cross-multiply’.

remember

7CEXCEL

Spreadsheet

Equationsolverax + b = c


5a

Mathca

d

Solving two-step equations


5b

GCpro

gram



C h a p t e r 7 S o l v i n g l i n e a r e q u a t i o n s 2233 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 − x g 9 − x = 0.1 h 140 − x = 121 i −30 − x = −4j −5 = −6 − x k −x + 1 = 2 l −x − 1 = 0

4 Solve each of the following linear equations.

a − = 5 b − = −8 c − = 11

d = 3 e = −1 f =

g = 6 h = −3 i − = −7

j − = 6 k = −2 l − = −


a = 5 b = 8 c = −4

d = 0 e = 6 f = −2

g = −3 h = 1.2 i = 0

j = −0.5 k = −4.6 l = 1


a = 2 b = 3 c = −1

d = −5 e = 8 f = −10

g = 3 h = 1 i = 8.8

j = −3.1 k = −2.5 l =

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

a = b = 7 c =

d = e = f = 1

g = h = i =

j = −1 k = l =


6a


6b x3--- x

2--- x

6---

x4--- x

3--- x

8--- 1

2---

2x3

------ 5x2

------ 3x4

------

8x3

------ 2x7

------ 3x10------ 1

5---


7

Mathcad

Solvingmulti-stepequations

z 1–3

----------- z 1+4

----------- z 4–2

-----------

6 z–7

----------- 3 z–2

----------- –z 50–22

-----------------

z 4.4–2.1

--------------- z 2+7.4

----------- 140 z–150

-----------------

–z 0.4–2

------------------- z 6–9

----------- z 65+73

--------------

5x 1+3

--------------- 2x 5–7

--------------- 3x 4+2

---------------

4x 13–9

------------------ 4 3x–2

--------------- 1 2x–6

---------------

–5x 3–9

------------------ –10x 4–3

--------------------- 4x 2.6+5

--------------------

5x 0.7–0.3

------------------- 1 0.5x–4

------------------- –3x 8–14

------------------ 12---


82x--- 1

2--- 3

x--- 4–

x------ 7

2---

5x--- 3–

4------ 0.4

x------- 9

2--- 8

x---

4–x

------ 23--- 6–

x------ 4–

5------ 1.7

x------- 1

3---

6x--- 4

x--- 15–

22--------- 50

x------ 35–

43---------



8a The solution to the equation 82 − x = 44 is:

b What is the solution to the equation 5x − 12 = −62?

c What is the solution to the equation = 5.3?


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

d = 0 e 5 − 3e = −10 f = 8

g 100 = 6g + 4.2 h = 5.5 i 452i − 124 = −98

j = 0 k = 4 l = 1.5

Solve the following equations.

1 x + 12 = 9

2 6x = 31

3 = −4

4 5.3 − x = 2.6

5 5x − 4 = 10

6 = −3

7 =

8 7x − 9.5 = 0

9 = 5

10 = −4

A x = 126 B x = −126 C x = 38 D x = −38 E x = 1.86

A x = −14.8 B x = 14.8 C x = 10 D x = −10 E None of these

A x = 9.6 B x = 10.6 C x = 11.6 D x = 2 E x = 1


x 1–2

-----------

d 4–67

------------ 2 f3

------

h 2+6

------------

WorkS

HEET 7.1

6 j 1–17

-------------- 12 k–5

-------------- l 5.2–3.4

---------------

1

x3---

x 15+9

---------------

47--- 3

x---

2x 1+3

---------------

–3x 8–2

------------------



Solving equations with pronumerals on both sides

There are times when an equation will be given where there is a particular pronumeralon both sides of the equals sign. The extra step is in ‘combining’ the two pronumeralsinto one. We perform an operation to both sides of the equation that will remove thepronumeral from one side.

Solve each of the following linear equations.

a 5y = 3y + 3 b 7x + 5 = 2 − 4x

THINK WRITE

a Write the equation. a 5y = 3y + 3

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

5y − 3y = 3y + 3 − 3y 2y = 3

Divide both sides by 2 to find the value of y.

=

y =

b Write the equation. b 7x + 5 = 2 − 4x

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

7x + 5 + 4x = 2 − 4x + 4x 11x + 5 = 2


11x + 5 − 5 = 2 − 5 11x = −3

Divide both sides by 11 to find the value of x.

