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ALGEBRA
ALGEBRA
1. Identify terms which are common 2. Put common terms together
1. Multiply the term outside the brackets by everything inside the brackets
2. Simplify
EXPANDING BRACKETS p 5SIMPLIFYING ALGEBRAIC EQUATIONS p 3
1. Multiply the First two terms of each bracket
2. Multiply the Outside two terms of each bracket
3. Multiply the Inside two terms of each bracket
4. Multiply the Last two terms of each bracket
5. Write each answer together6. Simplify (usually by adding the two
middle terms)
EXPANDING TWO BRACKETS: FOIL p 7
1. Simplify the number fraction
2. To simplify the x values, subtract the smaller power from the larger power and remove the x value with the small power.
SIMPLIFYING ALGEBRAIC FRACTIONS p 9
1. Write out a set of brackets2. Identify the greatest common
number in each term. Put this outside the brackets
3. Identify any common letters in each term. Put these outside the brackets
4. Put the original term into the brackets with the common factors cancelled out (powers cancel too)
5. Simplify (anything completely crossed out will have a 1 in its place)
FACTORISING -COMMON FACTOR p 11
1. Look for all of the combinations of number that multiply to make the last number
2. See which of these combinations can add to make the middle number
3. Write out two sets of brackets with an x at the start of each one
4. Put your two numbers in the brackets with the x
FACTORISING - QUADRATICS p 13
1. Factorise the top2. Factorise the bottom3. Cancel the common factors
SIMPLIFYING FRACTIONS WITH QUADRATICS p 15
THE SKILLS YOU NEED TO KNOW:
APPLY ALGEBRAIC PROCEDURES IN SOLVING EQUATIONS4 CREDITS (91027)
PAGE 2
1. Make the equation = 02. Factorise3. Make each bracket = 04. Solve for x (twice)
SOLVING QUADRATICS p 19
1. Factorise top and bottom2. Cancel common factors3. Multiply both sides by the bottom4. Solve as a linear equation
SOLVING FRACTIONS WITH QUADRATICS p 21
1. Simplify with the following rules:• If the base is the same add the
powers. e.g. x x xm n m n= +
• If the exponent is raised to a power multiply the exponent and power. e.g. ( )x xm n m n= ×
2. Trial with 23. Trial with 3 Etc.4. If the power is even numbered you
have both +ve and -ve answers
SOLVING EXPONENTS p 23
1. Multiply away any fractions2. Expand any brackets3. Put x’s on the same side4. Put numbers on the opposite side5. Simplify6. Divide by the last number
(if negative reverse the sign)
SOLVING INEQUALITIES p 25
1. Write out the two equations if they haven’t been given
2. Rearrange one equation so that a variable is the subject
3. Substitute into the other equation and solve for one variable
4. Substitute answer into rearranged equation and solve for other variable
SIMULTANEOUS EQUATIONS p 32
1. Put the numbers in place of the letters
2. Simplify or solve the term
SUBSTITUTION p 28
1. Rearrange to make sure that the subject is a numerator not a denominator
2. Get rid of all other symbols/numbers around the subject
REARRANGING p 28
1. Multiply away any fractions2. Expand any brackets3. Put x’s on the same side4. Put numbers on the opposite side5. Simplify6. Divide by the last number
SOLVING LINEAR EQUATIONS p 17
Note: Problems may rely on knowledge from earlier algebra sections
1. Give each variable a letter, the letter will ALWAYS represent a number.2. If calculations are needed, put the values from the question into the equations3. There is now enough info. to either solve the problem or explain your answer
WORD QUESTIONS p 35
PAGE 3
ALGEBRA
SIMPLIFYING ALGEBRAIC EQUATIONS
OLD NCEA QUESTIONSSimplify the following equations:
1. 4 32 2 2ab a b a b+ − 2. 3 22 2 2xy x y xy+ −
3. 3 22 2 2ab a b ab+ −
STEPS: EXAMPLE: 9 4 22 2 2 2x y xy x y xy x− + + −1. Identify terms
which are common 9 22 2x y x y x− + + −4xy xy2 2
2. Put common terms together 11 2x y x− −3xy2
For a complete tutorial on this topic visit www.learncoach.co.nz
PRACTICE QUESTIONSSimplify the following equations:
4. 5 3 6 8x x+ − + 5. 6 3 9 11a b a b− − +
6. 3 8 19 15w e w e− − + 7. 7 9 11a a− +
8. 3 12 11 66x x− − − 9. 4 3 5 62 2x y x y+ + + −
10. 2 4 3 6a b c b a+ − − − 11. 4 2 4 7 62 2xy y xy− + − +
12. − + + − +3 6 12 3 53 2 2 3a a a a 13. 5 6 4 3 6b c a b c a+ − + + −
PAGE 4
ANSWERSNCEA1. 4
4 2
2
2 2
abab a b
+ −+3a b a b2 2
(Achieved) h
2. 3 22 2
2
2 2
xy xy2 2+ −+x y
xy x y
(Achieved) h
3. 3ab ab2 2+ −+
22 2
2
2 2
a bab a b
(Achieved) h
PRACTICE4. 5x 8x+ − +
−3 6
13 3x
(Achieved) h
5. 6 93 8a aa b− − +
− +3b 11b
(Achieved) h
6. 3 1916 7w ww e− − +
− +8e 15e
(Achieved) h
7. 7a 9a− +− +
112 11a
(Achieved) h
8. 3x 11x− − −− −
12 668 78x
(Achieved) h9. 4x 6x2 2+ + + −
+ +3 5
10 2 52
y yx y
(Achieved) h
10. 2a a+ − − −− −
4 3 62 3b c b
a b c
(Achieved) h
11. 4xy 7xy2 2− + − +− − +
2 4 63 2 102
yxy y
(Achieved) h
12. − + + − ++ +
3a 5a3 36 12 32 3 12
2 2
23a a
a a
(Achieved) h
13. 5b 3b+ − + + −+ −
6 4 68 12 5
c a c ab c a
(Achieved) h
Study Tip:
Quality over QuantityGreat news! It is the quality of study that makes more of an impact than the quantity.Interesting fact: the amount of time students spend on home-work has risen over the last 3 decades – but the level of educa-tional attainment has not risen with the increased workload. Work Smarter, not harder.
