algebra (notation(notation and equations) and equations) · pdf fileopening problem 0 12...

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Contents:

Algebra(notation and equations)

Algebra(notation and equations)

A

B

C

D

E

F

G

H

Algebraic notation

Algebraic substitution

Linear equations

Rational equations

Linear inequations

Problem solving

Money and investment problems

Motion problems (Extension)

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Y:\HAESE\SA_10-6ed\SA10-6_01\011SA10-6_01.CDR Monday, 4 September 2006 2:15:47 PM PETERDELL

OPENING PROBLEM

0

12 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)

Algebra is a very powerful tool which is used to make problem solving easier. Algebra

involves using pronumerals (letters) to represent unknown values, or values which can vary

depending on the situation.

Many worded problems can be converted to algebraic symbols to make algebraic equations.

We learn how to solve equations in order to find solutions to the problems.

Algebra can also be used to construct formulae, which are equations that connect two or more

variables. Many people use formulae as part of their jobs, so an understanding of how to

substitute into and rearrange formulae is essential. Builders, nurses, pharmacists, engineers,

financial planners and computer programmers all use formulae which rely on algebra.

Consider the following:

²

²

²

The ability to convert worded sentences and problems into algebraic symbols, and to under-

stand algebraic notation, is essential in the problem solving process.

Notice that: ² x2 + 3x is an algebraic expression, whereas

² x2 + 3x = 8 is an equation, and

² x2 + 3x > 28 is an inequality (sometimes called an inequation).

Recall that when we simplify repeated sums, we use product notation:

i.e., x+ x

= 2 ‘lots’ of x

= 2£ x

= 2x

and x+ x+ x

= 3 ‘lots’ of x

= 3£ x

= 3x

Also, when we simplify repeated products, we use index notation:

i.e., x£ x = x2 and x£ x£ x = x3

ALGEBRAIC NOTATIONA

Holly bought XBC shares for

$ each and NGL shares for

$ each.

2 50

4 00

:

:

What did Holly pay, in total, for

XBC shares and NGL shares?

500

600

What did Holly pay in total for XBC

shares and NGL shares?

x

x( + 100)

Bob knows that Holly paid, in total, $ for her XBC and NGL shares. He also

knows that she bought more NGL shares than XBC shares. How can Bob use

algebra to find how many of each share type Holly bought?

5925

100


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Y:\HAESE\SA_10-6ed\SA10-6_01\012SA10-6_01.CDR Wednesday, 30 August 2006 11:11:19 AM DAVID3

ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) 13

1 Write, in words, the meaning of:

a 2a b pq cpm d a2

e a¡ 3 f b+ c g 2x+ c h (2a)2

i 2a2 j a¡ c2 k a+ b2 l (a+ b)2

Write the following as algebraic expressions:

a the sum of p and the square of q b the square of the sum of p and q

c b less than double a

a p+ q2 b (p+ q)2 c 2a¡ b

2 Write the following as algebraic expressions:

a the sum of a and c b the sum of p, q and r

c the product of a and b d the sum of r and the square of s

e the square of the sum of r and s f the sum of the squares of r and s

g the sum of twice a and b h the difference between p and q, if p > q

i a less than the square of b j half the sum of a and b

k the sum of a and a quarter of b l the square root of the sum of m and n

m the sum of x and its reciprocal n a quarter of the sum of a and b

o the square root of the sum of the squares of x and y

Write, in words, the meaning of:

a x¡ 5 b a+ b c 3x2 + 7

a x¡ 5 is “5 less than x”.

b a+ b is “the sum of a and b”, or “b more than a”.

c 3x2 + 7 is “7 more than three times the square of x”.

Self TutorExample 1

EXERCISE 1A

Self TutorExample 2

Write, in sentence form, the meaning of:

a D = ct b A =b+ c

2

a D is equal to the product of c and t.

b A is equal to a half of the sum of b and c,

or, A is the average of b and c.

Self TutorExample 3


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Y:\HAESE\SA_10-6ed\SA10-6_01\013SA10-6_01.CDR Tuesday, 25 July 2006 3:35:10 PM PETERDELL


3 Write, in sentence form, the meaning of:

a L = a+ b b K =a+ b

2c M = 3d

d N = bc e T = bc2 f F = ma

g K =

rn

th c =

pa2 + b2 i A =

a+ b+ c

3

4 Write the following as algebraic equations:

a S is the sum of p and r

b D is the difference between a and b where b > a

c A is the average of k and l

d M is the sum of a and its reciprocal

e K is the sum of t and the square of s

f N is the product of g and h

g y is the sum of x and the square of x

h P is the square root of the sum of d and e

Consider the number crunching machine alongside:

If we place any number x, into the machine, it calculates 5x¡ 7, i.e., x is multiplied by

5 and then 7 is subtracted.