=

x = −

1

2

32y2

------ 32---

32---

1

2

3

411x11

--------- 3–11------

311------

9WORKEDExample

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

remember



Solving equations with pronumerals on both sides

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

2a To solve the equation 3x + 5 = −4 − 2x, the first step is to:

b To solve the equation 6x − 4 = 4x + 5, the first step is to:

3 Solve each of the following linear equations.a 2x + 3 = 8 − 3x b 4x + 11 = 1 − x c x − 3 = 6 − 2xd 4x − 5 = 2x + 3 e 3x − 2 = 2x + 7 f 7x + 1 = 4x + 10g 5x + 3 = x − 5 h 6x + 2 = 3x + 14 i 2x − 5 = x − 9j 10x − 1 = −2x + 5 k 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 + 5 c 2x + 9 = 7x − 1d −2x + 7 = 4x + 19 e −3x + 2 = −2x − 11 f 11 − 6x = 18 − 5xg 6 − 9x = 4 + 3x h x − 3 = 18x − 1 i 5x + 13 = 15x + 3

5a The solution to 5x + 2 = 2x + 23 is:

b The solution to 3x − 4 = 11 − 2x is:

A add 3x to both sides B add 5 to both sidesC add 2x to both sides D subtract 2x from both sidesE subtract 4 from both sides

A subtract 4x from both sides B add 4x to both sidesC subtract 4 from both sides D add 5 to both sidesE add 6x to both sides

A x = 3 B x = −3 C x = 5 D x = 7 E x = −7

A x = 15 B x = 7 C x = 3 D x = 5 E x = 11

7D

SkillSH

EET 7.4 WWORKEDORKEDEExample

9a

EXCEL

Spreadsheet

Equation solver:ax + b = cx + d


GCpro

gram



9b


MA

TH

SQUEST

C H A L L

EN

GE

MA

TH

SQUEST

C H A L L

EN

GE

1 Julie wants to find the mass of a can of fruit. She has a balance and afew metal weights of known mass. After experimenting with the cans offruit and the weights, she discovers that by placing 5 cans on one side ofthe balance and 2 cans and a 2-kg weight on the other, the two sidesbalance. What is the mass of a can of fruit?

2 Find x in the magic square shown. Rememberthat in a magic square, each row, column anddiagonal add to the same amount.

x x – 1 8

x + 5 5 x – 3

x – 2 x + 3 x + 2



The values of thepronumerals in the equations below give

the puzzle’s answer code.

HoHow long does light takw long does light take to re to reach each the Ear the Earthth fr from .om . . . . .

2A + 1 = A + 10

7E + 4 = 76 – 2E

I + 2 = 6I – 48

O + 14 = 2O + 12

5S – 6 = 72 – S

W + 3 + 2W = 67 – W

8 + 3V = 5V – 205U = 3U + 6

x + 6 = 5x – 70 y + 18 = 6y – 42

50 – 2P = P – 43R – 12 = 8 – 2R

4T + 6 = 39 + T

3H + 10 = 5H2F – 6 = F – 5

3C – 35 = 2C – 20

3N = 5N – 12M – 15 = 19 – M

5 + D = 2D – 2

1

6 10 6 6 768 8 8 811 12 13 1313 15 214

2 3 4 5 3 6 7 4 8 7 79 6

11

11 16 8 116 10 3 1311 12 17 86

5 4 8 8 5 3 6 7 4 8 6 7 7 9

2 6 8 18 2 10 6 11 11 16 2 19 13 13 10 8 15 2 6 7 13



Solving linear equations with bracketsThere are many equations that are expressed in factorised form; that is, equations con-taining brackets. Sometimes this makes them easier to solve; sometimes an extra stepwill be involved — expanding the brackets.

Consider a linear equation such as 6(x + 3) = 7. With an equation like this you havetwo choices. Start with either:1. dividing both sides by 6 to get (x + 3) = , or2. expanding the brackets to get 6x + 18 = 7.

Although the first method will lead to the solution faster, its obvious disadvantage isthat, generally, a fraction is created ( ), making the next step potentially more difficult.Therefore, it is recommended that this method be used only if there is no fractioncreated. For example: if 6(x + 3) = 18, dividing both sides by 6 gives (x + 3) = 3. Wewill most often use the second method (that is, expanding brackets).

Note: The equation 7(x − 5) = 28 in the preceding worked example could also be solvedby expanding the brackets; however, it would take an unnecessary extra step.

76---

76---

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

THINK WRITE

a Write the equation. a 7(x − 5) = 28

Observe that 7 goes into 28 exactly, so divide both sides of the equation by 7.

=

Add 5 to both sides to solve for x. x − 5 = 4 x − 5 + 5 = 4 + 5 x = 9

b Write the equation. b 6(x + 3) = 7Observe that 6 does not divide into 7 exactly, so expand the brackets.