PAGE 5
ALGEBRA
EXPANDING BRACKETSSTEPS: EXAMPLE: 5 2x x( )−
1. Multiply the term outside the brackets by everything inside the brackets
5 2 5x x x× − ×
2. Simplify 10 5 2x x−For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONSExpand the following equations:
1. 2 3x x( )− 2. 2 7( )x −
3. 2 2( )x − 4. 5 3( )x −
5. 2 1 3 2( ) ( )x x− − + 6. 3 4 2 5( ) ( )x x+ − +
PRACTICE QUESTIONSExpand the following equations:
7. 4 3 2 2( ) ( )y x+ + − 8. 6 11 4 3( ) ( )x x+ − +
9. 8 9( )x + 10. 3 15x x( )−
11. 4 12( )t + 12. 6 7 5( )− x
13. 7 12( )− n 14. x x x( ) ( )4 3 2 5− + −
15. 2 3 2 4( ) ( )x x− + − 16. 3 9 4x x( )−
PAGE 6
ANSWERSNCEA1. 2 3
2 3 26 2 2
x xx x xx x
( )−× − ×−
(Achieved) h
2. 2 72 2 72 14
( )xx
x
−× + ×−−
(Achieved) h
3. 2 22 2 22 4
( )xx
x
−× + ×−−
(Achieved) h
4. 5 35 5 35 15
( )xx
x
−× + ×−−
(Achieved) h
5. 2 1 3 22 2 1 3 3 22 2 3 6
8
( ) ( )x xx x
x xx
− − +× + ×− − × − ×− − −
− −
(Achieved) h
6. 3 4 2 53 3 4 2 2 53 12 2 10
2
( ) ( )x xx x
x xx
+ − +× + × − × − ×+ − −+
(Achieved) h
PRACTICE7. 4 3 2 2
4 4 3 2 2 24 12 2 44 2 8
( ) ( )y xy x
y xy x
+ + −× + × + × + ×−+ + −+ +
(Achieved) h
8. 6 11 4 36 6 11 4 4 36 66 4 122 54
( ) ( )x xx x
x xx
+ − +× + × − × − ×+ − −+
(Achieved) h
9. 8 98 8 98 72
( )xx
x
+× + ×+
(Achieved) h
10. 3 153 3 153 452
x xx x xx x
( )−× + ×−−
(Achieved) h
11. 4 124 4 124 48
( )tt
t
+× + ×+
(Achieved) h
12. 6 7 56 7 6 542 30
( )−× + ×−−
xx
x
(Achieved) h
13. 7 127 12 784 7
( )−× + ×−−
nn
n
(Achieved) h
14. x x xx x x x
x x xx x
( ) ( )4 3 2 54 3 2 3 5
4 6 1510 15
2
2
− + −× + ×− + × + ×−− + −− −
(Achieved) h
15. 2 3 2 42 2 3 2 4 22 6 8 22
( ) ( )x xx x
x x
− + −× + ×− + × + ×−− + −
(Achieved) h
16. 3 9 43 9 3 427 12 2
x xx x x
x x
( )−× + ×−−
(Achieved) h
PAGE 7
ALGEBRA
EXPANDING TWO BRACKETS
STEPS: FOIL EXAMPLE: ( )( )3 4 2x x− + 1. Multiply the First two terms of
each bracket( )( )3 3 2x x x− + ⇒4 2
2. Multiply the Outside two terms of each bracket
( )( )3 2 6x x− + ⇒4 x3. Multiply the Inside two terms of
each bracket( )( )3 2x − + ⇒ −4 4x x
4. Multiply the Last two terms of each bracket
( )( )3x x− + ⇒4 2 8
5. Write each answer together 3 6 4 82x x x+ − −6. Simplify (usually by adding the two
middle terms) 3 2 82x x+ −For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONSExpand the following equations:
1. ( )( )2 5 2x x+ − 2. ( )( )2 3 4x x+ −
3. ( )( )x x+ −5 7 4. ( )( )1 2 3− +x x
5. ( )( )2 1 3x x− + 6. ( )( )4 5 2x x− +
PRACTICE QUESTIONSExpand the following equations:
7. ( )( )3 7 9j j+ − 8. ( )4 1 2n +
9. ( )( )4 3 4x x+ − 10. ( )( )x x+ −8 9
11. ( )x + 8 2 12. ( )( )6 7 2x x− −
13. ( )( )2 9 4 7x x+ − 14. ( )( )x x+ −4 4
15. ( )( )3 2 1x x+ + 16. ( )( )x x− +7 3
PAGE 8
ANSWERSNCEA1. ( )( )
( )( )( )( )( )( )
2 2 F2 2 -4 O
5 5 I5 2
2xx xxx x
+ − ⇒+ − ⇒+ − ⇒+ −
5 25
2 22
x xx
x x⇒⇒ −
− + −+ −
10 L2 4 5 102 10
2
2
x x xx x
(Achieved)
2. ( )( )( )( )( )( )( )( )
2 2 F2 4 8 O
3 3 I3 4
2x x xx x
x x
+ − ⇒+ − ⇒ −+ − ⇒+ −
3 43
2 42
xxx x ⇒⇒ −− + −− −
12 L2 8 3 122 5 12
2
2
x x xx x (Achieved)
3. ( )( )( )( )( )( )( )( )
x x xx x
x x
+ − ⇒+ − ⇒ −+ − ⇒+ − ⇒ −
5 75
7
2 F7 7 O
5 5 I5 7 35 L
xxx x
xx x xx x
2
27 5 352 35
− + −− −
(Achieved)
4. ( )( )( )( )( )( )( )( )
1 F1 3 3 O2 2 I2 3
− + ⇒− + ⇒− + ⇒ −− + ⇒ −
2 32
1 31
2
xx x
x
x x
x x xx 66 Lx
x x xx x
+ − −− −
3 2 63 5 2
2
2
(Achieved)
5. ( )( )( )( )( )( )( )( )
2 2 F2 3 6 O
1 I1 3
x x xx x
x x
− + ⇒− + ⇒− + ⇒ −− + ⇒
1 31
2 32
2
xxx x −−+ − −+ −
3 L2 6 32 5 3
2
2
x x xx x
(Achieved)
6. ( )( )( )( )( )( )( )( )
4 4 F4 2 8 O
5 5 I5 2
2x x xx x
x x
− + ⇒− + ⇒− + ⇒ −− +
5 25
4 24
xxx x ⇒⇒ −+ − −+ −
10 L4 8 5 104 3 10
2
2x x xx x
(Achieved)
PRACTICE7. ( )( )
( )( )( )( )( )( )
3 j jj j
j
+ − ⇒+ − ⇒ −+ − ⇒+ − ⇒ −
73 7 933 9
9 273
7 9 637
2j
jj jj 77 j
27 3 63 720 3 63
2
2j j jj j− + −− +
(Achieved)
8. ( )( )( )( )( )( )( )(
4 4 164 1 4
1 4 41 1
2n n nn n
n n
+ + ⇒+ + ⇒+ + ⇒+ +
1 11 4
4 14 4
nnn n ))⇒+ + ++ +
116 4 4 116 8 1
2
2n n nn n
(Achieved)
9. ( )( )( )( )( )( )( )( )
4 44 4 16
3 33 4
2x x xx x
x x
+ − ⇒+ − ⇒ −+ − ⇒+ − ⇒ −
3 43
4 44
xxx x 112
4 16 3 124 13 12
2
2
x x xx x− + −− −
(Achieved)
10. ( )( )( )( ) -( )( )( )( ) -
x x xx x
x x
+ − ⇒+ − ⇒+ − ⇒+ − ⇒−
8 98
9
92
2
9 98 88 9 72
xxx x
x xx xx x
+ −− −
8 72722
(Achieved)
11. ( )( )( )( )( )( )( )( )
x x xx x
x x
+ + ⇒+ + ⇒+ + ⇒+ + ⇒+ +
8 88
8
82
2
8 88 88 8 64
xxx x
x x 88 6416 642
xx x
++ +
(Achieved)
12. ( )( )( )( )( )( )( )( )
6 66 2 12
7 77 2
2x x xx x
x x
− − ⇒− − ⇒ −− − ⇒ −− − ⇒
7 27
6 26
xxx x 114
6 12 7 146 19 14
2
2
x x xx x− − +− +
(Achieved)
13. ( )( )( )( )( )( )( )(
2 4 82 7 14
9 4 369
2x x xx x
x x
+ − ⇒+ − ⇒ −+ − ⇒+
9 79 4
2 72 4
xxx x −− ⇒ −− + −+ −
7 63)8 14 36 638 22 63
2
2
x x xx x
(Achieved)
14. ( )( )( )( )( )( )( )( )
x x xx x
x x
+ − ⇒+ − ⇒ −+ − ⇒+ − ⇒ −−
4 44
4
42
2
4 44 44 4 16
xxx x
x xx xx
+ −−
4 16162
(Achieved)
15. ( )( )( )( )( )( )( )( )
3 33 1 3
2 22 1 2
2x x xx x
x x
+ + ⇒+ + ⇒+ + ⇒+ + ⇒
2 12
3 13
3
xxx x
x22
2
3 2 23 5 2
+ + ++ +
x xx x
(Achieved)
16. ( )( )( )( )( )( )( )( )
x xx x x
x x
− + ⇒− + ⇒− + ⇒ −− + ⇒ −+
77 3
33
3 3
7 3 217 7
2x
x xxx x22
221 7
4 21− −
− −x
x x (Achieved)
PAGE 9
ALGEBRA
SIMPLIFYING ALGEBRAIC FRACTIONS
STEPS: EXAMPLE: 82
2
5
xx
1. Simplify the number fraction4 2
5
xx
2. To simplify the x values, subtract the smaller power from the larger power and remove the x value with the small power.
4 42
35 2
xx x−
=
For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONSSimplify the following fractions:
1. 84
2xx
2. 12
3
2aa
3. 10
2
2aa
4. 9
12
5
3
xx
PRACTICE QUESTIONSSimplify the following fractions:
5. 47
7
3
aa
6. 11 9
4
xx
7. 124
8
3
xx
8. 36
6
5
ee
9. 93
4
7
xx
10. 126
3
2
xx
11. 244
7
4
xx
12. 5
10
3
4
xx
13. 28
5
4
xx
14. 124
9
9
xx
PAGE 10
ANSWERSNCEA
1. 84
2 2 22 2
2 1xx
xx
x x= = =− (Achieved) 2.