For example: if x = 2, 5x¡ 7

= 5£ 2 ¡ 7

= 10¡ 7

= 3

and if x = ¡2, 5x¡ 7

= 5£¡2 ¡ 7

= ¡10¡ 7

= ¡17

To evaluate an algebraic expression, we substitute numerical values for the unknown, then

calculate the result.

Write ‘S is the sum of a and the product of g and t’ as an equation.

The product of g and t is gt.

The sum of a and gt is a+ gt, ) the equation is S = a+ gt.

Self TutorExample 4

The

between two

numbers is the

larger one minus

the smaller one.

difference

ALGEBRAIC SUBSTITUTIONB

5 – 7 calculatorx

input, x

output


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Y:\HAESE\SA_10-6ed\SA10-6_01\014SA10-6_01.CDR Thursday, 3 August 2006 10:14:31 AM PETERDELL


If p = 4, q = ¡2 and r = 3, find the value of:

a 3q ¡ 2r b 2pq ¡ r cp¡ 2q + 2r

p+ r

a 3q ¡ 2r

= 3£¡2 ¡ 2£ 3

= ¡6 ¡ 6

= ¡12

b 2pq ¡ r= 2£ 4£¡2 ¡ 3

= ¡16 ¡ 3

= ¡19

cp¡ 2q + 2r

p+ r

=4 ¡ 2£¡2 + 2£ 3

4 + 3

=4 + 4 + 6

4 + 3

=14

7

= 2

1 If p = 5, q = 3 and r = ¡4 find the value of:

a 5p

e 3p¡ 2q

2 If w = 3, x = 1 and y = ¡2, evaluate:

ay

w

ey ¡ x+ w

2(y ¡ w)

b 4q

f 5r ¡ 4q

by +w

x

fxy +w

y ¡ x

c 3pq

g 4q ¡ 2r

c3x¡ yw

gx¡wy

y + w ¡ 2x

d pqr

h 2pr + 5q

d5w ¡ 2x

y ¡ xh

y

x¡w

If a = 3, b = ¡2 and c = ¡1, evaluate:

a b2 b ab¡ c3

a b2

= (¡2)2

= ¡2£¡2

= 4

b ab¡ c3= 3£¡2 ¡ (¡1)3

= ¡6 ¡ ¡1

= ¡6 + 1

= ¡5

3 If a = ¡3, b = ¡4 and c = ¡1, evaluate:

a c2

e b3 + c3b b3

f (b+ c)3c a2 + b2

g (2a)2d (a+ b)2

h 2a2

Self TutorExample 5

EXERCISE 1B

Notice the use

of brackets!

Self TutorExample 6


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INVESTIGATION SOLVING EQUATIONS


4 If p = 4, q = ¡1 and r = 2, evaluate:

app+ q

eppr ¡ q

bpp+ q

fpp2 + q2

cpr ¡ q

gpp+ r + 2q

dpp¡ pq

hp

2q ¡ 5r

1 Enter the function Y1 = 5X ¡ 3 into your calculator.

2 Set up a table that calculates the value of y = 5x¡ 3

for x values from ¡5 to 5.

3 View the table and scroll down until you find the value

of x that makes Y1 equal to 12.

As we can see, the solution is x = 3.

4 Use your calculator and the method given above to solve the following equations:

a 7x+ 1 = ¡20 b 8¡ 3x = ¡4

cx

4+ 2 = 1 d

1

3(2x¡ 1) = 3

5

a 2x¡ 3 = ¡6 b 6¡ 4x = 8 c x¡ 5 = ¡3:5

6 Use a calculator to solve the following equations:

a 3x+ 2 = 41 b 5¡ 4x = 70 c2x

3+ 5 = 22

3

If p = 4, q = ¡3 and r = 2, evaluate:

app¡ q + r b

pp+ q2

app¡ q + r

=p

4¡¡3 + 2

=p

4 + 3 + 2

=p

9

= 3

bpp+ q2

=p

4 + (¡3)2

=p

4 + 9

=p

13

+ 3:61

Self TutorExample 7

Linear equations like can be solved usinga table of values on a .

We try to find the value of which makes the expressionequal to upon . This is the

to the equation.

5 3 = 12

5 3 12

x

x

x

¡

¡

graphics calculator

solutionsubstitution

What to do:

The solutions to the following equations are , so change your table toinvestigate values from to in intervals of :

not integers

x :¡5 5 0 5


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Y:\HAESE\SA_10-6ed\SA10-6_01\016SA10-6_01.CDR Monday, 31 July 2006 9:31:44 AM PETERDELL


Linear equations are equations which can be written in the form ax+ b = 0, where x

is the unknown (variable) and a, b are constants.

The following steps should be followed when solving linear equations:

Step 1: Decide how the expression containing the unknown has been ‘built up’.