6x + 18 = 7


6x + 18 − 18 = 7 − 18 6x = −11

Divide both sides by 6. =

x = − (or −1 )

1

27 x 5–( )

7-------------------- 28

7------

3

1

2

3

46x6

------ 11–6

---------

116------ 5

6---

10WORKEDExample

rememberIf an equation contains brackets, expand the brackets first, unless dividing both sides by the coefficient in front of the brackets does not create a fraction.

remember



Solving linear equations with brackets

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

2 Solve each of the following equations.a 6(b − 1) = 1 b 2(m − 3) = 3 c 2(a + 5) = 7d 3(m + 2) = 2 e 5(p − 2) = −7 f 6(m − 4) = −8g −10(a + 1) = 5 h −12(p − 2) = 6 i −9(a − 3) = −3j −2(m + 3) = −1 k 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) = 14 c 7(a − 1) = 28d 4(b − 6) = 4 e 3(y − 7) = 0 f −3(x + 1) = 7 g −6(m + 1) = −30 h −4(y + 2) = −12 i −3(a − 6) = 3j −2(p + 9) = −14 k 3(2m − 7) = −3 l 2(4p + 5) = 18

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

5a The best first step in solving the equation 7(x − 6) = 23 would be to:

b The solution to the equation 84(x − 21) = 782 is:

6 Solve each of the following.a 5(x − 2) = 2x + 5 b 7(x + 1) = x − 11 c 2(x − 8) = 4xd 3(x + 5) = x e 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)

A add 6 to both sides B subtract 7 from both sidesC divide both sides by 23 D expand the bracketsE multiply both sides by 0.7

A x = 9.31 B x = 9.56 C x = 30.31 D x = −11.69 E x = 21

7E

SkillSH

EET 7.5WWORKEDORKEDEExample

10a


10b

EXCEL Spreadsheet

Equation solver:a(bx + c) = d

Mathcad

Solving linearequations

with brackets

GC program

Solvinglinear

equations


GAMEtime

Solvinglinear

equations— 001

MA

TH

SQUEST

C H A L L

EN

GE

MA

TH

SQUEST

C H A L L

EN

GE

1 How many whole numbers between 100 and1000 contain only 9, 8 or 7 as digits?

2 Find x in the magic square shown at right.Remember that each row, column and diagonaladd to the same amount.

3x 2x – 1 2(x + 2)

2x + 1 15 3x – 1

14 3x + 1 2x


Solve the equations given andcolour in the block containing each answer.

The remaining letters in the blocks left will spellout the puzzle’s answer.

The driest place!The driest place!

18 – 2x = 10

3(7+ 5x ) = –9

4(15 – 3a ) = 0

–3 = 3 + 2(5 – x )

6 – 5w + 2w = –27

5x + 8 – 7x = 26

105 – 12e = 21

7 – 8f = 95

17 + 4x = 41

–1 = 5 –

2(7 – 2b ) = 34

25 – 6c = 13

5 = –7 + 4f

1 – 7y = 85

8 + 3e19

= 2 2x3

17 + 8x7

= –1

5 – m4

3 =

M7

I5

D–4

E21

A4

S17

H2

E–20

R12

T19

I–10

N–16

Y6

C–13

H –1

I14

L–15

D9

R10

E–17

N–5

A3

T1

H–8

E16

M–3

A0

T13

C11

S–2

B–11

A18

C–6

K8

S–7

L–9

A–14

M15

A20

I–12



C h a p t e r 7 S o l v i n g l i n e a r e q u a t i o n s 231Solving problems with linear equations

At this point in our exploration of the world of linearequations, it is important to indicate that the ability tosolve these equations is a real-world skill. All kinds ofpeople, from economists to engineers and computerprogrammers must have superior equation-solving skillsin order to complete even the simplest of tasks. Linear equations are commonly used by these people.

There is no one single technique or set of rules which will enable us to convert aworded problem into an algebraic one. It is the search for keywords in the statement ofthe problem that should indicate the steps needed for the solution.

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

Define a pronumeral, representing the certain number.

Let x = a certain number.

Follow the instructions, to build up the equation.Three 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

Solve the equation. Since 12 is a factor of 108, divide both sides by 12 first. (You can omit the brackets.)

=

x + 3 = 9

Subtract 3 from both sides. x + 3 − 3 = 9 − 3 x = 6

Give a worded answer. The original number was 6.

1

2

312 x 3+( )

12----------------------- 108

12---------

4

5

11WORKEDExample

MQ9 Vic ch 07 1 Monday, September 17, 2001 9:24 AM


Taxi charges are $3.40 plus $0.35 per kilometre for any trip in Melbourne. If Elena’s taxi fare was $12.15, how far did she travel?

THINK WRITEDefine a pronumeral, representing what isn’t known.

Let x = distance travelled.

