# 1 Algebra (notation and equations)

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Algebra
(notation and equations)
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Algebraic notation
Algebraic substitution
Linear equations
Rational equations
Linear inequations
Problem solving
Money and investment problems
Motion problems (Extension)
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Contents:
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SA_10-6
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.
OPENING PROBLEM
Holly bought XBC shares for
\$2:50 each and NGL shares for
\$4:00 each.
0
Consider the following:
²
What did Holly pay, in total, for 500
XBC shares and 600 NGL shares?
²
What did Holly pay in total for x XBC
shares and (x + 100) NGL shares?
²
Bob knows that Holly paid, in total, \$5925 for her XBC and NGL shares. He also
knows that she bought 100 more NGL shares than XBC shares. How can Bob use
algebra to find how many of each share type Holly bought?
A
ALGEBRAIC NOTATION
The ability to convert worded sentences and problems into algebraic symbols, and to understand algebraic notation, is essential in the problem solving process.
x2 + 3x
x2 + 3x = 8
x2 + 3x > 28
²
²
²
Notice that:
is an algebraic expression, whereas
is an equation, and
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
x+x+x
= 3 ‘lots’ of x
=3£x
= 3x
and
Also, when we simplify repeated products, we use index notation:
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x £ x £ x = x3
and
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x £ x = x2
i.e.,
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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
SA_10-6
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
Example 1
Self Tutor
Write, in words, the meaning of:
a x¡5
b a+b
x¡5
a+b
3x2 + 7
a
b
c
13
c
3x2 + 7
is “5 less than x”.
is “the sum of a and b”, or “b more than a”.
is “7 more than three times the square of x”.
EXERCISE 1A
1 Write, in words, the meaning of:
a 2a
b pq
c
p
m
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
Example 2
Self Tutor
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
2 Write
a
c
e
g
i
k
m
o
p + q2
b (p + q)2
c
2a ¡ b
the following as algebraic expressions:
the sum of a and c
b the sum of p, q and r
the product of a and b
d the sum of r and the square of s
the square of the sum of r and s f the sum of the squares of r and s
the sum of twice a and b
h the difference between p and q, if p > q
a less than the square of b
j half the sum of a and b
the sum of a and a quarter of b
l the square root of the sum of m and n
the sum of x and its reciprocal
n a quarter of the sum of a and b
the square root of the sum of the squares of x and y
Example 3
Self Tutor
Write, in sentence form, the meaning of:
b+c
a D = ct
b A=
2
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D is equal to the product of c and t.
A is equal to a half of the sum of b and c,
or, A is the average of b and c.
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SA_10-6
14
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
3 Write, in sentence form, the meaning of:
a
L=a+b
b
d
N = bc
r
n
K=
t
e
a+b
2
2
T = bc
h
c=
g
K=
p
a2 + b2
Example 4
c
M = 3d
f
F = ma
i
A=
a+b+c
3
Self Tutor
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.
The difference
between two
numbers is the
larger one minus
the smaller one.
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
B
ALGEBRAIC SUBSTITUTION
input, x
Consider the number crunching machine alongside:
5x – 7 calculator
output
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
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To evaluate an algebraic expression, we substitute numerical values for the unknown, then
calculate the result.
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SA_10-6
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
Example 5
15
Self Tutor
If p = 4, q = ¡2 and r = 3, find the value of:
a 3q ¡ 2r
a
b 2pq ¡ r
3q ¡ 2r
= 3 £ ¡2 ¡ 2 £ 3
= ¡6 ¡ 6
= ¡12
b
p ¡ 2q + 2r
p+r
c
2pq ¡ r
= 2 £ 4 £ ¡2 ¡ 3
= ¡16 ¡ 3
= ¡19
p ¡ 2q + 2r
p+r
c
=
4 ¡ 2 £ ¡2 + 2 £ 3
4+3
=
4+4+6
4+3
14
7
=2
=
EXERCISE 1B
1 If p = 5, q = 3 and r = ¡4 find the value of:
5p
3p ¡ 2q
a
e
4q
5r ¡ 4q
b
f
c
g
2 If w = 3, x = 1 and y = ¡2, evaluate:
y
y+w
a
b
c
w
x
y¡x+w
xy + w
e
f
g
2(y ¡ w)
y¡x
Example 6
3pq
4q ¡ 2r
3x ¡ y
w
x ¡ wy
y + w ¡ 2x
b2
= (¡2)2
= ¡2 £ ¡2
=4
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h
5w ¡ 2x
y¡x
y
¡w
x
Notice the use
of brackets!
