Download Week 2

Introduction to Algebra
On completion of this module you will be able to:
• understand what algebra is
• identify and define a variable in a simple
word problem
• perform some basic mathematical operations
with variables
• work with algebraic fractions
• understand exponents and radicals involving
variables
• do some simple factoring
1
What is algebra?
Algebra:
• is a branch of maths where symbols are
used to represent unknown quantities
• is a symbolic language designed to solve
problems more easily
• involves manipulating expressions involving
symbols and numbers
• usually uses letters, called variables, to
represent unknown quantities
• uses +, -, ,  etc (just as last week)
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Example
Imagine Allan needs to save $1000 to buy a
new stereo.
He wants to purchase it in six weeks time
and is going to save an equal amount of
money each week.
Let’s name the amount he saves each week
using a variable, x.
Then he will save $x each week, so
6  weekly savings  1000
3
If we write this using the variable then
6  x  1000
Be careful that your ‘x’ looks different from
your ‘’.
The ‘’ is implied if we write:
6x  1000
This is an algebraic equation that summarises
the word problem.
4
Next we want to determine the value of the
variable that makes this expression true.
6x  1000
Divide both sides of the equations by 6 cancels
out the 6 in front of the x.
6x 1000

6
6
x  $166.67 ( to the nearest cent)
5
So Allan needs to save $166.67 each week for
the next six weeks in order to have enough
money for the stereo.
• QMA this week will focus on identifying
variables and manipulating algebraic
expressions.
• In Week 3 we will look more at solving
equations using algebra.
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Identifying the variable
• The first step in solving a word problem is to
identify the unknown (variable).
• Next choose a letter to represent this
variable – often x is used, but many times we
use something more meaningful: t for time,
p for price, q for quantity, r for (interest) rate
etc.
• We’ll look at a couple of examples of
identifying a variable.
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Example
Rhonda owns a small health food shop.
If she hires another employee to serve behind
the counter, it will cost her $18 per hour in
wages.
Based on her current sales revenue she can
afford to pay an extra $250 per week in
wages.
For how many hours each week can she
afford to employ an extra person?
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Solution
The clue to finding the variable is in the
question of the last sentence “how many
hours…”.
There are many ways we could express our
variable but one example is:
“Let x be the number of hours per week that
Rhonda can afford to employ an extra
person.”
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Example
Matthew received 45% on his first science test
for the year.
Because he was concerned about that result,
he studied hard and received 65% and 82%
respectively on the next two tests.
There is one final test for the year.
Matthew’s final grade will be awarded based
on the average mark for the four tests.
If he wants to receive at least 65% overall,
what is the minimum mark he must receive on
the final test to reach his goal?
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Solution
The clue is again in the last sentence “what is
the minimum mark he must receive on the
final tests to reach his goal?”.
“Let m be the minimum mark Matthew must
receive on the final test in order to attain 65%
overall.”
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Example
John rides his bike from his house to work each
day.
If he travels an average speed of 32km per hour
and it takes him 20 minutes to get to work, how
far does he travel?
Solution
Let d be the distance from John’s house to his
work.
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Example
1. Write an algebraic expression to represent
the following ideas:
a) Twice a number.
b) A number minus five.
c) Half a number plus one.
d) The area of a rectangle is the length
times the width.
Solutions
n
a) 2n
c)  1
d) A  lw
b) n-5
2
13
Example
2. Write a phrase to describe the following
expressions:
a) n  2
b) n
4
c) 4n  5
Solutions
a) A number plus two.
b) A quarter of a number or a number divided
by 4.
c) Four times a number minus five.
14
Evaluation
• Evaluation (or substitution) means
evaluating an expression at a certain value
of a variable.
Example
1. Evaluate A = lw when l = 10 and w = 5.
A  lw  10  5  50
2. Substitute a = 3, b = 2 and c = 6 into b2  4ac
b2  4ac  22  4 36   4  72  68
15
Algebraic Operations
• With algebra we can add or subtract ‘like
terms’.
• For example, 3t and 2t are ‘like terms’
because they both contain t ’s.
• We add like terms by adding the coefficients.
• So with 3t +2t, the 3 and 2 are coefficients
and 3t  2t  3  2  t  5t
• Similarly
6y  2y  4y
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Example
1. 4r  r  5r
(The coefficient of r is 1)
2. 2p  p  5p  2  1  5 p  6 p
3. 2a  3a  6  2  3 a  6  5a  6
The 6 is not
a like term
4. 2a  3b  4a  2  4  a  3b  6a  3b
The a and b
terms are not
like terms
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Examples
Collect like terms and simplify: 4x  5.4x  6x  7y
4 x  5.4 x  6 x  7y   4  5.4  6  x  7y
 3.4x  7y
Simplify 2x 2  5x  9x 2  x  5  8
 2 x 2  9 x 2  5x  x  5  8
 11x 2  4x  3
First group like
terms together so it
is easier to add and
subtract
18
Example
Simplify 8x  7xy  x  9xy  3
8x  7xy  x  9xy  3  8x  x  7xy  9xy  3
 7x  2xy  3
19
• Recall the rules for multiplying and
dividing negative numbers (see Week 1).
• The same rules apply to variables.
Example
2  y   2y
2  t   2t
 8 x  8x
  x   y   xy
3y
 3 (the y's cancel)
y
x
x 
1 