123
4 4 42 2
2 1aa
aa
a a= = =− (Achieved)
3. 10
25 5 5
2 22 1a
aaa
a a= = =− (Achieved) 4.
912
34
34
34
5
3
5
3
5 3 2xx
xx
x x= = =
−
(Achieved)
PRACTICE
5. 47
47
47
47
7
3
7
3
7 34a
aaa
a a= = =−
(Achieved) 6. 11 11 11 11
9
4
9
49 4 5x
xx
xx x= = =−
(Achieved)
7. 124
3 3 38
3
8
38 3 5x
xxx
x x= = =− (Achieved) 8.
36 2 2 2
6
5
6
5
6 5ee
ee
e e= = =
−
(Achieved)
9. 93
3 3 34
7
4
7 7 4 3
xx
xx x x
= = =− (Achieved) 10. 126
2 2 23
2
3
23 2x
xx
xx x= = =−
(Achieved)
11. 244
6 6 67
4
7
47 4 3x
xxx
x x= = =− (Achieved) 12.
510 2
12
12
3
4
3
4 4 3
xx
xx x x
= = =− (Achieved)
13. 28 4 4 4
5
4
5
4
5 4xx
xx
x x= = =
−
(Achieved) 14. 124
3 3 39
9
9
99 9x
xxx
x= = =− (Achieved)
Study Tip:
The Beginning and the EndFact: most information that sticks in our brains is learnt either at the beginning or end of a study session. (‘Beginning’ and ‘End’ are about 15 - 20 mins each.)Solution: Cut out the middle part! So each hour: Study 35 mins ⇒Break 5 mins ⇒Revise 10 mins ⇒Rest 10 mins
PAGE 11
ALGEBRA
FACTORISING - COMMON FACTOR
STEPS: EXAMPLE: 2 82 2xy x y−1. Write out a set of brackets ( )2. Identify the greatest common
number in each term. Put this outside the brackets
2( )
3. Identify any common letters in each term. Put these outside the brackets
2xy( )
4. Put the original term into the brackets with the common factors cancelled out (powers cancel too)
2 2 8 42 2xy x y x y( ( ) )−
5. Simplify (anything completely crossed out will have a 1 in its place)
2 4xy y x( )−
For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONSFactorise the following equations:
1. ab a b2 2+ 2. 2 22 2ab a b+
PRACTICE QUESTIONSFactorise the following equations:
3. 3 3n n− 4. 6 182x x−
5. 7 49 2x x− 6. 8 27 3d d−
7. 6 95 3 3 6x y x y− 8. 3 62 2ab b+
9. 24 18 3g g− 10. x y xy3 2 43+
11. 15 102 3 5a b ab− 12. 3 43 3 2xy x y−
PAGE 12
ANSWERSNCEA1. ab a b
abab ab a bab b a
2 2
2 2
+
++
( )( )( )
(Achieved) h
2. 2 222 2 22
2 2
2 2
ab a babab ab a bab b a
+
++
( )( )( )
(Achieved) h
PRACTICE3. 3
33 1
3
3 2
2
n
n n nn n
nn
−
−−
( )( )( )
( )
(Achieved) h
4. 6
18 36 3
1866 6
2
2
x
xx x
xxx x
−
−−
( )( ( ) )( )
(Achieved) h
5. 7 4977 7 49 77 1 7
2
2
x xxx x xx x
−
−−
( )( ( ) )( )
(Achieved) h
6. 8 222 8 4 22 4 1
7 3
3
3 7 4 3
3 4
d ddd d dd d
−
−−
( )( ( ) )( )
( )
(Achieved) h
7. 6 933 6 2 9 3
5 3 3 6
3 3
3 3 5 2 3 3 6
x y x yx yx y x y x y
−
−( )( ( ) ( )( ) (
33
3 3 2 33 2 3
) )( )x y x y−
(Achieved) h
8. 3 633 33 2
6 2
2 2
2
2 2 2
2
ab bbb a b bb a
+
++
( )( )( )
( )
(Achieved) h
9. 24 1866 24 4 18 36 4 3
3
3 2
2
g ggg g gg g
−
−−
( )( ( ) ( ) )( )
( )
(Achieved) h
10. x y xyxyxy x y x yxy x y
3 2 4
2
2 3 2 2 4 2
2 2 2
3
33
+
++
( )( )( )
( ) ( )
(Achieved) h
11. 15 1055 15 3 10 25
2 3 5
3
3 2 3 5 2
a b ababab a b abab
−
−( )( ( ) ( ) )( )
33 23 2( )a b−
(Achieved) h
12. 3 4
3 43 4
3 3 2
2
2 3 3 2 2
22
xy x yxyxy x y x yxy y x
−
−−
( )( )( )
( )
(Achieved) h
PAGE 13
ALGEBRA
FACTORISING - QUADRATICS
STEPS: EXAMPLE: x x2 3 10− −1. Look for all of the combinations of
number that multiply to make the last number
10 1 2× × and 5
2. See which of these combinations can add to make the middle number
10 and 1 cannot make -3, 2–5 does make -3 so 2 and -5 are the numbers
3. Write out two sets of brackets with an x at the start of each one ( )( )x x
4. Put your two numbers in the brackets with the x ( )( )x x+ −2 5
For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONSFactorise the following equations:
1. x x2 2 3− − 2. x x2 6 7− −
3. x x2 4 5− − 4. x x2 7 10+ +
5. x x2 8 12− + 6. x x2 7 60+ −
PRACTICE QUESTIONSFactorise the following equations:
7. x x2 5 14− − 8. x x2 18 32− +
9. x x2 8 9+ − 10. x x2 2 8− −
11. x x2 7 8+ − 12. x x2 7 6+ +
13. x2 4− 14. x x2 2 8+ −
15. x x2 5 6− − 16. x x2 9 18+ +
PAGE 14
ANSWERSNCEA1. x x
x x
2 2 31 3 2
1 3
− −− =
+ −1× 3 ( - )( )( )
(Achieved) h
2. x x
x x
2 6 71 7 6
1 7
− −=
+ −1× 7 ( - )( )( )
(Achieved) h
3. x
x x
x2 4 51 5 4
1 5
− −− =
+ −1× 5 ( - )( )( )
(Achieved) h
4. x x
x x
2 7 101 10 2 5 7
2 5
+ +× + =+ +
, ( )( )( )
2× 5
(Achieved) h
5. x x
x x
2 8 121 12 3 4 2 6 8
2 6
− +× × − =− −
, , (- - )( )( )
2× 6
(Achieved) h
6. x x
x x
2 7 601 60 2 30 3 20 4 15
6 10 12 5 712
+ −× × × ×
× − =+
, , , ,, ( )
( )(
5×12
−− 5)
(Achieved) h
PRACTICE7. x x
x x
2 5 141 14 2 7 5
2 7
− −× − =+ −
, ( - )( )( )
2× 7
(Achieved) h
8. x x
x x
2 18 321 32 4 8 2 16 18
2 16
− +× × =− −
, (- - - )( )( )
2×16,
(Achieved) h
9. x x
x x
2 8 93 9 1 8
9 1
+ −× − =
+ −1× 9, ( )( )( )
3
(Achieved) h
10. x x
x x
2 2 81 8 2 4 2
2 4
− −× − =+ −
, ( - )( )( )
2× 4
(Achieved) h
11. x x
x x
2 7 82 4 8 1 7
8 1
+ −× − =
+ −1× 8, ( )( )( )
(Achieved) h
12. x x
x x
2 7 62 3 6 1 7
6 1
+ +× + =
+ +1× 6, ( )( )( )
(Achieved) h
13. x x x
x x
2 24 0 41 4 2 2 0
2 2
− = + −× − =+ −
, ( )( )( )
2× 2
(Achieved) h
14. x x
x x
2 2 81 8 4 2 2
4 2
+ −× − =+ −
, ( )( )( )
2× 4
(Achieved) h
15. x x2 5 62 3 1 6 5
− −× − =1× 6, ( - )
*
It is not -2 and -3 as they do nnot multiply to -6( )( )x x+ −1 6
(Achieved) h
16. x x
x x
2 9 181 18 2 9 3 6 9
3 6
+ +× × + =+ +
, , ( )( )( )
3× 6
(Achieved) h
PAGE 15
ALGEBRA
SIMPLIFY FRACTIONS WITH QUADRATICS
STEPS: EXAMPLE: x xx x
2
28 12
6+ ++
1. Factorise the top( )( )x x
x x+ +
+2 6
62
2. Factorise the bottom( )( )
( )x x
x x+ +
+2 6
6
3. Cancel the common factors( ) ( )
( )x x
x xx
x+ +
+=
+2 66
2
For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONSSimplify the following fractions:
1. x xx x
2
2
4 56 5
− −+ +
2. x xx x
2
2
6 75 4
− −+ +
3. x xx x
2
2
2 37 12− −− +
4. x x
x x
2
2
7 102
+ ++
PRACTICE QUESTIONSSimplify the following fractions:
5. x xx x
2
2
66 8− −+ +
6. x
x x
2
2
98 15−
− +
7. x x
x x
2
2
2 83 12+ −+
8. x xx x
2
2
122 3
+ −− −
9. x xx x
2
2
2 15 4
+ ++ +
10. 4 16
2 82
xx x
−− −
11. x x
x x
2
2
14 322 32− −
−12.