Step 2: Isolate the unknown by performing inverse operations on both sides of

the equation to ‘undo’ the ‘build up’ in reverse order.

Step 3: Check your solution by substitution.

1 Solve for x:

a x+ 9 = 4

e 2x+ 5 = 17

b 5x = 45

f 3x¡ 2 = ¡14

c ¡24 = ¡6x

g 3¡ 4x = ¡17

d 3¡ x = 12

h 8 = 9¡ 2x

LINEAR EQUATIONSC

SOLVING EQUATIONS

The inverse ofis

andis

+ ¡

£ ¥

EXERCISE 1C

Solve for x: a 4x¡ 1 = 7 b 5¡ 3x = 6

a 4x¡ 1 = 7

) 4x¡ 1 + 1 = 7 + 1 fadding 1 to both sidesg) 4x = 8

)4x

4=

8

4fdivide both sides by 4g

) x = 2

Check: 4£ 2 ¡ 1 = 8¡ 1 = 7 X

b 5¡ 3x = 6

) 5¡ 3x¡ 5 = 6¡ 5 fsubtracting 5 from both sidesg) ¡3x = 1

)¡3x

¡3=

1

¡3fdividing both sides by ¡3g

) x = ¡1

3

Check: 5 ¡ 3£¡1

3= 5 + 1 = 6 X

Example 8 Self Tutor


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2 Solve for x:

ax

4= 12 b

1

2x = 6 c 5 =

x

¡2d

x

3+ 4 = ¡2

ex+ 3

5= ¡2 f

1

3(x+ 2) = 3 g

2x¡ 1

3= 7 h

1

2(5¡ x) = ¡2

If the unknown appears more than once, expand any brackets, collect like terms, and

then solve the equation.

Solve for x: ax

5¡ 3 = ¡1 b

1

5(x¡ 3) = ¡1

ax

5¡ 3 = ¡1

)x

5¡ 3 + 3 = ¡1 + 3 fadding 3 to both sidesg

)x

5= 2

)x

5£ 5 = 2£ 5 fmultiplying both sides by 5g

) x = 10 Check: 10

5¡ 3 = 2¡ 3 = ¡1 X

b1

5(x¡ 3) = ¡1

) 1

5(x¡ 3)£ 5 = ¡1£ 5 fmultiplying both sides by 5g) x¡ 3 = ¡5

) x¡ 3 + 3 = ¡5 + 3 fadding 3 to both sidesg) x = ¡2

Check: 1

5(¡2¡ 3) = 1

5£¡5 = ¡1 X

Self TutorExample 9

Solve for x: 3(2x¡ 5)¡ 2(x¡ 1) = 3

3(2x¡ 5)¡ 2(x¡ 1) = 3

) 6x¡ 15¡ 2x+ 2 = 3 fexpanding bracketsg) 4x¡ 13 = 3 fcollecting like termsg

) 4x¡ 13 + 13 = 3 + 13 fadding 13 to both sidesg) 4x = 16

) x = 4 fdividing both sides by 4gCheck: 3(2£ 4 ¡ 5)¡ 2(4¡ 1) = 3£ 3 ¡ 2£ 3 = 3 X

Self TutorExample 10


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Y:\HAESE\SA_10-6ed\SA10-6_01\018SA10-6_01.CDR Wednesday, 26 July 2006 10:02:41 AM PETERDELL


3 Solve for x:

a 2(x+ 8) + 5(x¡ 1) = 60 b 2(x¡ 3) + 3(x+ 2) = ¡5

c 3(x+ 3)¡ 2(x+ 1) = 0 d 4(2x¡ 3) + 2(x+ 2) = 32

e 3(4x+ 1)¡ 2(3x¡ 4) = ¡7 f 5(x+ 2)¡ 2(3¡ 2x) = ¡14

If the unknown appears on both sides of the equation, we

² expand any brackets and collect like terms

² move the unknown to one side of the equation and the remaining terms to

the other side

² simplify and solve the equation.

Solve for x:

a 3x¡ 4 = 2x+ 6 b 4¡ 3(2 + x) = x

a 3x¡ 4 = 2x+ 6

) 3x¡ 4¡ 2x = 2x+ 6¡ 2x fsubtracting 2x from both sidesg) x¡ 4 = 6

) x¡ 4 + 4 = 6 + 4 fadding 4 to both sidesg) x = 10

Check: LHS = 3£ 10¡ 4 = 26, RHS = 2£ 10 + 6 = 26. X

b 4¡ 3(2 + x) = x

) 4¡ 6¡ 3x = x fexpandingg) ¡2¡ 3x = x

) ¡2¡ 3x+ 3x = x+ 3x fadding 3x to both sidesg) ¡2 = 4x

)¡2

4=

4x

4fdividing both sides by 4g

) ¡1

2= x

i.e., x = ¡1

2

Check: LHS = 4 ¡ 3(2 +¡1

2) = 4 ¡ 3£ 3

2= 4¡ 41

2= ¡1

2= RHS X

4 Solve for x:

a 2x¡ 3 = 3x+ 6 b 3x¡ 4 = 5¡ xc 4¡ 5x = 3x¡ 8 d ¡x = 2x+ 4

e 12¡ 7x = 3x+ 7 f 5x¡ 9 = 1¡ 3x

g 4¡ x¡ 2(2¡ x) = 6 + x h 5¡ 3(1¡ x) = 2¡ 3x

i 5¡ 2x¡ (2x+ 1) = ¡6 j 3(4x+ 2)¡ x = ¡7 + x



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Sometimes when more complicated equations are expanded a linear equation results.

Solve for x: (x¡ 3)2 = (4 + x)(2 + x)

(x¡ 3)2 = (4 + x)(2 + x)

) x2 ¡ 6x+ 9 = 8 + 4x+ 2x+ x2 fexpanding each sideg) x2 ¡ 6x+ 9¡ x2 = 8 + 4x+ 2x+ x2 ¡ x2 fsubtracting x2 from both sidesg

) ¡6x+ 9 = 8 + 6x

) ¡6x+ 9 + 6x = 8 + 6x+ 6x fadding 6x to both sidesg) 9 = 12x+ 8

) 9¡ 8 = 12x+ 8¡ 8 fsubtracting 8 from both sidesg) 1 = 12x

)1

12=

12x


) x = 1

12

5 Solve for x:

a x(x+ 5) = (x¡ 2)(x¡ 3) b x(2x+ 1)¡ 2(x¡ 3) = 2x(x+ 1)

c (x+ 1)(x¡ 2) = (4¡ x)2 d x2 ¡ 3 = (2 + x)(1 + x)

e (x¡ 2)(2x¡ 1) = 2x(x+ 3) f (x+ 4)2 = (x+ 1)(x¡ 3)

6 Solve for x:

a 2(3x+ 1)¡ 3 = 6x¡ 1 b 3(4x+ 1) = 6(2x+ 1)

c Comment on your solutions to a and b.

Rational equations are equations involving fractions. We simplify them by writing all frac-

tions with the same least common denominator (LCD), and then equating the numerators.

Consider the following rational equations:

x

2=x

3LCD is 2£ 3 i.e., 6

5

2x=

3x

5LCD is 2x£ 5 i.e., 10x

x¡ 7

3=

x

2x¡ 1LCD is 3£ (2x¡ 1) i.e., 3(2x¡ 1)


RATIONAL EQUATIONSD


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Solve for x:x

2=

3 + x

5

x

2=

3 + x

5has LCD = 10

)x

2£ 5

5=

2

2£µ

3 + x

5

¶fto create a common denominatorg

) 5x = 2(3 + x) fequating numeratorsg) 5x = 6 + 2x fexpanding bracketsg

) 5x¡ 2x = 6 + 2x¡ 2x ftaking 2x from both sidesg) 3x = 6

) x = 2 fdividing both sides by 3g

1 Solve for x:

ax

2=

4

7

dx+ 1

3=

2x¡ 1

4

g2x¡ 1

3=

4¡ x6

b5

8=x

6

e2x

3=

5¡ x2

h4x+ 7

7=

5¡ x2

cx

2=x¡ 2

3

f3x+ 2

5=

2x¡ 1

2

i3x+ 1

6=

4x¡ 1

¡2

2 Solve for x:

a5

x=

2

3b

6

x=

3

5c

4

3=

5

xd

3

2x=

7

6

e3

2x=

7

3f

7

3x= ¡1

6g

5

4x= ¡ 1

12h

4

7x=

3

2x

Notice the

insertion of

brackets here.


EXERCISE 1D

Solve for x:4

x=

3

4

4

x=

3

4has LCD = 4x

)4

x£ 4

4=

3

4£ xx

fto create a common denominatorg

) 16 = 3x fequating numeratorsg) x = 16




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Solve for x:2x + 1

3 ¡ x=

3

4

2x + 1

3 ¡ x=

3

4has LCD = 4(3 ¡ x)

)4

4£µ

2x + 1

3 ¡ x

¶=

3

4£µ

3 ¡ x

3 ¡ x

¶fto create a common denominatorg

) 4(2x + 1) = 3(3 ¡ x) fequating numeratorsg) 8x + 4 = 9 ¡ 3x fexpanding the bracketsg