ab ¡ c3
= 3 £ ¡2 ¡ (¡1)3
= ¡6 ¡ ¡1
= ¡6 + 1
= ¡5
b
3 If a = ¡3, b = ¡4 and c = ¡1, evaluate:
a c2
b b3
c
3
3
3
e b +c
f (b + c)
g
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pqr
2pr + 5q
Self Tutor
If a = 3, b = ¡2 and c = ¡1, evaluate:
a b2
b ab ¡ c3
a
d
h
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a2 + b2
(2a)2
d
h
(a + b)2
2a2
SA_10-6
16
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
Example 7
Self Tutor
If p = 4, q = ¡3 and r = 2, evaluate:
p
p
a
p¡q+r
b
p + q2
p
p¡q+r
p
= 4 ¡ ¡3 + 2
p
= 4+3+2
p
= 9
=3
a
b
p
p + q2
p
= 4 + (¡3)2
p
= 4+9
p
= 13
+ 3:61
4 If p = 4, q = ¡1 and r = 2, evaluate:
p
p
a
p+q
b
p+q
c
p
p
e
pr ¡ q
f
p2 + q 2
g
p
r¡q
p
p + r + 2q
INVESTIGATION
d
h
p
p ¡ pq
p
2q ¡ 5r
SOLVING EQUATIONS
Linear equations like 5x ¡ 3 = 12 can be solved using
a table of values on a graphics calculator.
We try to find the value of x which makes the expression
5x ¡ 3 equal to 12 upon substitution. This is the solution
to the equation.
What to do:
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
x
+2=1
d 13 (2x ¡ 1) = 3
c
4
5 The solutions to the following equations are not integers, so change your table to
investigate x values from ¡5 to 5 in intervals of 0:5:
a 2x ¡ 3 = ¡6
b 6 ¡ 4x = 8
c x ¡ 5 = ¡3:5
6 Use a calculator to solve the following equations:
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5 ¡ 4x = 70
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3x + 2 = 41
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2x
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SA_10-6
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
C
17
LINEAR EQUATIONS
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.
SOLVING EQUATIONS
The following steps should be followed when solving linear equations:
Step 1:
Step 2:
Decide how the expression containing the unknown has been ‘built up’.
Isolate the unknown by performing inverse operations on both sides of
the equation to ‘undo’ the ‘build up’ in reverse order.
Check your solution by substitution.
Step 3:
Example 8
Self Tutor
a 4x ¡ 1 = 7
Solve for x:
a
)
4x ¡ 1 = 7
4x ¡ 1 + 1 = 7 + 1
) 4x = 8
)
The inverse of
+ is ¡
and
£ is ¥
b 5 ¡ 3x = 6
fadding 1 to both sidesg
8
4x
=
4
4
) x=2
fdivide both sides by 4g
Check: 4 £ 2 ¡ 1 = 8 ¡ 1 = 7 X
b
)
5 ¡ 3x = 6
5 ¡ 3x ¡ 5 = 6 ¡ 5
) ¡3x = 1
fsubtracting 5 from both sidesg
1
¡3x
=
¡3
¡3
)
fdividing both sides by ¡3g
x = ¡ 13
)
Check: 5 ¡ 3 £ ¡ 13 = 5 + 1 = 6 X
EXERCISE 1C
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5x = 45
3x ¡ 2 = ¡14
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1 Solve for x:
a x+9=4
e 2x + 5 = 17
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¡24 = ¡6x
3 ¡ 4x = ¡17
d
h
3 ¡ x = 12
8 = 9 ¡ 2x
SA_10-6
18
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
Example 9
Self Tutor
x
¡ 3 = ¡1
5
Solve for x:
a
a
x
¡ 3 = ¡1
5
x
¡ 3 + 3 = ¡1 + 3
5
x
)
=2
5
x
£5 = 2£5
)
5
) x = 10
)
1
5 (x
b
Check:
¡ 3) = ¡1
fmultiplying both sides by 5g
Check:
10
5
¡ 3 = 2 ¡ 3 = ¡1 X
¡ 3) = ¡1
¡ 3) £ 5 = ¡1 £ 5
) x ¡ 3 = ¡5
x ¡ 3 + 3 = ¡5 + 3
) x = ¡2
)
1
5 (x
fadding 3 to both sidesg
1
5 (x
)
1
5 (¡2
¡ 3) =
2 Solve for x:
x
a
= 12
4
x+3
= ¡2
5
e
b
1
5
b
1
2x
f
1
3 (x
fmultiplying both sides by 5g
fadding 3 to both sidesg
£ ¡5 = ¡1 X
=6
+ 2) = 3
c
5=
x
¡2
d
x
+ 4 = ¡2
3
g
2x ¡ 1
=7
3
h
1
2 (5
¡ x) = ¡2
If the unknown appears more than once, expand any brackets, collect like terms, and
then solve the equation.