x xx x
2
2
3014 48− −− +
13. x x
xx
2
2
4 530
− −+ −
14. x x
xx
2
2
73 28−
− −
PAGE 16
ANSWERSNCEA
1. x
x x
x x
x
xx x
x x
x x
x
2
2
2
4 56 5
6 5
1 5
5
1 5
1 5
5
− −+ +
+ +
+ +
+
+ −
+ −
−
( )( )
( )( )( )( )
(Merit)
2. x xx xx xx xx xx x
xx
2
2
2
6 75 41 7
5 41 71 4
74
− −+ ++ −+ ++ −+ +−+
( )( )
( )( )( )( )
(Merit)
3. x xx xx xx xx xx x
xx
2
2
2
2 37 121 37 121 33 4
1
− −− ++ −− ++ −− −+−
( )( )
( )( )( )( )
44
(Merit)
4. x
x x
x x
x
xx x
x x
x x
x
2
2
2
7 102
2
2
2 5
2 5
5
+ ++
+
+
+ +
+ +
+
( )( )
( )( )( )
(Merit)
PRACTICE
5. x xx xx xx xx xx x
xx
2
2
2
66 82 3
6 82 32 4
34
− −+ ++ −+ ++ −+ +−+
( )( )
( )( )( )( )
(Merit)
6. x
x xx xx xx xx x
xx
2
2
2
98 153 38 15
3 33 5
35
−− ++ −− ++ −− −+−
( )( )
( )( )( )( )
(Merit)
7. x x
x xx x
x xx x
x xx
x
2
2
2
2 83 12
2 43 12
2 43 4
23
+ −
− +
− ++
−
+
+( )( )
( )( )( )
(Merit)
8. x xx xx xx xx xx x
xx
2
2
2
122 33 4
2 33 41 3
41
+ −− −− +− −− ++ −++
( )( )
( )( )( )( )
(Merit)
9. x xx x
xx x
xx xxx
2
2
2
2
2
2 15 4
15 4
11 414
+ ++ ++
+ ++
+ +++
( )
( )( )( )
(Merit)
10. 4 16
4 4
4 4
4
2 8
2 8
2 4
2
2
2
x
x
x
x x
x x
x x
x
−
−
−
− −
− −
+ −
+
( )
( )( )( )
(Merit)
11. x x
x xx x
x xx x
x xx
x
2
2
2
14 322 32
16 22 32
16 22 16
22
− −−
− +−
− +−
+
( )( )
( )( )( )
(Merit)
12. x xx xx x
x xx xx x
x
2
2
2
3014 485 614 485 66 8
5
− −− ++ −− ++ −− −+
( )( )
( )( )( )( )
xx −8
(Merit)
13. x x
xx x
xx x
x
x
x
x x
x
2
2
2
4 530
1 530
1 5
16 5
6
− −+ −+ −+ −+ −
++ −
+
( )( )
( )( )( )( )
(Merit)
14. x x
xx x
xx x
x xx
x
x
x
2
2
2
73 28
73 28
74 7
4
−− −
−− −
−+ −
+
( )
( )( )( )
(Merit)
PAGE 17
ALGEBRA
SOLVING LINEAR EQUATIONS
STEPS: EXAMPLE: 5 7 2x x+ = − These are all of the steps, you will ever need to perform all in one question. So skip the steps which are not applicable:
1. Multiply away any fractions ( )none skip−
2. Expand any brackets ( )none skip−
3. Put x’s on the same side 5 7 2x x+ − = − −x x4. Put numbers on the opposite side
to the x’s 5 7 2x + − − = − −x 7 75. Simplify 4 9x = −
6. Divide by the last number x = −94
For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONSSolve the following equations:
1. 5 2 7 10x x− = + 2. 2 7 20( )x − =
3. 2 54
3−=
x4.
3 45
7−=
x
5. 6 2 8 6x x− = + 6. 2 2 24( )x − =
PRACTICE QUESTIONSSolve the following equations:
7. 8 9 27x − = 8. 6 8 3x x+ = −
9. 53
7 13x− = 10.
7 28
9x +=
11. 9 1 4 7x x− = + 12. 4 6 9x x+ = −
13. 45
132
x= 14.
43
2 14x− =
15. 6 1 7( )− =e 16. 7 3 3 9x x− = +
PAGE 18
ANSWERSNCEA1. 5 2 7 10
5 2 7 102 2 102 12
122
6
x xx x
xx
x
− = +− − = + −
− − + = +− =
=−
= −
7 72 2x x
(Achieved)
2. 2 7 202 2 7 202 14 202 34
342
17
( )xx
xx
x
− =× − × =− + = +=
= =
14 14
(Achieved)
3. 2 54
3
2 5 3 42 5 12
5 1010
52
−=
− = ×− − = −
− =
=−
= −
x
xx
x
x
2 2
(Achieved)
4. 3 45
7
3 4 7 53 4 35
4 3232
48
−=
− = ×− − = −− =
=−
= −
x
xx
x
x
3 3
(Achieved)
5. 6 2 8 66 2 8 6
2 2 62 8
82
4
x xx x
xx
x
− = +− − = + −
− − + = +− =
=−
= −
8 82 2x x
(Achieved)
6. 2 2 242 2 2 242 4 242 28
282
14
( )xx
xx
x
− =× − × =− + = +=
= =
4 4
(Achieved)
PRACTICE7. 8 9 27
8 9 278 36
368
92
xxx
x
− =− + = +=
= =
9 9
(Achieved)
8. 6 8 36 8 35 8 35 11
115
x xx xxx
x
+ = −+ − = − −+ − = − −= −
=−
x x8 8
(Achieved)
9. 53
7 13
5 7 3 13 35 21 395 60
605
12
x
xxx
x
− =
− × = ×− + = +=
= =
21 21
(Achieved)
10. 7 28
9
7 2 9 87 2 727 70
707
10
x
xxx
x
+=
+ = ×+ − = −=
= =
2 2
(Achieved)
11. 9 1 4 79 1 4 75 1 75 8
85
x xx xxx
x
− = +− − = + −− + = +=
=
4 41 1
x x
(Achieved)
12. 4 6 94 6 93 6 93 15
153
5
x xx xxx
x
+ = −+ − = − −+ − = − −= −
=−
= −
x x6 6
(Achieved)
13. 45
132
4 2 5 138 65
658
x
xx
x
=
× = ×=
= (Achieved)
14. 43
2 14
4 2 3 14 34 6 424 48
484
12
x
xxx
x
− =
− × = ×− + = +=
= =
6 6
(Achieved)
15. 6 1 76 6 76 6 7
6 116
( )− =− =− − = −
− =
= −
eee
e
e
6 6
(Achieved)
16. 7 3 3 97 3 3 94 3 94 12
124
3
x xx xxx
x
− = +− − = + −− + = +=
= =
3 33 3
x x
(Achieved)
PAGE 19
ALGEBRA
SOLVING QUADRATICSSTEPS: EXAMPLE: x x2 3 10− =
1. Make the equation = 0 x x2 3 10 0− − =2. Factorise (if unsure see prev. steps) ( )( )x x+ − =2 5 03. Make each bracket = 0 x x+ = − =2 0 5 0 and 4. Solve for x (twice) x x= =- and 2 5
For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONSSolve the following equations:
1. x x2 2 3 0− − = 2. x x2 6 7 0− − =
3. x x2 4 5 0− − = 4. x x2 7 6 0− + =
PRACTICE QUESTIONSSolve the following equations:
5. ( )( )x x− + =8 5 0 6. 4 8 0x x( )− =