or  x 

6
6 
6 
2t
2t 
2 

or  t 

7
7 
7 
6 xy 6 xy

 3xy
2
2
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Expanding Brackets
•The same rules apply to variables as did
with numbers.
Examples
2  t  1  2  t  2  1  2t  2
3  x  7    3 x    37   3  x   3 7 
 3x  21
2 3x  1  2 3x   2 1  6 x  2
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Examples
 x  3  x  2  x?22  2x  3x  6
 x 2  5x  6
x  y
2
 x  yx  y
 x x  y  y x  y
 x 2  xy  yx  y 2
 x 2  xy  xy  y 2
 x 2  2 xy  y 2
22
Examples
 x  1  x  5  x  x  5  1  x  5
 x 2  5x  x  5
 x 2  4x  5
 t  3  t  7   t  t  7   3  t  7 
 t 2  7t  3t  21
 t 2  4t  21
23
The following identities are important:
2
2
1. (x  y )  x  2 xy  y
2
2. (x  y )2  x 2  2 xy  y 2
2
3. ( x  y )(x  y )  x  y
2
Expand the left hand sides to show how the
left hand sides are equal to the right hand
sides…
24
Fractions
Adding and subtracting fractions
• To add or subtract fractions we convert to
equivalent fractions with the same
denominator (as with number fractions).
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Example
2x x 8x
 
4
7
2
Multiplying the denominators together gives:
4  7  2  56 (or use LCD=28)
2x x 8x 2x 7 2   x  4 2   8x  4 7 
 

4
7
2
 472
28x  8x  224x

56
Since both
188x
47


x
numerator and
56
14
denominator divide
by 4.
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Example
6
2

x x 1
The common denominator is found by multiplying
denominators together:
6  x  1  2x
6
2


x x 1
x  x  1
6 x  6  2x

x2  x
4x  6
 2
x x
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Multiplying and dividing fractions
• To multiply fractions multiply numerators
together and multiply denominators
together (as in Week 1).
x  x2
 x   1  x  x 1  x 
 5   x  4   5 x  4  5x  20


 

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• To divide fractions, multiply by the reciprocal.
5 y  x 
4x
3x
4x



x  y 5 y  x  x  y
3x
4x  5  y  x 


 x  y  3x
20x  y  x 

3x  x  y 
20  y  x  Since the x ’s

cancel.
3x  y
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Exponents and Radicals
• As before, y 4 means y  y  y  y
• Any variable raised to the power of zero
always equals one, so
x 0  1, t 0  1 etc.
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Multiplying Powers
A more general form of the rules from Week 1
is given here.
If you are:
1. multiplying bases with the same exponent,
then multiply the bases and raise them to this
exponent. m m
m
a b
  ab 
2. multiplying the same base with different
exponents, then add the exponents.
am  an  am n
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3. Raising a number to an exponent and then
to another exponent, then multiply the
exponents.
n
m
a
 amn
 