Example 10
Self Tutor
3(2x ¡ 5) ¡ 2(x ¡ 1) = 3
Solve for x:
3(2x ¡ 5) ¡ 2(x ¡ 1) = 3
) 6x ¡ 15 ¡ 2x + 2 = 3
) 4x ¡ 13 = 3
) 4x ¡ 13 + 13 = 3 + 13
) 4x = 16
) x=4
fexpanding bracketsg
fcollecting like termsg
fadding 13 to both sidesg
fdividing both sides by 4g
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Check: 3(2 £ 4 ¡ 5) ¡ 2(4 ¡ 1) = 3 £ 3 ¡ 2 £ 3 = 3 X
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SA_10-6
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
3 Solve for x:
a 2(x + 8) + 5(x ¡ 1) = 60
c 3(x + 3) ¡ 2(x + 1) = 0
e 3(4x + 1) ¡ 2(3x ¡ 4) = ¡7
b
d
f
19
2(x ¡ 3) + 3(x + 2) = ¡5
4(2x ¡ 3) + 2(x + 2) = 32
5(x + 2) ¡ 2(3 ¡ 2x) = ¡14
If the unknown appears on both sides of the equation, we
²
²
expand any brackets and collect like terms
move the unknown to one side of the equation and the remaining terms to
the other side
simplify and solve the equation.
²
Example 11
Self Tutor
Solve for x:
a 3x ¡ 4 = 2x + 6
b 4 ¡ 3(2 + x) = x
3x ¡ 4 = 2x + 6
) 3x ¡ 4 ¡ 2x = 2x + 6 ¡ 2x
) x¡4 = 6
) x¡4+4 = 6+4
) x = 10
a
fsubtracting 2x from both sidesg
fadding 4 to both sidesg
Check: LHS = 3 £ 10 ¡ 4 = 26, RHS = 2 £ 10 + 6 = 26. X
4 ¡ 3(2 + x) = x
) 4 ¡ 6 ¡ 3x = x
) ¡2 ¡ 3x = x
) ¡2 ¡ 3x + 3x = x + 3x
) ¡2 = 4x
4x
¡2
=
)
4
4
1
) ¡2 = x
b
fexpandingg
fadding 3x to both sidesg
fdividing both sides by 4g
i.e., x = ¡ 12
Check: LHS = 4 ¡ 3(2 + ¡ 12 ) = 4 ¡ 3 £
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= 4 ¡ 4 12 = ¡ 12 = RHS X
3x ¡ 4 = 5 ¡ x
¡x = 2x + 4
5x ¡ 9 = 1 ¡ 3x
5 ¡ 3(1 ¡ x) = 2 ¡ 3x
3(4x + 2) ¡ x = ¡7 + x
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4 Solve for x:
a 2x ¡ 3 = 3x + 6
c 4 ¡ 5x = 3x ¡ 8
e 12 ¡ 7x = 3x + 7
g 4 ¡ x ¡ 2(2 ¡ x) = 6 + x
i 5 ¡ 2x ¡ (2x + 1) = ¡6
3
2
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SA_10-6
20
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
Sometimes when more complicated equations are expanded a linear equation results.
Example 12
Self Tutor
(x ¡ 3)2 = (4 + x)(2 + x)
Solve for 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
) ¡6x + 9 = 8 + 6x
) ¡6x + 9 + 6x = 8 + 6x + 6x
) 9 = 12x + 8
) 9 ¡ 8 = 12x + 8 ¡ 8
) 1 = 12x
12x
1
=
)
12
12
)
x=
fsubtracting x2 from both sidesg
fadding 6x to both sidesg
fsubtracting 8 from both sidesg
fdividing both sides by 12g
1
12
5 Solve for x:
a x(x + 5) = (x ¡ 2)(x ¡ 3)
c (x + 1)(x ¡ 2) = (4 ¡ x)2
e (x ¡ 2)(2x ¡ 1) = 2x(x + 3)
x(2x + 1) ¡ 2(x ¡ 3) = 2x(x + 1)
x2 ¡ 3 = (2 + x)(1 + x)
(x + 4)2 = (x + 1)(x ¡ 3)
b
d
f
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.