) 8x + 4 + 3x = 9 ¡ 3x + 3x fadding 3x to both sidesg) 11x + 4 = 9

) 11x + 4 ¡ 4 = 9 ¡ 4 fsubtracting 4 from both sidesg) 11x = 5

) x = 5


3 Solve for x:

a2x + 3

x + 1=

5

3

dx + 3

2x¡ 1=

1

3

g6x¡ 1

3 ¡ 2x= 5

bx + 1

1 ¡ 2x=

2

5

e4x + 3

2x¡ 1= 3

h5x + 1

x + 4= 4

c2x¡ 1

4 ¡ 3x= ¡3

4

f3x¡ 2

x + 4= ¡3

i 2 +2x + 5

x¡ 1= ¡3

Solve for x:x

3¡ 1 ¡ 2x

6= ¡4

x

3¡ 1 ¡ 2x

6= ¡4 has LCD of 6

)x

3£ 2

2¡µ

1 ¡ 2x

6

¶= ¡4 £ 6

6fto create a common denominatorg

) 2x¡ (1 ¡ 2x) = ¡24 fequating numeratorsg) 2x¡ 1 + 2x = ¡24 fexpandingg

) 4x¡ 1 = ¡24

) 4x¡ 1 + 1 = ¡24 + 1 fadding 1 to both sidesg) 4x = ¡23

) x = ¡23


Notice the use ofbrackets here.




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Y:\HAESE\SA_10-6ed\SA10-6_01\022SA10-6_01.CDR Friday, 18 August 2006 1:28:14 PM PETERDELL


4 Solve for x:

ax

2¡ x

6= 4 b

x

4¡ 3 =

2x

3

cx

8+

x + 2

2= ¡1 d

x + 2

3+

x¡ 3

4= 1

e2x¡ 1

3¡ 5x¡ 6

6= ¡2 f

x

4= 4 ¡ x + 2

3

g2x¡ 7

3¡ 1 =

x¡ 4

6h

x + 1

3¡ x

6=

2x¡ 3

2

ix

5¡ 2x¡ 5

3=

3

4j

x + 1

3+

x¡ 2

6=

x + 4

12

kx¡ 6

5¡ 2x¡ 1

10=

x¡ 1

2l

2x + 1

4¡ 1 ¡ 4x

2=

3x + 7

6

The speed limit (s km/h) when passing roadworks is often 25 km/h.

This can be written as a linear inequation,

i.e., s 6 25 (reads ‘s is less than or equal to 25’).

We can also represent the allowable speeds on a number line,

i.e.,

The number line shows that any speed of 25 km/h or less (24, 23, 18, 71

2etc), is an acceptable

speed. We say these are solutions of the inequation.

Notice that 5 > 3 and 3 < 5, and

similarly ¡3 < 2 and 2 > ¡3

This suggests that if we interchange the LHS and RHS of an inequation, then we must

reverse the inequation sign.

Note: > is the reverse of <; > is the reverse of 6.

Recall also that:

² If we add or subtract the same number to both sides the inequation sign is

maintained, for example, if 5 > 3, then 5 + 2 > 3 + 2.

² If we multiply or divide both sides by a positive number the inequation sign is

maintained, for example, if 5 > 3, then 5 £ 2 > 3 £ 2:

² If we multiply or divide both sides by a negative number the inequation sign is

reversed, for example, if 5 > 3, then 5 £¡1 < 3 £¡1:

LINEAR INEQUATIONSE

250 s

25

km/h

RULES FOR HANDLING INEQUATIONS


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Y:\HAESE\SA_10-6ed\SA10-6_01\023SA10-6_01.CDR Tuesday, 5 September 2006 2:25:56 PM PETERDELL

4

The filled-in circle indicatesthat 4 included.is

The arrowhead indicates that all numbers onthe number line in this direction are included.

x

5

The hollow circle indicates that 5 included.is not

x

Notice the reversal of

the inequation sign in

line as we are

dividing by

4

2¡ :

b

2 x

�2 x


Consequently, we can say that the method of solution of linear inequalities is identical to that

of linear equations with the exceptions that:

Suppose our solution to an inequation is x > 4, so every number which is 4 or greater than

4 is a possible value for x. We could represent this on a number line by

Likewise if our solution is x < 5 our representation would be

Solve for x and graph the solutions: a 3x¡ 4 6 2 b 3 ¡ 2x < 7

a 3x¡ 4 6 2

) 3x¡ 4 + 4 6 2 + 4 fadding 4 to both sidesg) 3x 6 6

)3x

36

6


) x 6 2

Check: Let x = 1 ) 3x¡ 4 = 3 £ 1 ¡ 4 = ¡1

and ¡1 < 2 is true.

b 3 ¡ 2x < 7

) 3 ¡ 2x¡ 3 < 7 ¡ 3 ftaking 3 from both sidesg) ¡2x < 4

)¡2x

¡2>

4

¡2fdividing both sides by

¡2, so reverse the signg) x > ¡2

Check: Let x = 3 ) 3 ¡ 2x = 3 ¡ 2 £ 3 = ¡3

and ¡3 < 7 is true.