Another useful rule for dividing numbers with
the same base but different exponents –
subtract the exponents:
am
m n

a
an
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Taking the root of a number (fractional exponents):
The square root of a number is the reverse of
squaring.
(x)2 
2
2

x

x

x

( x)2 
Another way of writing roots is using fractional
powers:
3
x x
1
3
11
t t
1
11
33
Negative exponents
A power with a negative exponent can be
rewritten as one over the same number with a
positive exponent:
x
2
1
 2
x
2t
7
1
2
 2 7  7
t
t
2
1
y
x 3y 2  3 y 2  3
x
x
34
Examples of negative fractional exponents:
x
t
1
1
2
y


1
x
1
2
1
t
1
y
1


x
1
y
t
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Example
a3bc 2 a3
b
c2
 2  4 
2 4
c
ab c
a
b
1 1
cc
4
 2 3  bb 
c
a a
1
 23  b1 4  c
a
1 5
b5c
 5bc 5
a
a
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Example (alternate solution)
a3bc 2
1  4 
3 2
2 1

a

b

c
a2b4c
 a5  b5  c1
1
 5  b5  c
a
b5c
 5
a
37
Example
Simplify and express your solution in terms
of positive exponents:  x 2y 1z 2
 xy 
2 4
x y z
2 4
 xy 
2
1
2
x 4y 2 z 2

x 4y 8
x 4 y 2
 4  8  z 2
x
y
38
x 4 y 2
 4  8  z 2
x
y
2
8
y y z
y
28
2
1
z2
y 10
 2
z
39
Alternate solution
x y z
2 4
 xy 
2
1
2
x 4y 2 z 2

x 4y 8
 x 4 4y 2 8 z 2
 x 0y10 z 2
y 10
 2
z
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Factoring
• Look for common factor in two or more
expressions
• E.g. in ab + ac, the common factor is a.
• If we divide each term by a we get
ab
a
 b and ac
a
c .
• We can therefore rewrite ab + ac as a(b + c).
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Examples
1. 2 x  2  2  x  1
2. 5x  10  5  x  2 
3. 2abc  ab  abc  ab 2c  1  c 
 ab 3c  1
4. x 2  3x  x  x  3
42
Putting it all together
Examples
2 xy  4y
1. Simplify
.
4 xy  8y
4 xy  8y
2. Simplify
.
2y  4xy
3. Simplify and express as a single fraction:
 2st s  6u
 3  q  q .


43
Solutions
2 xy  4y
1. Simplify
.
4 xy  8y
2xy  4y 2y  x  2  2 y  x  2 
2 1


 
4xy  8y 4y  x  2  4 y  x  2 
4 2
4 xy  8y
2. Simplify
.
2y  4xy
4y  x  2 
2  x  2
4xy  8y


2y  4xy 2y 1  2x  1  2x 
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3. Simplify and express as a single fraction:
 2st s  6u
 3  q  q .


 2st s  6u  2st q  6u
 3  q   q   3  s   q




2qst 6u 2qt 6u




3s
q
3
q
12qtu 4tu


 4tu
3q
1
45
Examples
1. Expand the brackets and collect like terms:
x
2

 1  x  2 .
2
3
t
 6t
2. Simplify
.
3t
46
Solutions
1. Expand the brackets and collect like terms:


x 2  1  x  2 .


x 2  1  x  2  x 2  x  2  1  x  2
 x 3  2x 2  x  2
3t 2  6t
2. Simplify
.
3t
3t 2  6t 3t  t  2 

 t 2
3t
3t
47
Factoring Trinomials (Optional)
• This section will help with solving quadratic
equations in later weeks, but is not essential
(you will not lose any marks for not using this
technique).
• If you are familiar with factoring trinomials
(from school) then it would be wise to work
through this section.
• If factoring trinomials is new to you and you
are interested, then work through this section.
48