Download When there are only two fraction terms it is easiest to cross multiply

Unit 1. Algebraic Reasoning.
Part 1.
• Alternative methods for solving equations with one variable.
• Solve equations with fractional coefficients
• Solve absolute value equations
• Solve inequalities
• Graph inequalities in one variable
• Transpose formulae.
• Solve mixture, motion, age, number and work word problems
• Solve word problems that involves inequalities
• Simplify expressions with zero and negative exponents
• Use scientific notation
Be aware of alternative methods for solving an equation.
2m
48
5
2m
48
5
2m
48
5
4 on both sides
5 on both sides
2 on both sides
2m  4
5
2m  20  40
m 24
5
5 on both sides
20 on both sides
2 on both sides
2m  20
2m  20
2 on both sides
2 on both sides
m 2
5
5 on both sides
m  10
m  10
m  10
2(2m  8)  6  14
2(2m  8)  6  14
2(2m  8)  6  14
6 on both sides
6 on both sides
2 on both sides
2(2m  8)  8
2(2m  8)  8
2m  8  3  7
2 on both sides
distribute
simplify
2m  8  4
4m  16  8
2m  11  7
8 on both sides
16 on both sides
11 on both sides
2m  4
4m  8
2m  4
2 on both sides
4 on both sides
2 on both sides
m  2
m  2
m  2
5n  4  3n  8
5n  4  3n  8
Subtract 3n on both sides
Subtract 5n on both sides
2n  4  8
4  2n  8
Subtract 4 on both sides
Add 8 on both sides
Divide by 2 on both sides.
Divide by  2 on both sides.
2n  12
n  6
12  2n
6  n
1  5x  11  7x
1  5x  11  7x
Add 7x on both sides
Add 5x on both sides
1  2x  11
1  11  2x
Subtract 1 on both sides
Subtract 11 on both sides
Divide by 2 on both sides.
Divide by  2 on both sides.
2x  10
x 5
10  2x
x 5
Solving Equations With Fractions:
3y 7

4 5
3
4

2w 7
When there are only two fraction terms it is easiest to cross multiply.
5  3y  4  7
15y  28
7  3  4  2w
21  8w
28
21
y
w
15
8
check your answer with your calculator
3y 7 2y
 
4 6 3
When there are more than two terms including some fractions you should multiply
by the lowest common denominator to get rid of the fractions
3y 12 7 12 2y 12
   

4 1 6 1
3 1
9y  14  8y
y  14
check your answer with your calculator
3m
7m
3
5
2
Make sure you multiply all terms by the LCD
3m 10
7m 10
  3  10 

5
1
2
1
6m  30  35m
30  29m
30
m
29
check your answer with your calculator
3 7 2
5
 

4x 8 3x 12
3 24x 7 24x 2 24x 5 24x

 
 
 
4x
1
8
1
3x
1
12 1
18  21x  16  10x
2  31x
2
x
31
check your answer with your calculator
1
3
5

x2 4
1
4(x  2) 3 4(x  2)
5  4(x  2) 

 
x2
1
4
1
As you get more experience you should be able to do the previous step mentally
20(x  2)  4  3(x  2)
20x  40  4  3x  6
17x  30
30
x
17
check your answer with your calculator
Special Cases:
1
4
1


2x  6 5x  15 6
You need to recognise that there is a factoring step.
1
4
1


2  x  3 5(x  3) 6
15  24  5(x  3)
9  5x  15
24  5x
24
 x
5
3
2
1


x 1 1  x 2
You need to recognise that 1-x is the negative of x-1.
3
2
1


x 1 x 1 2
This is a case where it is easier if you don’t multiply to get rid of
the fraction first but it doesn’t matter if you do.
1
1

x 1 2
x 1  2
x 3
1 5  2x

3 6x  15
6x  15  15  6x
Note: The first step would be different if you realised that you could have done some
factoring of 6x-15
12x  30
30 5
x

12 2
Check with your calculator.
You should find that there is no
a denominator of zero.
valid solution because the algebraic solution gives
1
3
1