7. x x2 4 12 0− − = 8. x x2 13 42− = -
9. 5 2 7 0x x( )− = 10. ( )( )7 9 2 7 0x x+ + =
11. x x2 12 35 0+ + = 12. x x2 8 7 0− + =
13. x x2 52 100 0− + = 14. x x2 13 78 36+ + =
PAGE 20
ANSWERSNCEA1. x x
x xx xx x
2 2 3 01 3 0
1 0 3 01 3
− − =+ − =+ = − == =
( )( ) and
- and (Merit)
2. x xx x
x xx x
2 6 7 07 1 0
7 0 1 07 1
− − =− + =− = + == =
( )( ) and
and - (Merit)
3. x xx x
x xx x
2 4 5 05 1 0
5 0 1 05 1
− − =− + =− = + == =
( )( ) and
and - (Merit)
4. x xx x
x xx x
2 7 6 06 1 0
6 0 1 06 1
− + =− − =− = − == =
( )( ) and
and (Merit)
PRACTICE5. ( )( )x x
x xx x
− + =− = + == =
8 5 08 0 5 08 5
and and - (Merit)
6. 4 8 04 0 8 0
0 8
x xx x
x x
( )− == − == =
and and (Merit)
7. x xx x
x xx x
2 4 12 06 2 0
6 0 2 06 2
− − =− + =− = + == =
( )( ) and
and - (Merit)
8. x xx xx x
x xx x
2
213 4213 42 06 7 0
6 0 7 06
− =− + =− − =− = − == =
-
( )( ) and
and 77 (Merit)
9. 5 2 7 05 0 2 7 0
0 72
x xx x
x x
( )− == − =
= =
and
and (Merit)
10. ( )( )7 9 2 7 07 9 0 2 7 0
97
72
x xx x
x x
+ + =+ = + =
= =
and
- and - (Merit)
11. x xx x
x xx x
2 12 35 05 7 0
5 0 7 05 7
+ + =+ + =+ = + == =
( )( ) and
- and - (Merit)
12. x xx x
x xx x
2 8 7 07 1 0
7 0 1 07 1
− + =− − =− = − == =
( )( ) and
and (Merit)
13. x xx x
x xx x
2 52 100 050 2 0
50 0 2 050 2
− + =− − =− = − == =
( )( ) and
and (Merit)
14. x xx xx x
x xx
2
213 78 3613 42 06 7 0
6 0 7 06
+ + =+ + =+ + =+ = + ==
( )( ) and
- and -x = 7 (Merit)
PAGE 21
ALGEBRA
SOLVING FRACTIONS WITH QUADRATICS
STEPS: EXAMPLE: x xx x
2
28 12
62+ +
+=
1. Factorise top and bottom( )( )
( )x x
x x+ +
+=
2 66
2
2. Cancel common factorsx
x+
=2 2
3. Multiply both sides by the bottom x x+ =2 2
4. Solve as a linear equation x x x xx
+ − = −=
2 22
For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONSSolve the following fractions:
1. x xx x
2
2
4 56 5
2− −+ +
= 2. x xx x
2
2
6 75 4
2− −+ +
=
3. x xx x
2
2
2 37 12
2− −− +
=
PRACTICE QUESTIONSSolve the following fractions:
4. x x
x
2 126 18
1+ −−
= 5. x x
x x
2
2
26
2+− −
=
6. x x
x x
2
2
4214 48
2+ −− +
= 7. x x
x
2
2
9 2025
10+ +−
=
8.
x xx x
2
2
6 77
5− −−
= 9. x xx x
2
2
122 8
2− −− −
=
10. x xx x
2
2
3 45 4
3+ −+ +
= 11. x xx x
2
2
66 8
3+ −− +
=
12. x xx x
2
2
2 33 18
4+ −− +
= 13. x xx x
2
2
7 106
3− +−+
=
PAGE 22
ANSWERSNCEA
1. x xx xx xx x
xxx xx
2
2
4 56 5
2
1 51 5
2
55
2
5 2 5
− −+ +
=
+ −+ +
=
−+
=
− = +−
( )( )( )( )
( )55 2 10
15= +=
xx-
(Excellence)
2. x xx xx xx x
xxx xx
2
2
6 75 4
2
1 71 4
2
74
2
7 2 4
− −+ +
=
+ −+ +
=
−+
=
− = +−
( )( )( )( )
( )77 2 8
15= +
− =x
x
(Excellence)
3. x
x xx x
xxx
xx x
xx
2
2
2 37 12
2
2
2
2 4
1 33 4
141
− −− +
=
=
=
= −
+ −− −+−+
( )( )( )( )
( )++ = −=
1 2 89
xx
(
(Excellence)
PRACTICE
4. x xx
xx x
x
xx
2 126 18
6 3
6
1
4 3 1
4 1
4 62
+ −−
−
=
+ −=
+=
+ ==
( )( )( )
(Excellence)
5. x xx x
x xx xx
xx xx x
x
2
2
26
2
22 3
2
32
2 32 6
6
+− −
=
++ −
=
−=
= −= −=
( )( )( )
( )
(Excellence)
6. x xx xx xx x
xxx x
2
2
4214 48
2
7 66 8
2
78
2
7 2 8
+ −− +
=
+ −− −
=
+−
=
+ = −
( )( )( )( )
( )xx x
x+ = −=7 2 16
23
(Excellence)
7. x xx
x xx x
xxx x
2
2
9 2025
10
5 45 5
10
45
10
4 10 5
+ +−
=
+ ++ −
=
+−
=
+ = −
( )( )( )( )
( ))54 96
==
xx
(Excellence)
8. x x
x x
x
x x
x x
xx x
x
x
2
2
6 7
1 7
1
75
75
5
1 51 414
− −
+ −
+
−=
−=
=
+ ==
=
( )( )( )
(Excellence)
9. x xx xx xx x
xxx xx
2
2
122 8
2
3 42 4
2
32
2
3 2 2
− −− −
=
+ −+ −
=
++
=
+ = ++
( )( )( )( )
( )33 2 4
1= +
− =x
x
(Excellence)
10. x xx xx xx x
xxx xx
2
2
3 45 4
3
1 44 1
3
11
3
1 3 1
+ −+ +
=
− ++ +
=
−+
=
− = +−
( )( )( )( )
( )11 3 3
4 22
= +− ==
xx
x-
(Excellence)
11. x xx xx xx x
xxx xx
2
2
66 8
3
3 22 4
3
34
3
3 3 43
+ −− +
=
+ −− −
=
+−
=
+ = −+
( )( )( )( )
( )== −
=
=
3 1215 2152
xx
x
(Excellence)
12. x xx xx xx x
xxx xx
2
2
2 33 18
4
3 13 6
4
16
4
1 4 6
+ −− +
=
+ −+ −
=
−−
=
− = −
( )( )( )( )
( )−− = −=
=
1 4 2423 3233
xx
x (Excellence)
13. x xx xx xx x
xxx xx
2
2
7 106
3
5 23 2
3
53
3
5 3 3
− +−+
=
− −+ −
=
−+
=
− = +−
( )( )( )( )
( )55 3 9
14 27
= +==
xx
x-- (Excellence)
PAGE 23
ALGEBRA
SOLVING EXPONENTSSTEPS: EXAMPLE: ( )y2 2 81=
1. Simplify with the following rules:• If the base is the same add the
powers. e.g. x x xm n m n= +
• If the exponent is raised to a power multiply the exponent and power. e.g. ( )x xm n m n= ×
yy y y y
4 8181
=× × × =
2. Trial with 2 2 2 2 2 81× × × ≠ No3. Trial with 3 3 3 3 3 81× × × = Yes!4. Trial with 4
Etc. Unnecessary.