D
RATIONAL EQUATIONS
Rational equations are equations involving fractions. We simplify them by writing all fractions with the same least common denominator (LCD), and then equating the numerators.
Consider the following rational equations:
x
x
=
2
3
LCD is 2 £ 3 i.e., 6
3x
5
=
2x
5
LCD is 2x £ 5 i.e., 10x
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LCD is 3 £ (2x ¡ 1) i.e., 3(2x ¡ 1)
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x¡7
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SA_10-6
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
Example 13
Self Tutor
x
3+x
=
2
5
Solve for x:
3+x
x
=
2
5
µ
¶
2
3+x
x 5
£ = £
2 5
2
5
)
has LCD = 10
Notice the
insertion of
brackets here.
fto create a common denominatorg
) 5x = 2(3 + x)
) 5x = 6 + 2x
5x ¡ 2x = 6 + 2x ¡ 2x
) 3x = 6
) x=2
)
21
fequating numeratorsg
fexpanding bracketsg
ftaking 2x from both sidesg
fdividing both sides by 3g
EXERCISE 1D
1 Solve for x:
a
x
4
=
2
7
b
5
x
=
8
6
c
x
x¡2
=
2
3
d
2x ¡ 1
x+1
=
3
4
e
5¡x
2x
=
3
2
f
2x ¡ 1
3x + 2
=
5
2
g
4¡x
2x ¡ 1
=
3
6
h
5¡x
4x + 7
=
7
2
i
4x ¡ 1
3x + 1
=
6
¡2
Example 14
Self Tutor
3
4
=
x
4
Solve for x:
4
3
=
x
4
3 x
4 4
£ = £
x 4
4 x
) 16 = 3x
)
)
x=
has LCD = 4x
fto create a common denominatorg
fequating numeratorsg
16
3
fdividing both sides by 3g
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3
=
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2x
75
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5
1
=¡
4x
12
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g
5
7
1
=¡
3x
6
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3
7
=
2x
3
50
e
75
3
7
=
2x
6
25
d
0
4
5
=
3
x
5
c
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3
=
x
5
100
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50
5
2
=
x
3
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2 Solve for x:
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SA_10-6
22
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
Example 15
Self Tutor
Solve for x:
3
2x + 1
=
3¡x
4
2x + 1
=
3¡x
¶
µ
4
2x + 1
=
£
4
3¡x
)
)
)
3
4
has LCD = 4(3 ¡ x)
3
£
4
µ
3¡x
3¡x
¶
fto create a common denominatorg
4(2x + 1) = 3(3 ¡ x)
)
)
Notice the use of
brackets here.
fequating numeratorsg
8x + 4 = 9 ¡ 3x
fexpanding the bracketsg
8x + 4 + 3x = 9 ¡ 3x + 3x
) 11x + 4 = 9
11x + 4 ¡ 4 = 9 ¡ 4
) 11x = 5
fadding 3x to both sidesg
)
x=
fsubtracting 4 from both sidesg
5
11
fdividing both sides by 11g
3 Solve for x:
a
5
2x + 3
=
x+1
3
b
2
x+1
=
1 ¡ 2x
5
c
3
2x ¡ 1
=¡
4 ¡ 3x
4
d
x+3
1
=
2x ¡ 1
3
e
4x + 3
=3
2x ¡ 1
f
3x ¡ 2
= ¡3
x+4
g
6x ¡ 1
=5
3 ¡ 2x
h
5x + 1
=4
x+4
i
2+
Example 16
Self Tutor
x 1 ¡ 2x
¡
= ¡4
3
6
x 1 ¡ 2x
¡
= ¡4
3
6
µ
¶
6
x 2
1 ¡ 2x
£ ¡
= ¡4 £
3 2
6
6
fto create a common denominatorg
2x ¡ (1 ¡ 2x) = ¡24
2x ¡ 1 + 2x = ¡24
) 4x ¡ 1 = ¡24
) 4x ¡ 1 + 1 = ¡24 + 1
) 4x = ¡23
fequating numeratorsg
fexpandingg
)
95
yellow
Y:\HAESE\SA_10-6ed\SA10-6_01\022SA10-6_01.CDR Friday, 18 August 2006 1:28:14 PM PETERDELL
100
50
75
0
95
fdividing both sides by 4g
100
50
75
25
0
95
50
75
25
0
5
95
100
50
75
25
0
5
100
magenta
5
x = ¡ 23
4
)
cyan
fadding 1 to both sidesg
25
)
has LCD of 6
5
Solve for x:
)
2x + 5
= ¡3
x¡1
black
SA_10-6