² interchanging the sides reverses the inequation sign

² multiplying dividing negative

reverses

or both sides by a

number, the inequation sign.

GRAPHING SOLUTIONS



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Y:\HAESE\SA_10-6ed\SA10-6_01\024SA10-6_01.CDR Monday, 21 August 2006 10:06:54 AM PETERDELL


1 Solve for x and graph the solutions:

a 3x+ 2 < 0

d 5¡ 2x 6 11

b 5x¡ 7 > 2

e 2(3x¡ 1) < 4

c 2¡ 3x > 1

f 5(1¡ 3x) > 8


a 7 > 2x¡ 1

d ¡3 > 4¡ 3x

b ¡13 < 3x+ 2

e 3 < 5¡ 2x

c 20 > ¡5x

f 2 6 5(1¡ x)

EXERCISE 1E

Solve for x and graph the solutions: 3¡ 5x > 2x+ 7

3¡ 5x > 2x+ 7

) 3¡ 5x¡ 2x > 2x+ 7¡ 2x fsubtract 2x from both sidesg) 3¡ 7x > 7

) 3¡ 7x¡ 3 > 7¡ 3 fsubtract 3 from both sidesg) ¡7x > 4

)¡7x

¡76

4

¡7fdivide both sides by ¡7;

reverse the signg

) x 6 ¡4

7

Check: if x = ¡1, say, 3 ¡ 5£¡1 > 2£¡1 + 7, i.e., 8 > 5 is true.

¡4

7x

Solve for x and graph the solutions: ¡5 < 9¡ 2x

¡5 < 9¡ 2x

) ¡5 + 2x < 9¡ 2x+ 2x fadding 2x to both sidesg) 2x¡ 5 < 9

) 2x¡ 5 + 5 < 9 + 5 fadding 5 to both sidesg) 2x < 14

)2x

2<

14


i.e., x < 7

Check: if x = 5, say ¡5 < 9¡ 2£ 5, i.e., ¡5 < ¡1 is true.

Example 18

7 x


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a 3x+ 2 > x¡ 5 b 2x¡ 3 < 5x¡ 7

c 5¡ 2x > x+ 4 d 7¡ 3x 6 5¡ xe 3x¡ 2 > 2(x¡ 1) + 5x f 1¡ (x¡ 3) > 2(x+ 5)¡ 1

4 Solve for x:

a 3x+ 1 > 3(x+ 2) b 5x+ 2 < 5(x+ 1) c 2x¡ 4 > 2(x¡ 2)

d Comment on your solutions to a, b and c.

Many problems can be translated into algebraic equations. When problems are solved using

algebra, we follow these steps:

Step 1: Decide the unknown quantity and allocate a pronumeral.

Step 2: Decide which operations are involved.

Step 3:

Step 4:

Step 5: Check that your solution does satisfy the original problem.

Step 6: Write your answer in sentence form. Remember, there is usually no pronumeral

in the original problem.

1 When three times a certain number is subtracted from 15, the result is ¡6. Find the

number.

2 Five times a certain number, minus 5, is equal to 7 more than three times the number.

What is the number?

PROBLEM SOLVINGF

When a number is trebled and subtracted from 7 the result is ¡11.

Find the number.

Let x be the number.

) 3x is the number trebled.

) 7¡ 3x is this number subtracted from 7.

So, 7¡ 3x = ¡11

) 7¡ 3x¡ 7 = ¡11¡ 7 fsubtracting 7 from both sidesg) ¡3x = ¡18

) x = 6 fdividing both sides by ¡3gSo, the number is 6:

Check: 7 ¡ 3£ 6 = 7¡ 18 = ¡11 X

Example 20

EXERCISE 1F

Translate the problem into an equation and check your translation is correct.

Solve the equation by isolating the pronumeral.

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3 Three times the result of subtracting a certain number from 7 gives the same answer as

adding eleven to the number. Find the number.

4 I think of a number. If I divide the sum of 6 and the number by 3, the result is 4 more

than one quarter of the number. Find the number.

5 The sum of two numbers is 15. When one of these numbers is added to three times the

other, the result is 27. What are the numbers?

What number must be added to both the numerator and the denominator of the

fraction 1

3to get the fraction 7

8?

Let x be the number.

)1 + x

3 + x=

7

8where the LCD is 8(3 + x)

)8

8£µ1 + x

3 + x

¶=

7

8£µ3 + x

3 + x

¶fto get a common denominatorg

) 8(1 + x) = 7(3 + x) fequating numeratorsg) 8 + 8x = 21 + 7x fexpanding bracketsg

) 8 + 8x¡ 7x = 21 + 7x¡ 7x fsubtracting 7x from both sidesg) 8 + x = 21

) x = 13 So, 13 is added to both.