If it costs $0.35 to travel 1 kilometre, write the distance-based cost (i.e. the cost to travel x kilometres).

Distance-based cost = $0.35 × x= $0.35x

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

Constant charge = $3.40

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

Total cost = $3.40 + $0.35x

Write down the total cost of Elena's trip, as given in the question.

Total cost = $12.15

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

3.40 + 0.35x = 12.15

Solve the equation: first subtract 3.40 from both sides.

3.40 + 0.35x − 3.40 = 12.15 − 3.40 0.35x = 8.75

Divide both sides by 0.35 to find the value of x.

x =

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

1

2

3

4

5

6

7

88.750.35----------

9

12WORKEDExample



Note: In the previous problem the solution could be verified by finding the total cost oftravelling 200 km over the three-day period, when renting from each company. Thiscan be easily done by substituting 200 for x in each of the two expressions for the totalcost. Check these calculations:

Total cost (SIVA) = 150 + 1.20 × 200 Total cost (HURTS) = 120 + 1.35 × 200 = 150 + 240 = 120 + 270 = 390 = 390

We can now see that the cost of travelling 200 km over the three-day period is indeedthe same for both companies and is equal to $390.

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?

THINK WRITE

Define a pronumeral, representing the distance travelled.

Let x = distance travelled in km.

Write an expression for the total cost of renting the car.

Total cost = flat fee × 3 + cost per km × x

Write the amount of the flat fee and cost per km, charged by SIVA.

SIVA: flat fee = $50; cost per km = $1.20

Determine the total cost of renting from SIVA.

Total cost = 50 × 3 + 1.20 × x = 150 + 1.2x

Write the amount of the flat fee and the cost per km, charged by HURTS.

HURTS: flat fee = $40; cost per km = $1.35

Determine the total cost of renting from HURTS.

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

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

Subtract 1.2x from both sides. 150 + 1.2x − 1.2x = 120 + 1.35x − 1.2x 150 = 120 + 0.15x

Subtract 120 from both sides. 150 − 120 = 120 + 0.15x − 120 30 = 0.15x

Divide both sides of the equation by 0.15.

=

200 = xGive a worded answer to the question. If Nathan travels 200 km over 3 days the cost

will be the same.

1

2

3

4

5

6

7

8

9

1030

0.15---------- 0.15x

0.15-------------

11

13WORKEDExample



Solving problems with linear equations

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

2 One half my age is 10 years more than one-third my age. How old am I?

3 The Green Cab taxi company charges $3.25 plus $0.72 per kilometre. Michael has $12 left after going out for dinner.a How far can he go in the taxi?b If he lives 13 km from the

restaurant, will he make it home in the taxi? If not,how far will he have to walk?

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

5 A maker of an orange drink can purchase her raw materials from two sources.The first source provides liquid with 6% orange juice, while the second sourceprovides liquid with 3% orange juice. She wishes to make 1 litre of drink with 5%orange juice. Let x = 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?

rememberHow 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.

remember

7F

EXCEL

Spreadsheet

Equationsolver


11

GCpro

gram

Solvinglinearequations


12


6 The cost of producing computer CD-ROM disks is quoted as $1200 plus $0.95 per disk. If Maya’s small software company has a budget of $2100, how many CDs can she get made?

7 Joseph wishes to have some flyers delivered for his grocery business. Post quickdistributor quotes a price of $200 plus 50 cents per flyer, while another, 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?

8 Rachel, the bushwalker, goes ona 4 day journey. She travels acertain distance on the first day,half that distance on the secondday, a third that distance on thethird day and a fourth of thatdistance on the fourth day. If thetotal journey is 50 km, how fardid she walk on the first day?

9 Svetlana, another bushwalkergoes on a 5-day journey, usingthe same pattern as Rachel inthe previous question (a certainamount, then half that amount,then one third, one fourth andone fifth). If her journey is also50 km, how far did she travel onthe first day?

10 Nile.Com, the Internet bookstore, advertises its shipping cost to Australia as aflat rate of $20 for up to 10 books; while Sheds & Meager, their competitor,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?

11 A new Internet bookstore, Mississippi.com is starting up and offersthe following shipping costs: a flat rate of $10 plus $1.50 for eachbook over 3 books plus an additional $1 for each book over 6 booksdelivered.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 10) and find the ‘break-even’point.