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
4 Solve for x:
x x
a
¡ =4
2
6
b
x
2x
¡3 =
4
3
c
x x+2
+
= ¡1
8
2
d
x+2 x¡3
+
=1
3
4
e
2x ¡ 1 5x ¡ 6
¡
= ¡2
3
6
f
x
x+2
=4¡
4
3
g
x¡4
2x ¡ 7
¡1=
3
6
h
2x ¡ 3
x+1 x
¡ =
3
6
2
i
x 2x ¡ 5
3
¡
=
5
3
4
j
x+1 x¡2
x+4
+
=
3
6
12
k
x¡1
x ¡ 6 2x ¡ 1
¡
=
5
10
2
l
3x + 7
2x + 1 1 ¡ 4x
¡
=
4
2
6
E
23
LINEAR INEQUATIONS
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,
25
km/h
i.e.,
0
s
25
The number line shows that any speed of 25 km/h or less (24, 23, 18, 7 12 etc), is an acceptable
speed. We say these are solutions of the inequation.
RULES FOR HANDLING INEQUATIONS
Notice that
similarly
5>3
¡3 < 2
3 < 5, and
2 > ¡3
and
and
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
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100
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5
95
100
50
75
25
0
5
95
50
25
0
5
95
100
50
75
25
0
5
²
100
²
75
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.
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:
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:
black
Y:\HAESE\SA_10-6ed\SA10-6_01\023SA10-6_01.CDR Tuesday, 5 September 2006 2:25:56 PM PETERDELL
SA_10-6
24
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
Consequently, we can say that the method of solution of linear inequalities is identical to that
of linear equations with the exceptions that:
²
²
interchanging the sides reverses the inequation sign
multiplying or dividing both sides by a negative
number, reverses the inequation sign.
GRAPHING SOLUTIONS
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
x
4
The filled-in circle indicates
that 4 is included.
The arrowhead indicates that all numbers on
the number line in this direction are included.
Likewise if our solution is x < 5 our representation would be
x
5
The hollow circle indicates that 5 is not included.
Example 17
Self Tutor
Solve for x and graph the solutions:
3x ¡ 4 6 2
3x ¡ 4 + 4 6 2 + 4
) 3x 6 6
6
3x
6
3
3
) x62
magenta
ftaking 3 from both sidesg
Notice the reversal of
the inequation sign in
b line 4 as we are
dividing by ¡2:
fdividing both sides by
¡2, so reverse the signg
x
-2
yellow
95
100
50
75
3 ¡ 2x = 3 ¡ 2 £ 3 = ¡3
and ¡3 < 7 is true.
0
50
)
75
0
5
95
50
25
0
5
95
100
50
75
25
0
75
Let x = 3
Check:
5
3x ¡ 4 = 3 £ 1 ¡ 4 = ¡1
and ¡1 < 2 is true.
x > ¡2
25
)
100
)
cyan
)
3 ¡ 2x < 7
3 ¡ 2x ¡ 3 < 7 ¡ 3
) ¡2x < 4
4
¡2x
>
)
¡2
¡2
b
x
2
25
Check: Let x = 1
fdividing both sides by 3g
5
)
b 3 ¡ 2x < 7
fadding 4 to both sidesg
95
)
100
a
a 3x ¡ 4 6 2
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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
SA_10-6
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
25
EXERCISE 1E
1 Solve for x and graph the solutions:
a 3x + 2 < 0
b
d 5 ¡ 2x 6 11
e
5x ¡ 7 > 2
2(3x ¡ 1) < 4
Example 18
2 ¡ 3x > 1
5(1 ¡ 3x) > 8
c
f
Self Tutor
Solve for x and graph the solutions: ¡5 < 9 ¡ 2x
¡5 < 9 ¡ 2x
) ¡5 + 2x < 9 ¡ 2x + 2x
) 2x ¡ 5 < 9
) 2x ¡ 5 + 5 < 9 + 5
) 2x < 14
fadding 2x to both sidesg
fadding 5 to both sidesg
14
2x
<
2
2
)
fdividing both sides by 2g
i.e., x < 7
x
7
Check: if x = 5, say ¡5 < 9 ¡ 2 £ 5, i.e., ¡5 < ¡1 is true.