6 What number must be added to both the numerator and the denominator of the fraction2


8?

7 What number must be subtracted from both the numerator and the denominator of the

fraction 3


3?

Example 21

Let Sarah’s present age be x years.

) father’s present age is 3x years.

Table of ages:

Now 13 years time

Sarah x x+ 13

Father 3x 3x+ 13

So, 3x+ 13 = 2(x+ 13)

) 3x+ 13 = 2x+ 26

) 3x¡ 2x = 26¡ 13

) x = 13

) Sarah’s present age is 13.

Sarah’s age is one third her father’s age and in years time her age will be a half

of her father’s age. How old is Sarah now?

13


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Y:\HAESE\SA_10-6ed\SA10-6_01\027SA10-6_01.CDR Wednesday, 30 August 2006 11:12:11 AM DAVID3


8 Eli is now one-quarter of his father’s age and in 5 years time his age will be one-third

the age of his father. How old is Eli now?

9 When Maria was born her mother was 24 years old.

At present Maria’s age is 20% of her mother’s age.

How old is Maria now?

10 Five years ago Jacob was one-sixth the age of his

brother. In three years time his age doubled will

match his brother’s age. How old is Jacob now?

Problems involving money are frequently made easier to understand by constructing a table

and placing the given information into it.

MONEY AND INVESTMENT PROBLEMSG

Brittney has only 2-cent and 5-cent stamps with a total value of $1:78 and there are

two more 5-cent stamps than there are 2-cent stamps. How many 2-cent stamps are

there?

If there are x 2-cent stamps then

there are (x+ 2) 5-cent stamps

Type Number Value

2-cent x 2x cents

5-cent x+ 2 5(x+ 2) cents

) 2x+ 5(x+ 2) = 178 fequating values in centsg) 2x+ 5x+ 10 = 178

) 7x+ 10 = 178

) 7x = 168

) x = 24 So, there are 24, 2-cent stamps.

1 Michaela has 5-cent and 10-cent stamps with a total value of $5:75 . If she has 5 more

10-cent stamps than 5-cent stamps, how many of each stamp does she have?

2 The school tuck-shop has milk in 600 mL and 1 litre cartons. If there are 54 cartons and

40 L of milk in total, how many 600 mL cartons are there?

3 Aaron has a collection of American coins. He has three times as many 10 cent coins as

25 cent coins, and he has some 5 cent coins as well. If he has 88 coins with total value

$11:40, how many of each type does he have?

4 Tickets at a football match cost $8, $15 or $20 each. The number of $15 tickets sold

was double the number of $8 tickets sold. 6000 more $20 tickets were sold than $15

tickets. If the total gate receipts were $783 000, how many of each type of ticket was

sold?

Example 23

EXERCISE 1G

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5 Kelly blends coffee. She mixes brand A costing $6 per

kilogram with brand B costing $8 per kilogram. How

many kilograms of each brand does she have to mix to

make 50 kg of coffee costing her $7:20 per kg?

6 Su Li has 13 kg of almonds costing $5 per kg. How many kg

of cashews costing $12 per kg should be added to get a mixture

of the two nut types which would cost $7:45 per kg?

I invest in oil shares which earn me 12% yearly, and in coal mining shares which

pay 10% yearly. If I invest $3000 more in oil shares than in coal mining shares and

my total yearly earnings amount to $910, how much did I invest in each type of

share?

Let the amount I invest in coal mining shares be $x. Draw up a table:

Type of Shares Amount invested ($) Interest Earnings ($)

Coal x 10% 10% of x

Oil (x+ 3000) 12% 12% of (x+ 3000)

Total 910

From the table we can write the equation from the information about earnings:

10% of x + 12% of (x+ 3000) = 910

) 0:1x + 0:12(x+ 3000) = 910

) 0:1x+ 0:12x+ 360 = 910

) 0:22x+ 360 = 910

) 0:22x = 550

) x =550

0:22

) x = 2500

) I invested $2500 in coal shares and $5500 in oil shares.

7 Qantas shares pay a yearly return of 9% and Telstra shares pay 11%. John invests $1500

more on Telstra shares than on Qantas shares and his total yearly earnings from the two

investments is $1475. How much did he invest in Qantas shares?

8 I invested twice as much money in technology shares as I invested in mining shares.

Technology shares earn me 10% yearly and mining shares earn me 9% yearly. My yearly

income from these shares is $1450: Find how much I invested in each type of share.

9 Wei has three types of shares; A, B and C. A shares pay 8%, B shares pay 6% and C

shares pay 11% dividends. Wei has twice as much invested in B shares as A shares and

has $50 000 invested altogether. The yearly return from the share dividends is $4850.