13

WorkS

HEET 7.2



1 Solve the equation = −23.

2 Solve the equation 2.2x + 3.2 = 4.2.

3 Solve the equation 9 − 2x = 15.

4 Solve the equation = .

5 Solve the equation 6x + 5 = 2x − 11.

6 Solve the equation 5(4 − x) = 100.

7 Solve the equation 3(6x − 11) = 31.

8 Solve the equation 5.1(4.3x + 2.7) = 17.2, correct to 3 decimal places.

9 Twenty years less than half my current age is equal to my age 43 years ago. How old am I?

10 The cost of printing a textbook is quoted as $500 plus $0.08 per page. A small pub-lisher has a budget of $10 000 to publish a mathematics textbook. If they wish to have1000 copies printed, how many pages long is the book?

Solving linear inequations An equation is a statement of equality such as x = 2, while an inequation is a statementsuch as x < 2 (x is less than 2). The solution to a linear equation is a single point on anumber line, while the solution to an inequation is a portion of the number line.

The following table shows four types of simple inequations and their correspondingrepresentation on a number line.

Note that an open circle placed over the 2 indicates that 2 is not included; that is, 2does not satisfy the inequality statement. A closed or solid circle indicates that 2 isincluded; that is, it does satisfy the inequality statement.

Mathematical statement English statement Number line diagram

x > 2 x is greater than 2

x ≥ 2 x is greater than or equal to 2

x < 2 x is less than 2

x ≤ 2 x is less than or equal to 2

2x4---

27--- 8

x---

–2 0 2 4–4–6–8–10 6 8 10

–2 0 2 4–4–6–8–10 6 8 10

–2 0 2 4–4–6–8–10 6 8 10

–2 0 2 4–4–6–8–10 6 8 10


C h a p t e r 7 S o l v i n g l i n e a r e q u a t i o n s 237The basic technique for solving inequations is to:

1. imagine that in place of the inequality sign, there is an equals sign.2. solve as if it were a linear equation, except that in place of the equals sign keep the

original inequality sign.There is one exception, which we will discuss in the next section. Although we could

solve inequations that are as complex as equations, we will generally work with thesimple inequations, leaving more difficult ones for future years.

The special case — multiplying or dividing by a negative numberConsider the inequality 6 > 5 (6 is greater than 5).

If we multiply both sides of the inequality by −1 we get:−6 > −5 which is not correct.

We know that in fact −6 < −5. Applying this to inequations generally, when we multiplyor divide an inequation by a negative number, the direction of the inequality sign mustchange.

When multiplying or dividing by a negative number, change the direction of the inequality sign; that is, change: < to >

> to <≤ to ≥≥ to ≤

Solve each of the following linear inequations.a x + 3 ≤ 4 b 4x − 1 < −2 c 6x − 7 ≥ 3x + 5

THINK WRITE

a Write the inequation. a x + 3 ≤ 4Obtain x by subtracting 3 from both sides of the inequation. Keep the inequality sign the same throughout.

x + 3 − 3 ≤ 4 − 3x ≤ 1

b Write the inequation. b 4x − 1 < −2Add 1 to both sides. 4x − 1 + 1 < −2 + 1

4x < −1

Obtain x by dividing both sides by 4. <

x < −

c Write the inequation. c 6x − 7 ≥ 3x + 5Combine the pronumeral terms by subtracting 3x from both sides.

6x − 7 − 3x ≥ 3x + 5 − 3x 3x − 7 ≥ 5

Add 7 to both sides. 3x − 7 + 7 ≥ 5 + 7 3x ≥ 12

Obtain x by dividing both sides by 3. ≥

x ≥ 4

1

2

1

2

34x4

------ 1–4

------

14---

1

2

3

43x3

------ 123

------

14WORKEDExample



Solving linear inequations

1 Solve each of the following inequations. a x + 1 > 3 b a + 2 > 1 c y − 3 ≥ 4 d m − 1 ≥ 3 e p + 4 < 5 f x + 2 < 9g m − 5 ≤ 4 h a − 2 ≤ 5 i x − 4 > −1j 5 + m ≥ 7 k 6 + q ≥ 2 l 5 + a > −3

Solve each of the following linear inequations.a −3m + 5 < −7 b 5(x − 2) ≥ 7(x + 3)

THINK WRITE

a Write the inequation. a −3m + 5 < −7Subtract 5 from both sides. (No change to the inequality sign.)

−3m + 5 − 5 < −7 − 5−3m < −12

Obtain m by dividing both sides by −3. Reverse the inequality sign, since you are dividing by a negative number.

>

m > 4

b Write the inequation. b 5(x − 2) ≥ 7(x + 3)Expand both brackets. 5x − 10 ≥ 7x + 21Combine the pronumeral terms by subtracting 7x from both sides of the inequation.

5x − 10 − 7x ≥ 7x + 21 − 7x −2x − 10 ≥ 21

Add 10 to both sides. −2x − 10 + 10 ≥ 21 + 10−2x ≥ 31

Obtain x by dividing both sides by −2. Because we need to divide by a negative number, reverse the direction of the inequality sign.