5. If the power is even numbered you have both +ve and -ve answers
y = +3 3 and - as there is an even power (4)
For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONSSolve the following exponents:
1. ( )c2 3 64= 2. x4 81=
3. ( )x3 2 64=
PRACTICE QUESTIONSSolve the following exponents:
4. x3 64= 5. x5 32=
6. ( )( )x x4 2 64= 7. ( )x2 2 81=
8. x3 125= 9. x3 27=
10. x x( )5 64= 11. ( ( ))x x3 2 32=
PAGE 24
ANSWERSNCEA1. ( )c
c
c
2 3
664
642 2 2 2 2 2 64
2
==× × × × × ==
Yes and -2
(Excellence)
2. x
x
4 812 2 2 2 813 3 3 3 81
3
=× × × ≠× × × ==
No Yes
and -3
(Excellence)
3. ( )xx
x
3 2
664
642 2 2 2 2 2 64
2 2
==× × × × × ==
Yes and -
(Excellence)
PRACTICE4. x
x
3 642 2 2 643 3 3 644 4 4 64
4
=× × ≠× × ≠× × ==
No No Yes
(Merit) h
5. x
x
5 322 2 2 2 2 32
2
=× × × × ==
Yes
(Merit) h
6. ( )( )x xx
x
4 2
664
642 2 2 2 2 2 64
2 2
==× × × × × ==
and -
Yes
(Excellence)
7. ( )xx
x
2 2
481
812 2 2 2 813 3 3 3 81
3 3
==× × × ≠× × × ==
NoYes
and -
(Excellence)
8. x
x
3 1252 2 2 1253 3 3 1254 4 4 1255 5 5 125
5
=× × ≠× × ≠× × ≠× × ==
No No No Yes
(Merit) h
9. x
x
3 272 2 2 273 3 3 27
3
=× × ≠× × ==
No Yes
(Merit) h
10. x x
x
x( )5
664
642 2 2 2 2 2
2 2
==× × × × ×=
Yes and -
(Excellence)
11. ( ( ))x xx
x
3
5
2 3232
2 2 2 2 2 322
==× × × × ==
Yes
(Merit) h
PAGE 25
ALGEBRA
SOLVING INEQUALITIESSTEPS: EXAMPLE: 2 2
43( )x −
>
1. Multiply away any fractions 2 2 3 4( )x − > ×
2. Expand any brackets 2 4 12x − >3. Put x’s on the same side ( )none skip−
4. Put numbers on the opposite side 2 4 12x − + > +4 45. Simplify 2 16x >6. Divide by the last number
(Remember if dividing by a negative number reverse the sign)
x > 8For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONSSolve the following inequalities:
1. 4 2
38−
>x 2. 2 5
43−
>x
3. 3 4
57−
>x
PRACTICE QUESTIONSSolve the following inequalities:
4. 3 14
5x −> 5.
4 65
2( )x x+<
6. 3 7 4x x+ > 7. 24
3−>
x
8. 6 3 2( )x − < 9. 14 4 6− <x
10. 5 8
37x x−
> 11. 9 3 4 2 1( ) ( )− < − −x x
12. 2 3 2
36( )−
>x
13. 5 7 2 3x x+ < +
PAGE 26
ANSWERSNCEA1. 4 2
38
4 2 8 34 2 24
2 2010
−>
− > ×− − > −
− >< −
x
xx
xx
4 4
(Merit)
2. 2 54
3
2 5 3 42 5 12
5 102
−>
− > ×− − > −
− >< −
x
xx
xx
2 2
(Merit)
3. 3 45
7
3 4 7 53 4 35
4 328
−>
− > ×− − > −
− >< −
x
xx
xx
3 3
(Merit)
PRACTICE4. 3 1
45
3 1 5 43 1 203 21
7
x
xxx
x
−>
− > ×− + > +>>
1 1
(Merit)
5. 4 65
2
4 4 6 5 24 24 1024 64
( )x x
x xx x
xx
+<
× + × < ×+ − < −<<
4 4x x
(Merit)
6. 3 7 43 7 47
x xx x
x
+ >+ − > −>
3 3x x
(Merit)
7. 24
3
2 3 42 12
1010
−>
− > ×− − > −
− >< −
x
xx
xx
2 2
(Merit)
8. 6 3 26 6 3 26 18 26 20
206
103
( )xx
xx
x
− <× − × <− + < +<
<
18 18
or
(Merit)
9. 14 4 614 4 6
4 82
− <− − < −
− < −>
xx
xx
14 14
(Merit)
10. 5 83
7
5 8 3 75 8 21
8 1612
x x
x xx x
x
x
−>
− > ×− − > −
− >
− >
5 5x x
(Merit)
11. 9 3 4 2 19 3 9 4 2 4 127 9 7 4
5
( ) ( )− < − −× − × < × − × −− + − < − + −
− <
x xx x
x xx
4 27 4 27x x−−
>20
4x
(Merit)
12. 2 3 23
6
2 3 2 2 6 36 4 18
4 123
( )−>
× − × > ×− − > −
− >< −
x
xx
xx
6 6
(Merit)
13. 5 7 2 35 7 2 33 4
43
x xx xx
x
+ < ++ − − < + − −< −
<−
2 7 2 7x x
(Merit)
PAGE 27
ALGEBRA
Study Tip:
RepetitionTo burn information into your brain:Don’t study the same topic for a long period of timeDo learn the information then go over it again:• Within 24 hours (retention goes from 20% to 80%) • Again after a week (retention goes up to 90%)• Again after a month (long term retention)
PAGE 28
SUBSTITUTION AND REARRANGING
STEPS: EXAMPLE: T lg
= 2π
Substitution: If l is 90 and g is 10. Give T in terms of π.
T = 2 9010
π1. Put the numbers in place of the letters
2. Simplify or solve the termTT
T
== ×=
2 92 3
6
πππ
Rearranging: Make g the subject.Tp
lg
Tp
lg
g Tp
l
2
2
2
2
2
=
=
=
1. Rearrange to make sure that the subject is a numerator (on the top) not a denominator (on the bottom)
2. Get rid of all other symbols/numbers around the subject
g lT
=
2
2
πFor a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONS1. Sara found the following equation:
A wg
= π
a. If W = 25 and g = 16, find A in terms of π.b. Sara now wants to use the formula to find W in
a different situation.Rewrite the formula with W as the subject.
2. The formula for the volume of a cone is v r h=π3
2
where r is the radius and h is the height of the cone.a. Write the formula for the radius, r, of the cone
in terms of V, h, and π.
b. Max has two cones that have the same volume. One cone is twice the height of the other. Give an expression for the radius, r, of the shorter cone in terms of R, the radius of the taller cone.Give your answer in the simplest form.
PAGE 29
ALGEBRA3. Mele is exploring the sequence of numbers given
by the rule:
3 2 12n n− +Give the rule for finding the difference between any two consecutive terms from the sequence 3 2 12n n− + in its simplest form.
Hint: consecutive terms follow each other, e.g. the 5th and 6th terms or the 17th and 18th terms or the nth and (n+1)th terms.You must show your working.
4. A square tile is surrounded by a 4 cm wide border. The area of the tile is x2. The area of the border is
x x+( ) −8 2 2. If the border covers an area of
640 cm2, what is the tile length in centimetres?
5. A formula for the perimeter, G, of a rectangle is G j s= +2 2 where j is the length of the base of the rectangle and s is the height of the rectangle. Make j the subject of this formula.
6. Make h the subject of 4 7 10l h= −( )7. Rearrange 2 3 450K H+ = to make H the
subject.
8. Jeans sizes are even numbers. The size J can be calculated from the waist measurement m using
the formula: J m=
−4 906
where m is in centimetres. a. Rearrange to make m the subject of the
formula. b. What size jeans should Henry order, if his waist
measurement is 71cm?
9. Posting parcels is expensive. It depends both on the weight of the parcel and the distance it has to travel. The rates are $3 per kg and 15 cents per km. The cost (P) of sending a parcel can be calculated by the formula P w d= +3 0 15. where w is the weight in kg and d is the distance in km. The distance from Auckland to Invercargill is 1572km. Find the cost of sending a parcel weighing 8kg from Auckland to Invercargill.
10. In a trampoline competition the score (T) for a trick is calculated using the formula T HF=1 5. where F is the difficulty of the trick and H is the height of the jump. John completes a trick of difficulty 8. The height of his jump was 1.5m. Calculate the score for John.
11. May prints butterflies on different shaped pillows. Her rule for calculating the total number of butterflies (B) she prints on the pillow is
B n n=
+( )12
where n is the number of edges
on the pillow. Find the total number of butterflies she prints on a pillow with 10 edges.