2 Solve for x and graph the solutions:
a 7 > 2x ¡ 1
b
d ¡3 > 4 ¡ 3x
e
¡13 < 3x + 2
3 < 5 ¡ 2x
c
f
Example 19
20 > ¡5x
2 6 5(1 ¡ x)
Self Tutor
Solve for x and graph the solutions: 3 ¡ 5x > 2x + 7
3 ¡ 5x > 2x + 7
3 ¡ 5x ¡ 2x > 2x + 7 ¡ 2x
) 3 ¡ 7x > 7
) 3 ¡ 7x ¡ 3 > 7 ¡ 3
) ¡7x > 4
fsubtract 2x from both sidesg
)
fsubtract 3 from both sidesg
4
¡7x
6
¡7
¡7
)
fdivide both sides by ¡7;
reverse the signg
x 6 ¡ 47
)
x
¡ 47
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100
50
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25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
Check: if x = ¡1, say, 3 ¡ 5 £ ¡1 > 2 £ ¡1 + 7, i.e., 8 > 5 is true.
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Y:\HAESE\SA_10-6ed\SA10-6_01\025SA10-6_01.CDR Wednesday, 26 July 2006 10:40:40 AM PETERDELL
SA_10-6
26
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
3 Solve for x and graph the solutions:
a 3x + 2 > x ¡ 5
c 5 ¡ 2x > x + 4
e 3x ¡ 2 > 2(x ¡ 1) + 5x
2x ¡ 3 < 5x ¡ 7
7 ¡ 3x 6 5 ¡ x
1 ¡ (x ¡ 3) > 2(x + 5) ¡ 1
b
d
f
4 Solve for x:
a 3x + 1 > 3(x + 2)
b 5x + 2 < 5(x + 1)
d Comment on your solutions to a, b and c.
F
c
2x ¡ 4 > 2(x ¡ 2)
PROBLEM SOLVING
Many problems can be translated into algebraic equations. When problems are solved using
algebra, we follow these steps:
Step
Step
Step
Step
Step
Step
1:
2:
3:
4:
5:
6:
Decide the unknown quantity and allocate a pronumeral.
Decide which operations are involved.
Translate the problem into an equation and check your translation is correct.
Solve the equation by isolating the pronumeral.
Check that your solution does satisfy the original problem.
Write your answer in sentence form. Remember, there is usually no pronumeral
in the original problem.
Example 20
Self Tutor
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
) ¡3x = ¡18
) x=6
fsubtracting 7 from both sidesg
fdividing both sides by ¡3g
So, the number is 6:
Check: 7 ¡ 3 £ 6 = 7 ¡ 18 = ¡11 X
EXERCISE 1F
1 When three times a certain number is subtracted from 15, the result is ¡6. Find the
number.
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5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
2 Five times a certain number, minus 5, is equal to 7 more than three times the number.
What is the number?
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Y:\HAESE\SA_10-6ed\SA10-6_01\026SA10-6_01.CDR Wednesday, 26 July 2006 10:18:08 AM PETERDELL
SA_10-6
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
27
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?
Example 21
Self Tutor
What number must be added to both the numerator and the denominator of the
fraction 13 to get the fraction 78 ?
Let x be the number.
1+x
=
3+x
µ
¶
8
1+x
£
=
8
3+x
)
)
)
7
8
where the LCD is 8(3 + x)
7
£
8
µ
3+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 fraction
2
7
5 to get the fraction 8 ?
7 What number must be subtracted from both the numerator and the denominator of the
fraction 34 to get the fraction 13 ?
Example 22
Self Tutor
Sarah’s age is one third her father’s age and in 13 years time her age will be a half
of her father’s age. How old is Sarah now?
Let Sarah’s present age be x years.
) father’s present age is 3x years.
So, 3x + 13 = 2(x + 13)
) 3x + 13 = 2x + 26
) 3x ¡ 2x = 26 ¡ 13
) x = 13
Table of ages:
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95
100
50
75
25
0
) Sarah’s present age is 13.