≤

x ≤ −15

1

2

33m–3–

---------- 12–3–

---------

1

2

3

4

52x–2–

--------- 312–

------

12---

15WORKEDExample

remember1. The solution to an inequation is a portion of the number line. (That is, there are

an infinite number of solutions to any given inequation.)

2. When solving inequations, imagine an equals sign in place of the inequality sign and solve as if it was a linear equation. Remember to keep writing the original inequality sign between the two sides of each step.

3. Special case: if in the process of the solution you need to multiply or divide by a negative number, reverse the inequality sign. That is, change: < to >, > to <, ≤ to ≥ and ≥ to ≤.

remember

7G

Mathca

d

Solvinglinearinequations

EXCEL

Spreadsheet

Solving inear inequations


14a


C h a p t e r 7 S o l v i n g l i n e a r e q u a t i o n s 2392 Solve each of the following inequations.

a 3m > 9 b 5p ≤ 10 c 2a < 8d 4x ≥ 20 e 5p > −25 f 3x ≤ −21

g 2m ≥ −1 h 4b > −2 i > 6

j < 4 k ≤ −2 l ≥ 5

3 Solve each of the following inequations.a 2m + 3 < 12 b 3x + 4 ≥ 13 c 5p − 9 > 11d 4n − 1 ≤ 7 e 2b − 6 < 4 f 8y − 2 > 14g 10m + 4 ≤ − 6 h 2a + 5 ≥ −5 i 3b + 2 < −11j 6c + 7 ≤ 1 k 4p − 2 > −10 l 3a − 7 ≥ −28

4 Solve each of the following inequations.a 2m + 1 > m + 4 b 2a − 3 ≥ a −1 c 5a − 3 < a − 7d 3a + 4 ≤ a − 2 e 5x − 2 > 40 − 2x f 7x − 5 ≤ 11 − xg 2b + 25 > 7b + 5 h 2(a + 4) > a + 13 i 3(m − 1) < m + 1j 5(2m − 3) ≤ 3m + 6 k 3(5b + 2) ≤ −10 + 4b l 5(3m + 1) ≥ 2(m + 9)

5 Solve each of the following inequations.

a ≤ 4 b ≥ −4 c < −1

d > 6 e ≥ 2 f < 0

6 Solve each of the following inequations. a −2m > 4 b −5p ≤ 15 c −2a ≥ −10 d −p − 3 ≤ 2 e 10 − y ≥ 13 f 14 − x < 7 g 1 − 6p > 1 h 2 − 10a ≤ 0 i 2(3 − x) < 12 j −4(a + 9) ≥ 8 k −15 ≤ −3(2 + b) l 2x − 3 > 5x + 6m k + 5 < 2k − 3 n 3(x − 4) < 5(x + 5) o 7(a + 4) ≥ 4(2a − 3)

7When solving the inequation −2x > −7 we need to:

8 Solve each of the following inequations.

a > 1 b ≥ 2 c < −4

d < −1 e ≤ 0 f ≤ 3

9 Solve each of the following inequations.a 3k > 6 b −a − 7 < −2 c 5 − 3m ≥ 0d x + 4 > 9 e 10 − y ≤ 3 f 5 + 3d < −1

g ≥ −2 h ≤ 2 i > 0

j 5a − 2 < 4a + 7 k 6p + 2 ≤ 7p − 1 l 2(3x + 1) > 2x − 16

A change the sign to ≥ B change the sign to <C change the sign to = D change the sign to ≤E keep the sign unchanged

m3----

x2--- a

7--- m

5----


14b


14c

x 1+2

------------ x 2–5

----------- x 7+3

------------

2x 3+4

--------------- 3x 1–7

--------------- 5x 9+6

---------------


15


2 x–3

----------- 5 m–4

------------- –3 x–5

---------------

3 8a–2

--------------- 4 3m–2

---------------- –2m 6+10

--------------------

7 p3

------ 1 x–3

----------- – 4 2m–5

--------------------

GAMEtime

Solvinglinear

equations— 002



Rearranging formulasA skill related to solving linear equations is the rearrangement of formulas. In this case,a formula is an algebraic equation with 2 (or more) pronumerals. It is traditional forone pronumeral to be on the left-hand side of the equal sign and the rest on the right-hand side. However, there are times when another of the pronumerals is required to beon the left-hand side. The pronumeral that needs to be left by itself (usually on the left-hand side) is called the subject of the formula.

Solving for another pronumeralConsider the well-known formula known as Ohm’s Law:

V = I R.In this case I is the current, R is the resistance and V is the voltage.In this form the formula is useful if you know I and R and wish to find V.