How much is invested in each type of share?



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10 Mrs Jones invests $4000 at 10% annual return and $6000 at 12% annual return. How

much should she invest at 15% return so that her total annual return is 13% of the total

amount she has invested?

1 Joe can run twice as fast as Pete. They start at the same point and run in opposite

directions for 40 minutes and the distance between them is then 16 km. How fast does

Joe run?

2 A car leaves a country town at 60 km per hour and 2 hours later a second car leaves the

town and catches the first car in 5 hours. Find the speed of the second car.

3 A boy cycles from his house to a friend’s house at 20 km/h and home again at 25 km/h.

If his round trip takes 9

10of an hour, how far is it to his friend’s house?

4 A motor cyclist makes a trip of 500 km. If he had increased his speed by 10 km/h, he

could have covered 600 km in the same time. What was his original speed?

5 Normally I drive to work at 60 km/h. If I drive at 72 km/h I cut 8 minutes off my time

for the trip. What distance do I travel?

Motion problems are problems concerned with speed, distance travelled and time taken.

Recall that speed is measured in kilometres per hour, i.e.,kilometres

hours:

So, speed =distance

time, distance = speed £ time and time =

distance

speed:

MOTION PROBLEMS (EXTENSION)H

EXERCISE 1H

Let the speed in the first 2 hours be s km/h. Draw up a table.

Speed (km/h) Time (h) Distance (km)

First section s 2 2s

Second section (s+ 10) 3 3(s+ 10)

Total 455

Using

distance

= speed £ time

So, 2s+ 3(s+ 10) = 455

) 2s+ 3s+ 30 = 455

) 5s = 425 and so s = 85

) the car’s speed in the first two hours was 85 km/h.

A car travels for hours at a certain speed and then hours more at a speed km/hfaster than this. If the entire distance travelled is km, find the car’s speed in thefirst two hours of travel.

2 3 10

455



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REVIEW SET 1A

6 My motor boat normally travels at 24 km/h in

still water. One day I travelled 36 km against

a constant current in a river and it took me the

same time to travel 48 km with the current.

How fast was the current?

1 Write in algebraic form:

a “3 more than the square of x” b “the square of the sum of 3 and x”


apa + 3 b

pa + 3

3 If p = 1, q = ¡2 and r = ¡3, find the value of4q ¡ p

r:

4 Solve for x:

a 5 ¡ 2x = 3x + 4 bx¡ 1

2¡ 2 ¡ 3x

7=

1

3

5 Solve the following and graph the solutions: 5 ¡ 2x > 3(x + 6)

6 If a number is increased by 5 and then trebled, the result is six more than two thirds

of the number. Find the number.

7

OPENING PROBLEM

�Revisit the Opening Problem on page 12. Answer the questions posed.

A drinks stall sells small, medium and large cups of fruit drink for $ , $ and

$ respectively. In one morning three times as many medium cups were sold as

small cups, and the number of large cups sold was less than the number of

medium cups. If the total of the drink sales was $ , how many of each size cup

was sold?

1 50 2

2 50

140

1360

:

:

Click on the icon to access a set of problems called .Mixture Problems PRINTABLE

PAGES

EXTENSION


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Y:\HAESE\SA_10-6ed\SA10-6_01\031SA10-6_01.CDR Tuesday, 5 September 2006 2:38:21 PM PETERDELL


REVIEW SET 1B


aa + b

2b a +

b

2

2 Write in algebraic form:

a “the sum of a and the square root of b”

b “the square root of the sum of a and b”

3 If a = ¡1, b = 4 and c = ¡6, find the value of2b + 3c

2a.

4 Solve the following inequation and graph the solution set: 5x + 2(3 ¡ x) < 8 ¡ x

5 Solve for x:

a 2(x¡ 3) + 4 = 5 b2x + 3

3¡ 3x + 1

4= 2

6 What number must be added to the numerator and denominator of 3

4in order to finish

with 1

3?

7 X shares pay 8% dividend and Y shares pay 9%. Reiko has invested $2000 more

on X shares than she has on Y shares. Her total earnings for the year for the two

investments is $2710. How much does she invest in X shares?

8 Carlos cycles for 2 hours at a fast speed, then

cycles for 3 more hours at a speed 10 km/h

slower than this. If the entire distance trav-

elled is 90 km, find Carlos’ speed in the first

two hours of travel.

8 Ray drives between towns A and B at an

average speed of km/h. Mahmoud

drives the same distance at km/h and

takes minutes less. What is the distance

between A and B?

75

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Y:\HAESE\SA_10-6ed\SA10-6_01\032SA10-6_01.CDR Monday, 21 August 2006 10:11:26 AM PETERDELL

algebra (notation(notation and equations) and equations) · pdf fileopening problem 0 12...

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