x  2 4(x  2) 4(x  2)
4  3 1
11
This is always true so the normal answer would be to say
X = all real numbers
But in this case there is an invalid solution.
Therefore x = all
real numbers except 2
See page 306 in the text book for this type of question.
Absolute Value:
The formal definition is “The distance between the number and zero on a number line”.
The common understanding is “the positive value or size of the number”
the symbol is x
Examples:
3  3
17 17
23  23
1234  1234
Equations with absolute values
if x  3 what are the possible values of x? x  3
x 1  5
x  1  5
x  1  5
x  4 or  6
2x  1  7
2x 1  7
1 7
x
2
x  4 or  3
2x 1  7
not possible
absolute value
can't be negative
3  2 3x  1  5
6  3 5  2y  3
2 3x  1  2
3 5  2y  9
3x  1  1
5  2y  3
3x 1  1
5  2y  3
3x  1 1
x  11
3
x  0 or  2
3
2y  5  3
y  53
2
y  1 or  4
4 5m  3  5  2
4 5m  3  3
no solution as absolute
value can't be negative
Manipulating Inequalities: see page 462 of the text book.
If you understand the logic of the manipulations using arithmetic you can apply the same
logic to algebraic inequalities.
Adding
6  11
+5 on both sides
11  16
Is it still true?
Yes. Therefore If you add
something on both sides of
an inequality it will still be
true.
x27
+2 on both sides
x9
Subtracting
16  9
2 on both sides
14  7
Is it still true?
Yes. Therefore If you
subtract on both sides of an
inequality it will still be
true.
x  5  14
5 on both sides
x9
Multiplying
by a positive number
25
4 on both sides
8  20
Is it still true?
Yes. Therefore If you multiply by
a positive number on both sides
of an inequality it will still be
true.
m 7
2
2 on both sides
m  14
Dividing
by a positive number
12  8
4 on both sides
32
Is it still true?
Yes. Therefore If you divide by a
positive number on both sides of
an inequality it will still be true.
4x  20
4 on both sides
x5
Multiplying
by a negative number
25
  4 on both sides
8  20
Is it still true?
No. Therefore If you multiply by
a negative number on both sides
of an inequality it will only be
true if you reverse the inequality.
m 7
2
  2 on both sides
m  14
Dividing
by a negative number
12  8
  4 on both sides
3  2
Is it still true?
No. Therefore If you divide by a
negative number on both sides of
an inequality it will only be true
if you reverse the inequality.
4x  20
  4 on both sides
x  5
5 m  7
2
3x  2  13
m2
2
3x  15
5 on both sides
2 on both sides
  2 on both sides
m  4
3 on both sides
x5
y
 4 1
2
4 on both sides
y
5
2
2 on both sides
y  10
7  3x  19
7 on both sides
 3x  12
  3 on both sides
x  4
Graphing Inequalities in One Variable – see page 464 in the textbook.
x4
the value 4 is not included in the inequality so there is an open circle at 4.
x2
the value 2 is included in the inequality so there is a solid circle at 2.
3  x  4 this is called a conjunction
x  3 or x  4 this is called a disjunction
Transposing Formulae or Equations:
Transpose the following to make “t” the subject. This follows
the same logic as solving equations.
tx w m


v xy 4y
multiply by 4xyv
x
3y
 3  transpose for x
2
4
multiply by 4
4tx 2 y  4vw  mvx
2x  12  3y
4tx y  mvx  4vw
2x  3y  12
2
mvx  4vw
t
4x 2 y
x  3y6
2
Mixture Problems: see page 324 and supplementary material.
How much 42% acid solution needs to be added to a 56% acid solution to make
21L of 48% acid solution.
The structure for these questions is based on the idea that the total amount
of acid before and after mixing must stay the same.
The amount of acid is given by amount = % times volume
let V  volume of 42% acid
amount of acid  amount of acid  amount of acid
0.42  V  0.56(21-V)  0.48  21
42  V  56(21-V)  48  21
42V  1176  56V  1008
14V  168
V  12
12L 0f 42% acid is needed.
Check your answer using the original language, not your equation.
Work Problems: see page 329 and supplementary material.
It takes Tom 45 minutes to mow the lawn while Harry can do it in 30 minutes.
How long will it take if they do the job together
The structure for these questions is based on the idea that the sum of the
fraction of the work done by each person has to equal the fraction done
together.
Let T  the time they take to do the job together
Tom  Harry  Together
1
1 1
multiply by 90T


45 30 T
2T  3T  90
5T  90
T  18
It takes them 18 minutes to do the job together
Consecutive Number Questions: see page 307 and supplementary material.
The sum of the three consecutive odd integers is five less than four time the first integer.
Let I  the first integer.
Therefore I  2 and I  4 are the next consecutive odd integers
I  (I  2)  (I  4)  4I  5
3I  6  4I  5
11  I
The three consecutive odd integers are 11,13,15
Check your answer with the words not with your equation.
Age Questions: see the supplementary material.
Sue’s age now is twice Anne’s age. Five years ago Sue’s age was three times Anne’s age.
What is Sue’s age now?
Let A  Anne's age now.
Therefore Sue's age now  2A.
Anne's age five years ago  A  5.
Sue's age five years ago  2A - 5
Five years ago
Sue  3  Anne
2A  5  3(A  5)
2A  5  3A  15
10  A
Sue  2A  2 10  20
Sue's age now is 20
Check your answer with the words not with your equation.
Motion Questions: see the supplementary material.
Tim made a round trip between two cities that were 450km apart. However his average
speed on the outward trip was 5 times more than on return. If his time for the return trip
was 10 hour more than for the outward trip, find his average speed for the outward trip.
distance
The basic relationship is speed  distance
so
time

time
speed
the return trip was slower so let S  average speed on the return trip.
Time Return - Time Outwards  10
dis tan ce dis tan ce