The cost of concreteAt the beginning of this chapter we looked at the problem John had of deciding for what volume of concrete he should use either of two companies to do his concrete pours.Angelico’s Concrete: charges $700 plus $20 per cubic metre of concrete.Baux Cementing: charges $1200 plus $15 per cubic metre of concrete.1 Write an algebraic expression for

the cost of using Angelico’s Concrete.

2 Write an algebraic expression for the cost of using Baux Cementing.

3 Write an inequation that, when solved, will tell you the volume of concrete for which it is cheaper to use Angelico’s Concrete.

4 For what volume of concrete will it be cheaper to use Baux Cementing?

5 For what volume of concrete will the cost be the same (if any)?


C h a p t e r 7 S o l v i n g l i n e a r e q u a t i o n s 241However, it is likely that an engineer might wish to find R, given I and V. So what is

needed is a ‘new’ or rearranged formula, where R is on the left-hand side of the equalssign and the rest of the pronumerals (that is, I and V) are on the right-hand side. Inother words, we want to make R the subject of the formula.

In this case, dividing both sides by I yields the formula

which when simplified, gives

= R

or R = .

So now, if electrical engineers wish to know R, all they have to do is divide V by I. Generally, we use the same technique as for linear equations to obtain the desired

pronumeral (the subject) by itself by performing additions, subtractions, multiplicationsand divisions.

Can you see how similar this is to solving linear equations? Just get the pronumeral ofinterest by itself, by performing appropriate inverse operations!

VI---- IR

I------=

VI----

VI----

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

THINK WRITE

a Write the equation. a y = kx + mAttempt to get x by itself. First subtract m from both sides.

y − m = kx + m − m y − m = kx

Divide both sides by k. =

= x

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

x =

b Write the equation. b 6(y + 1) = 7(x − 2)Expand the brackets first. 6y + 6 = 7x − 14Attempt to get x by itself. First add 14 to both sides of the equation.

6y + 6 + 14 = 7x − 14 + 14 6y + 20 = 7x


= x

Re-write so that x is on the left-hand side.

x =

1

2

3y m–

k------------- kx

k-----

y m–k

-------------

4y m–

k-------------

1

2

3

46y 20+

7------------------ 7x

7------

6y 20+7

------------------

56y 20+

7------------------

16WORKEDExample



Rearranging formulas

1 Rearrange each formula to make x the subject.a y = ax b y = ax + b c y = 2ax − bd y + 4 = 2x − 3 e 6(y + 2) = 5(4 − x) f x(y − 2) = 1g x(y − 2) = y + 1 h 5x − 4y = 1 i 6(x + 2) = 5(x − y)j 7(x − a) = 6x + 5a k 5(a − 2x) = 9(x + 1) l 8(9x − 2) + 3 = 7(2a − 3x)

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

a g = 6d − 3 (d) b a = (v)

THINK WRITE

a Write the equation. a g = 6d − 3

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

g + 3 = 6d − 3 + 3 g + 3 = 6d


= d

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

b Write the equation. b a =

Attempt to get v by itself. First multiply both sides by t.

at = × t

at = v − u

Add u to both sides. at + u = v − u + u at + u = v

Rewrite so that v is on the left-hand side. v = at + u

v u–t

------------

1

2

3

g 3+6

------------ 6d6

------

g 3+6

------------

4g 3+

6------------

1v u–

t-----------

2v u–

t-----------

3

4

17WORKEDExample

remember1. 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.

remember

7HWWORKEDORKEDEExample

16


C h a p t e r 7 S o l v i n g l i n e a r e q u a t i o n s 2432 For each of the following, make the variable shown in brackets the subject of the formula.

a g = 4P − 3 (P) b f = (c)

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

e v = u + at (t) f d = b2 − 4ac (c)

g m = (y) h m = (y)

i m = (a) j m = (x)

k C = 2πr (r) l f = ax + by (x)

m s = ut + at2 (a) n F = (G)

3 The cost to rent a car isgiven by the formula C = 50d + 0.2k, where d = the number ofdays rented and k = thenumber of km driven. Linhas $300 to spend on carrental for her 4-day holiday.How far can she travel onthis holiday?

4 The total surface area of a cylinder is given by the formula T = 2πr2h + 2πrh, wherer = radius and h = height. A car manufacturer wants the engine’s cylinders to have aradius of 4 cm and a total surface area of 400 cm2. What is the height of the cylinder?(Hint: Express h in terms of T and r.)