PRACTICE QUESTIONS
Study Tip:
Retrieving InformationDon’t just put information into your brain.Do practice getting the information out of your brain.Retrieving the stored information in your brain by answering lots of questions is more effective than reading or taking notes and makes us stronger in exams.
PAGE 30
ANSWERSNCEA1.
a. A wg
A
= =
= =
π π
π π
2516
2516
54
(Merit)
b. A wg
A wg
A wg
g A w
= → =
= →
=
ππ
π π
2 2
(Merit)2. a. V r h
Vh
r
r Vh
=
=
=
π
π
π
33
3
3
2
2
/
(Merit)
b. Let H be the height of the cone with radius R and let h be the height of the cone with radius r. Therefore H = 2h. Since the cones have the same volume:π π3 3
2 2
2 2
2 2
2 2
22 2
2
2
r h R H
r h R H
r R Hh
R hh
R
r R R
=
=
= = =
= = (Excellence)
3. The nth term of the sequence is 3 2 12n n− +the (n+1)th term of the sequence is 3 1 2 1 12( ) ( )n n+ − + + which simplifies to:
3 2 1 2 2 13 6 3 2 13 4 2
2
2
2
n n nn n nn n
+ +( ) − − += + + − −= + +
so the difference between consecutive terms is:
3 4 2 3 2 13 4 2 3 2 16 1
2 2
2 2n n n n
n n n nn
+ + − − +( )= + + − + −= + (Excellence)
PRACTICE4. x x
x x xxx
x
+( ) − =+ + − =+ == −
= =
8 64016 64 640
16 64 64016 640 64
57616
3
2 2
2 2
66 cm
(Merit)
5. G j sj G s
j G s
= += −
= −
2 22 2
2 (Merit)
6. 4 7 104 7 707 4 70
47
10
l hl hh l
h l
= −( )= −= +
= +
(Merit)
7. 2 3 4503 450 2
150 23
K HH K
H K
+ == −
= −
(Merit)
8. a. J m
J mm J
m J
=−
= −= +
= +
4 906
6 4 904 6 90
32
452
(Merit)
b. J m
J
=−
=× −
=
4 906
4 71 906
32 33.
The nearest even number size is 32. (Merit)
9. p w dpp
= += × + ×=
3 0 153 8 0 15 1572259 80
..
$ . (Merit)
10. T HFTT
== × ×=
1 51 5 1 5 818
.
. . (Merit)
11. B n n
B
=+
=+
=
( )
( )
12
10 10 12
55 (Merit)
PAGE 31
ALGEBRA
Study Tip:
Time ManagementThis is the key to successful study.You need to:• Decide how to best use the hours you have each day, week,
and term• Schedule study times and free time!• Stick to your schedule• After each study session, write down the next session time and
what you plan to study
PAGE 32
SIMULTANEOUS EQUATIONS
STEPS: EXAMPLE: 4 3 463 16
x yx y+ =+ =
1. Write out the two equations if they haven’t been given
4 3 463 16
x yx y+ =+ =
2. Rearrange one equation so that a variable is the subject
x y= −16 3
3. Substitute into the other equation and solve for one variable
4 16 3 3 4664 12 3 46
9 182
( )− + =− + =− = −
=
y yy yyy
4. Substitute answer into rearranged equation and solve for other variable
x yxx
= −= − ×
=
16 316 3 2
10For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONS1. Ari spent $45 buying some CDs in a sale. He bought
R Rock CDs and B Blues CDs. Ari writes an equation for the amount he spent as: 2 45R B+ = .Ari bought four times as many Rock CDs as Blues CDs.How many blues CDs did he buy all together?
2. A milk drink costs $1.50 more than a fruit drink.5 fruit drinks and 4 milk drinks cost a total of $24What is the cost of one milk drink?You must show at least one equation that can be used in solving this problem.
3. Scenic School is using two vans to take a group of students on a field trip.If two students move from van A to van B, then the two vans would have the same number of students in each, but if two students moved from van B to van A, then van B would have half the number of students that were then in van A.Use the information above to find the total number of students on the field trip.In your answer, you must give at least ONE equation that you would use to solve the problem.
4. A drink costs $2.50 more than a packet of chips.2 drinks and 4 packets of chips cost a total of $17.What is the cost of 1 drink?You must show at least one equation that can be used in solving this problem.
PAGE 33
ALGEBRA
PRACTICE QUESTIONS5. 370 people paid a total of $2330 to watch the
Cirque de Soleil. Adults pay $8 entry, while children pay less. These two simultaneous equations can be used to find how many adults paid to watch the circus. x yx y+ =+ =
3708 5 2330How many paying adults attended the circus?
6. It costs each adult $12 and each child $5 to get into a soccer game. An $80 group pass to the soccer game is available. This lets in a group of up to 8 people into the game. The group pass is not always the cheapest option. If a group of 8 people is to save money with the group pass, what is the minimum number of adults required?
7. Solve this pair of simultaneous equations for x and y.x yx y+ =+ =
107 8 200
8. Fiona bought tickets to events at the Olympic Games. She paid $1095 for 15 tickets. The tickets for the cycling events cost $65 and the tickets for the track and field events cost $85. Solve the simultaneous equations to find the number of tickets Fiona bought to each event. 65 85 1095
15c t
c t+ =
+ =9. In a sale at a market, all the seconds fruit were one
price and all the export grade fruit was another price. Zachariah bought 2 cartons of seconds fruit and four of export grade fruit and paid in total $39.70. Kirsty bought 3 cartons of seconds fruit and one of export quality and paid $39.80. Be aware the seconds cartons are much larger and therefore more expensive. Solve the following simultaneous equations to calculate the sale price of one export carton. 2 4 39 703 39 80s xs x+ =+ =
..
10. Daisy was a flower expert and had been studying roses. She discovered a new species early in 2005 and another later in the year. The trend line for each of her discoveries are: y x y x= + = +40 175 2 85 290 and . Y is
the number of thorns on their stalks and x is the number of years since 1985. Predict the year in which the roses will both have the same number of thorns.
11. Blue and Tasty cheeses are blended together to make new cheeses. Blue cheese has 20% fat and Tasty cheese, 35%. One of the blends of the cheeses is called Tasty Blue Cheese. The Tasty Blue cheese is to have a fat content of 30%. 800 grams of the new blend is to be made in trial. Let B = mass of Blue cheese and T = mass of Tasty cheese in grams. 20 35 24000
800B T
B T+ =
+ =Solve the simultaneous equations to find the mass of Tasty and Blue required. Show all your working.
12. The total annual health bill of the residents of Sloth Avenue is $480000. The health costs per adult, H, are $1500 per year more than for a child, C. 80 64 480000
1500H C
H C+ =
= +Solve the simultaneous equations to find the yearly health costs for a child.
Study Tip:
QuizzingBeing quizzed by a person is easier and more fun than learning from a textbook.But remember, if you are doing it with a friend, stay on task!
PAGE 34
NCEA1. R B R B
B BBB
= + =( ) + =
==
4 2 452 4 45
9 455
and Sub:
He bought 5 Blues CD’s. (Excellence)
2. m fm f f m
f
= == + + =
+
milk drink and fruit drink and
sub: 1 5 5 4 24
5.44 1 5 24
5 4 6 249 18
22 1 5 3 50
ff fffm
+( ) =+ + ==== + =
.
$. $ .
(Excellence)
3. A B B A
B AA B
B B
− = + − = +( )−( ) = +
= +− = +( ) +
2 2 2 12
2
2 2 24
2 4 4 2
and
or
Sub: 22 4 6
104
10 4 14
B BBA B
− = +== += + =
Total number of students is 24. (Excellence)
4. d cd c d c
c
= == + + =
+
drink and packet of chips and
sub: 2 5 2 4 17
2.
( 22 5 4 172 5 4 176 12
22 2 5 4 50
. )
$. $ .
+ =+ + ==== + =
cc cccd
(Excellence)
PRACTICE5. x y
x y x yy x
= =+ = + == −
adults and children and
sub:
370 8 5 23303708xx x
x xxx
+ −( ) =− − ===
5 370 23308 1850 5 23303 480
160160 adults paid for tickets. (Excellence)
6. let adults and children and
sub:
x yx y x yy x
= =+ = + == −
8 12 5 80812xx x
x xx
x
+ −( ) =+ − ==
=
5 8 8012 40 5 807 40
407
5 71 = .