 10
speed
speed
450  450  10
S
5S
450  90  10
S
S
360
S
simplify is better in this case
 10
S  36
Average speed on the outwards trip is 180km/hr
Motion Questions: with quadratics in solutions.
Tim made a round trip between two cities that were 210km apart. However his average
speed on the return trip was 5 km/hr more than on the way out ? If his time for the return
trip was 1 hour less than for the outward trip. Find his average speed for the outward
trip.
the trip out was slower so let S  speed on way out.
Time Outwards - Time Return  1.
dis tan ce dis tan ce

1
speed
speed
210 210

1
S
S5
multiply by S(S  5)
210(S  5)  210S  S(S  5)
210S  1050  210S  S2  5S
0  S2  5S  1050
0  (S  35)(S  30)
S  35 or 30
Average speed on the outwards trip is 30km/hr
Inequality Questions: Define your variable, write an equation then solve the equation.
The sum of two consecutive numbers is at most 25 more than half the first number.
Let n  the first number
n  (n  1)  n  25
2
2n  1  n  25
2
4n  2  n  50
3n  48
n  16
check by substituting
appropariate values
see page 472 in the text book
How many grams of copper must be alloyed with 387g of pure silver to produce and
alloy that is no more than 90% pure copper?
Let c  the amount of copper
c  0.1
c387
c  0.1c  38.7
0.9c  38.7
c  43
check by using the original words not the equation you wrote.
see page 473  474 in the text book
Zero Exponents.
A number or variable with an exponent of zero has a value of one.
5 1
0
x0  1
 2x 
0
1
In the following examples only the part with the zero exponent equals 1.
2m
 2 m
0
3  2m 
 3   2m 
0
 2 1  2
0
0
 3 1  3
3y  4(xy  456x )  3  y  4  (xy  456x )
0
2
9 0
0
 3 1  4 1  3  4  7
2
9 0
Negative Exponents.
A negative exponent is the same as the reciprocal of the positive exponent.
a
n
 1n
a
In the following examples only the parts with the negative exponent become the
reciprocal of the positive exponent.
 12
3
2
Evaluate 3
Simplify mn
2

mn
 19
2
 m  12
n
m
 2
n
Simplify
0 2
 1
1
y2
3 5

1
x3

1
y5

1
x3y5
2 3

x2
1

1
y3

x2
y3

1
x3

1
y5

1
x3y5
1. x y
2. x y
3. x y
x 3
4. 5
y
15x 5 y 5
5.
10x 3 y 7


3x3
2y12
4m 5n 3
m 2m 7
6.

3
12n 4 m 7
14a 4 b 7
7.
2a 3b 2
 79
ab
1
y2
Scientific Notation:
Scientific notation has the form c  10n
c is a number with one digit before the decimal point
n is an integer showing the exponent of ten you get from the product form.
32000  3.2  10000  3.2  10
4
0.00032  3.2  10000  3.2  104
Changing to scientific notation:
• Put the decimal point after the first digit
• Count the number or places you have to
move the decimal point to get to where it
started. That tells you the exponent of ten.
.
43500000
4 3500000 10000000
7
4.35  10
Scientific notation with small numbers:
Changing to scientific notation:
•
Put the decimal point after the first digit
•
Count the number or places you have to
move the decimal point to get to where it
started. That tells you the exponent of ten.
Examples.
6
4.5

10
4500000 
3
6700  6.7  10
4
7.8

10
0.00078 
2
1.23

10
0.0123 
0.000678
.
6 78  10000
6.78  10
4
Changing to standard form:
1. The exponent of ten tells you how many
places to move the decimal point.
2.7  105
2.7 100000
2. Count the number of places and add
zeros if needed. Then put the decimal
point if required.
Examples.
3.65  10  36500
7
7.5  10  75000000
4
3.6  10  0.00036
7
9  10  0.0000009
4
270000
Multiplying with scientific notation:
4.5  10  5  10
3
6 Multiply the numbers normally and think
about how many places the exponents will give.
103  106 is 1000  1000000 which is 109
What is the short cut?
 22.5  10
9
Add the exponents.
Make sure the answer is written in scientific notation
with only one digit before the decimal point.
22.5  109 move nine places so there are 8 zeros
If we make it 2.25 there will be 10 places
 2.25  10
10
4.5  10
6
5  10
3
Divide the numbers and exponents normally and then
check that the answer is in scientific notation.
DO NOT CHANGE INTO STANDARD FORM.
 0.9  103
 9  10
4
3
1.5  10  6  10
6
5
2  10  2.5  10
9
9  10

11
5  10
6
 1.8  10
17