Maximum viewing areaA rectangular window is to be constructed so that its perimeter is 200 cm. The window needs to have the largest possible area. Find the dimensions of the window. 1 Draw a diagram of the window, showing the width (w) and height (h).2 Express the area (A) in terms of w and h.3 Express the perimeter (P) in terms of w and h. 4 Form an equation by substituting 200 for P in the expression for the perimeter. 5 Rearrange the equation to make h the subject.6 Substitute the expression found for h into the area formula.7 Calculate the area for some selected values of w. (Try w = 10, 20, 30; and so on.)8 Which value of w gives the largest area? Try a few values either side of this w

value to confirm your result.9 Find the corresponding value of h.

10 Write the answer to the problem in words.


179c5

------9c5

------

y k–h

----------- y a–x b–-----------

y a–x b–----------- y a–

x b–-----------

12---

GMmr2

--------------

WorkS

HEET 7.3



Copy the sentences below. Fill in the gaps by choosing the correct word or phrase from the word list that follows.

1 A linear equation must have exactly one and one. The power of the pronumeral must be .

2 In solving equations, if you do something to the you must also do it to the right-hand side.

3 To solve a linear equation, perform appropriate inverse operations toboth sides of the equation until the pronumeral is left .

4 The inverse operations, which will allow us to get the pronumeral on itsown, must be performed in the .

5 If there is a pronumeral in the denominator of a fraction, either invert or.

6 If there is more than one term with the pronumeral in it, a term needs to be created.

7 If an equation contains brackets, the brackets first, unlessdividing both sides by the coefficient in front of the brackets does notcreate a fraction.

8 The solution to an inequation is a of the number line.

9 If in the process of the solution of an inequation, you need to divide ormultiply by a negative number, the inequality sign must be .

10 The pronumeral that is required to be left by itself on the left-hand side ofthe equals sign is called the of the formula.

11 To rearrange the , so that the required pronumeral is thesubject, use the same techniques as for linear equations, treating the

as numbers.

summary

W O R D L I S Tsubjectother

pronumeralssingle

reverse ordercross-multiplyequals signexpand

pronumeralby itselfoneportion

reversedformulaleft-hand side



1 Which of the following are linear equations?

a 5x + y = 0 b 2x + 3 = x − 2 c = 3

d x2 = 1 e + 1 = 3x f 8 = 5x − 2

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

2 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.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.

3 Solve each of the following linear equations. a y − 3 = 10 b y + 3 = 22 c 2y = 45

d y − 9 = −10 e = −6.4 f a + 1.2 = 2.1

g 3a = 8.4 h a + 2.3 = 1.7 i = −0.12

j b − 1.45 = 1.65 k b + 3.45 = 0 l 7.53b = 5.64

4 Solve each of the following linear equations. a 6x + 2 = 14 b −9x + 2 = 20 c 8 − x = 4

d − = 7 e 8x − 5 = 25 f 82 = 5x − 9

g −7x − 6 = 8 h = 8 i = 5

j = −4 k = 3 l = 5

m = n = −6 o = −4.1


a 6x + 2 = 5x b 7x = 5x − 2 c 11x = 15x − 2

d 3x + 4 = 16 − x e 5x + 2 = 3x + 8 f 8x − 9 = 7x − 4

g 2x + 5 = 8x − 7 h 3 − 4x = 6 − x i x − 15 = 20 − 4x

7A

CHAPTERreview

x2---

1x---

7A

7Ba2---

b21------

7Cx3---

x 1+2

------------ 2x 3–7

---------------

5 x–2

----------- 3x– 4–5

------------------- 6x---

4x--- 3

5--- 3x

4------ x 1.7+

2.3----------------

7D



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

7 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.

8 Samuel decides to go on a holiday. He travels a certain distance on the first 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?

9 Solve each of the following inequations. a 2x − 5 < 3 b 6x + 7 ≥ 12 c −x − 2 > 0

d > −2 e −5x ≥ 10 f 5x < −6

g 2 − x ≤ −22 h 8 + 3x > −17 i 7x + 3 < 2x + 1j 6x + 1 ≥ 5 − 4x k 6(x + 2) ≤ 5(x + 1) l 3(5 − 2x) > 6(1 − 2x)m 8 + 3x ≤ −1 n 0 ≥ 9 + 2x o 4(2 − 3x) < 1

p ≥ −2 q > −5 r ≤ 1

10 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 = t (t)

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

7E

7F

7F

7Gx7---

3 x–2

----------- 5 2x+3

--------------- 5 2x–3

---------------

7H

u v+2

----------- testtest

CHAPTERyyourselfourself

testyyourselfourself

7


Solving linear equations · Solving linear equations MQ9 Vic ch 07 Page 209 Monday, September 17,...

Documents

Transcript of Solving linear equations · Solving linear equations MQ9 Vic ch 07 Page 209 Monday, September 17,...