Round up from 5.33 to 6. A minimum of 6 adults is needed. (Excellence)
7. x y x yx y
y yy
+ = + == −
−( ) + =− + =
10 7 8 200107 10 8 200
70 7 8 200
and
sub: y y
=== − = −
13010 130 120x
(Excellence)
8. c tc t c t
c
= =+ = + =
=
cycling ticket, track ticket and 65 85 1095 15
15565 15 85 1095
975 65 85 109520 120
615 6
−−( ) + =
− + ==
== −
t
t tt
tc
sub: t t
== 9
(Excellence)
9. s xs x s xx s
= =+ = + == −
seconds, export and
sub
2 4 39 70 3 39 8039 80 3
. ..
:: s s2 4 39 80 3 39 7010 119 50
11 9539 8 3 11 95 3
+ −( ) =− = −== − × =
. ..
.. .
ssx ..95
(Excellence)
10. y x y xx x
x x
= + = +−( ) = +
+ =
40 175 2 85 2902 40 175 85 290
80 350 85
and sub:
++==
+ =
29060 5
122005 12 2017
xxYear the same is
(Excellence)
11. 20 35 24000 80080020 800 35 24000
16
B T B TB T
T T
+ = + == −
−( ) + =
and
sub: 0000 20 35 24000
15 8000533 3800 533 3 266 7
− + ==
= ( )= − =
T TT
T gB
.. .
1 dpgg 1 dp( )
(Excellence)
12. 80 64 480000 150080 1500 64 480000
80 1
H C H C+ = = ++( ) + =
+
and sub: C C
C 220000 64 480000144 360000 2500
2500 1500 4000
+ == =
= + =
CC C $
H $
(Excellence)
ANSWERS
PAGE 35
ALGEBRA
WORD QUESTIONSSTEPS:
1. Give each variable a letter, the letter will ALWAYS represent a number. E.g. R will represent the number of Rock CDs.
2. If a calculation is required, substitute the numbers given in the question into your equations. E.g. h is the height of the ball. The question tell you the ball is on the ground so substitute 0 in for h (height is zero).
3. You now have enough information to either solve the problem or explain your answer.
For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONS1. Ari spent $45 buying some CDs in a sale. He bought
R Rock CDs and B Blues CDs. Ari writes an equation for the amount he spent as: 2 45R B+ = .Explain the terms of the equation
2. Tom and his son Tane are throwing a ball to each other on the deck of their house.Tane misses the ball and it falls to the ground.The path of the ball can be modelled by the equation h t t= − + +2 2 8 where t is the time in seconds since the ball is thrown and h is the height in meters above the ground at any time t.a. How long after it is thrown will it hit take to hit
the ground?Explain what you are calculating at each step of your answer.
b. How much higher does the ball rise above the height of the point from which it is thrown?Explain what you are calculating at each step of your answer.
3. Joey needs to make a path from the front of his house to the back as shown in the diagram below.The width of the path is in metres.Jim only has enough to make a path with a total area of 9 m2.Form equations and use these to find the width of the concrete path around his house.
PAGE 36
PRACTICE QUESTIONS4. A person in a hot air balloon dropped a hat
accidently over the side of the basket. The height of the hat was modelled by the equation: h t= −180 5 2
Where h = height of the hat above the ground (in m) and t = time since the hat was dropped (in seconds). Use the equation to find:a. The height of the hot air balloon above the
ground when the hat was droppedExplain what you are calculating at each step of your answer.
b. The time, from when it was dropped, for the hat to hit the groundExplain what you are calculating at each step of your answer.
5. 678 people paid a total of $9870 to get into a volleyball game. If n represents the number of adults at the game, then we can write the expression: 8 5 678 9870n n+ − =( ) . Explain what the expression ( )678− n represents.
6. Mr and Mrs Cameron had a possum problem on their farm. The possums came from other farms to eat their trees and so a deadly disease was introduced to get rid of the nasty creatures. Each day they counted the number of possums up their trees. The number of possums was modelled by the equation:P n n= − + +2 13 44 , where P is the number of possums and n = day number. Explain why n cannot be greater than 15.
7. Henry saved $5000 for a snowboarding trip to France. He wanted to board for as many days as possible. Each day pass cost $125. Travel, food and accommodation cost $3100. Write an equation and use it to tell how many days Henry is able to board with that saved money.
8. Katie and Sarah are sisters. Katie is 10 and Sarah is 4 years younger. Their favourite number is 896. Form a relevant equation and use it to find how many years it will take until Katie’s and Sarah’s ages, when multiplied together make 896.
9. At the rowing world champs 40 years ago, the average number of rowing teams per discipline was three times the total number of disciplines. At the Karapiro world champs in 2010, there were 10 more disciplines than 40 years ago and three times as many teams/competitors per discipline. There were a total of 3375 teams/competitors at the 2010 world champs. Write at least one equation to model this situation. Use the model to find the number of disciplines that were contested 40 years ago.
Study Tip:
Challenge YourselfDon’t: practice questions that are too easy – your brain discounts the information as not important.Do: Challenge yourself – the information is more likely to stick (and you learn how to solve difficult problems!)
PAGE 37
ALGEBRA
ANSWERSNCEA1. 2R is the amount spent on rock CDs or $2.00 per
Rock CD1D is the amount spent on blues CDs or $1.00 per Blues CD$45 is the total amount spent. (Achieved - one correct) (Merit - two correct) (Excellence - three correct)
2. a. The ball will
hit the ground when h = 0 so solve:− + + =− − =+ − == −
t tt tt t
t
2
22 8 0
2 8 02 4 02 4
( )( ) or
As t cannot be negative, the ball must hit the ground after 4 seconds. (Merit)
b. The highest point is at t = − +=
2 42
1. This
is because 1 is half way between the -2 and 4 second solutions found in part (a). When t h= = − + + =1 1 2 1 8 92, ( ) ( ) so the ball rises 1 m higher than when it is thrown, from 8 m to 9 m. (Excellence)
3. Area
m
= × + × + − ×= + + −= +
+ =+ −
2 3 6 22 3 6 2
88 98
2 2
2
2 2
2
x x x x xx x x x
x xxx
xx
( )
99 01 9 0
1 0 9 01 9
=− + =− = + == = −
( )( )x xx xx x
and and
x = -9 is not a valid solution so Jim’s path must be 1 m wide. (Excellence)
PRACTICE4.
a. The hat was dropped at t = 0 and so the height of the balloon will be the same as the height of the hat at t = 0. h = − =180 5 0 1802( ) m (Merit)
b. The hat hit the ground when h = 0 so solving for that will give the time taken to fall.180 5 036 0
366
2
2
2
− =− ===
tt
tt seconds (Merit)
5. This looks like a substitution into a simultaneous equation. If n is the number of adults then it can be assumed that ( )678− n is the number of other people who are attending (children/students). 678 would represent the total number of people at the event. (Merit)
6. The equation; P n n= − + +2 13 44 shows that when n2 is greater than 13n + 44, P will be negative, as n2 is negative. This means that there will be a negative number of possums after 15 days and this is impossible. (Merit)
7. 5000 3100 125 01900 125 01900 12515 2
− − ≥− ≥≥≥
xxx
x.The most days he can board is 15. (Excellence)
8. Let x be the number of years to wait.10 6 896
10 6 60 89616 836 038 22 0
2
2
+( ) +( ) =+ + + =+ − =+( ) −( ) =
x xx x xx xx x
x == − =38 22 or x Time cannot be negative so 22 years is required before their ages multiply together to make 896. (Excellence)
9. c=no. of teams/competitors 40 years agox=no. of disciplines 40 years agos=no. of disciplines in 2010cx
x
x s
sx x
xx
x xx
=
+ =
= ( ) =
+=
= +( )= +
3
103375 3 3 9
337510
9
3375 9 103375 9 2 9909 90 3375 0
10 375 015 25 0
15 25
2
2
xx xx xx xx x
+ − =+ − =−( ) +( ) == = − or
40 years ago there were 15 disciplines at the rowing world champs (-25 is not valid in this situation) (Excellence)
10 6 89610 6 60 89616 836 038 22 0
2
2
+( ) +( ) =+ + + =+ − =+( ) −( ) =
x xx x xx xx x
x == − =38 22 or x
10 6 89610 6 60 89616 836 038 22 0
2
2
+( ) +( ) =+ + + =+ − =+( ) −( ) =
x xx x xx xx x
x == − =38 22 or x