5
95
100
50
75
25
0
5
95
13 years time
x + 13
3x + 13
Now
x
3x
100
50
75
25
0
5
95
100
50
75
25
0
5
Sarah
Father
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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
SA_10-6
28
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
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?
G
MONEY AND INVESTMENT PROBLEMS
Problems involving money are frequently made easier to understand by constructing a table
and placing the given information into it.
Example 23
Self Tutor
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?
Type
2-cent
5-cent
If there are x 2-cent stamps then
there are (x + 2) 5-cent stamps
Number
x
x+2
Value
2x cents
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.
EXERCISE 1G
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?
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yellow
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100
50
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25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
95
100
50
75
25
0
5
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?
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Y:\HAESE\SA_10-6ed\SA10-6_01\028SA10-6_01.CDR Wednesday, 26 July 2006 10:24:02 AM PETERDELL
SA_10-6
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
29
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?
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?
6
Example 24
Self Tutor
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
Coal
Oil
Amount invested (\$)
x
(x + 3000)
Interest
10%
12%
Total
Earnings (\$)
10% of x
12% of (x + 3000)
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
550
) x=
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.
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0
5
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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Y:\HAESE\SA_10-6ed\SA10-6_01\029SA10-6_01.CDR Wednesday, 26 July 2006 10:25:17 AM PETERDELL
SA_10-6
30
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
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?
H
MOTION PROBLEMS (EXTENSION)
Motion problems are problems concerned with speed, distance travelled and time taken.
kilometres
:
hours
Recall that speed is measured in kilometres per hour, i.e.,
speed =
So,
distance
,
time
distance = speed £ time
and
Example 25
time =
distance
:
speed
Self Tutor
A car travels for 2 hours at a certain speed and then 3 hours more at a speed 10 km/h
faster than this. If the entire distance travelled is 455 km, find the car’s speed in the
first two hours of travel.
Let the speed in the first 2 hours be s km/h. Draw up a table.
Speed (km/h)
s
(s + 10)
First section
Second section
Time (h)
2
3
Total
Distance (km)
2s
3(s + 10)
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.
EXERCISE 1H
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.
9
If his round trip takes 10
of 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?
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95
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5
95
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5
95
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5
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?
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Y:\HAESE\SA_10-6ed\SA10-6_01\030SA10-6_01.CDR Wednesday, 26 July 2006 10:25:59 AM PETERDELL
SA_10-6
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
31
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?
EXTENSION
Click on the icon to access a set of problems called Mixture Problems.
PRINTABLE
PAGES
OPENING PROBLEM
0
Revisit the Opening Problem on page 12. Answer the questions posed.
REVIEW SET 1A
1 Write in algebraic form:
a “3 more than the square of x”
“the square of the sum of 3 and x”
b
2 Write, in words, the meaning of:
p
p
a+3
b
a+3
a
3 If p = 1, q = ¡2 and r = ¡3, find the value of
4 Solve for x:
a
5 ¡ 2x = 3x + 4
4q ¡ p
:
r
1
x ¡ 1 2 ¡ 3x
¡
=
2
7
3
b
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.
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5
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7 A drinks stall sells small, medium and large cups of fruit drink for \$1:50, \$2 and
\$2:50 respectively. In one morning three times as many medium cups were sold as
small cups, and the number of large cups sold was 140 less than the number of
medium cups. If the total of the drink sales was \$1360, how many of each size cup
was sold?
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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
SA_10-6
32
ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1)
8 Ray drives between towns A and B at an
average speed of 75 km/h. Mahmoud
drives the same distance at 80 km/h and
takes 10 minutes less. What is the distance
between A and B?
REVIEW SET 1B
1 Write, in words, the meaning of:
b
a+b
b a+
a
2
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 of
2b + 3c
.
2a
4 Solve the following inequation and graph the solution set: 5x + 2(3 ¡ x) < 8 ¡ x
5 Solve for x:
2(x ¡ 3) + 4 = 5
a
2x + 3 3x + 1
¡
=2
3
4
b
6 What number must be added to the numerator and denominator of
with 13 ?
3
4
in order to finish
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?
cyan
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95
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5
95
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5
95
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25
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5
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 travelled is 90 km, find Carlos’ speed in the first
two hours of travel.
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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
SA_10-6
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