Algebra revision slides

2. Intermediate Tier - Algebra revision
Contents :
Collecting like terms
Multiplying terms together
Indices
Expanding single brackets
Expanding double brackets
Substitution
Solving equations
Finding nth term of a sequence
Simultaneous equations
Inequalities
Factorising – common factors
Factorising – quadratics
Other algebraic manipulations
Solving quadratic equations
Rearranging formulae
Curved graphs
Graphs of y = mx + c
Graphing inequalities
Graphing simultaneous equations
Solving other equations graphically

3. Collecting like terms You can only add or subtract terms in an
expression which have the same
combination of letters in them
e.g. 1. 3a + 4c + 5a + 2ac
= 8a + 4c + 2ac
e.g. 2. 3xy + 5yx – 2xy + yx – 3x2
= 7xy – 3x2
1. a + a + a + a + b + b + a =
Simplify each of these expressions:
2. x2 + 3x – 5x + 2 =
3. ab – 5a + 2b + b2 =
4. – 9x2 + 5x – 4x2 – 9x =
5. 4a2 – 7a + 6 + 4a =
6. 4ab + 3ba – 6ba =
5a + 2b
x2 – 2x + 2
cannot be simplified
– 13x2 – 4x
4a2 – 3a + 6
ab

4. Multiplying terms together Remember your negative numbers
rules for multiplying and dividing
Signs same  +ve answer
Signs different  -ve answer
e.g. 1. 2 x 4a
= 8a
e.g. 2. - 3b x 5c
= - 15bc
e.g. 3. (5p)2
= 5p x 5p
= 25p2
1. -6 x a =
Simplify each of these expressions:
2. 4 x -3d =
3. -5a x -6c =
4. 3s x 4s =
5. a x 3a =
6. (7a)2 =
-6a
-12d
30ac
12s2
3a2
49a2
7. -7f x 8a =
8. 4b2 x -9 =
9. -2t x -2s =
10. 5y x 7 =
11. 6a x -a =
12. (-9k)2 =
-56af
-36b2
4st
35y
-6a2
81k2

5. Indices
(F2)4
t2  t2
b
1
a2 x a3 c0 a4
x7  x4
2e7 x 3ef2
4xy3  2xy
5p5qr x 6p2q6r

6. Expanding single brackets
e.g. 4(2a + 3) =
x
8a
x
+ 12
Remember to multiply all the terms
inside the bracket by the term
immediately in front of the bracket
If there is no term in
front of the bracket,
multiply by 1 or -1
Expand these brackets and simplify wherever possible:
1. 3(a - 4) =
2. 6(2c + 5) =
3. -2(d + g) =
4. c(d + 4) =
5. -5(2a - 3) =
6. a(a - 6) =
3a - 12
12c + 30
-2d - 2g
cd + 4c
-10a + 15
a2 - 6a
7. 4r(2r + 3) =
8. - (4a + 2) =
9. 8 - 2(t + 5) =
10. 2(2a + 4) + 4(3a + 6) =
11. 2p(3p + 2) - 5(2p - 1) =
8r2 + 12r
-4a - 2
-2t - 2
16a + 32
6p2 - 6p + 5

7. Expanding double brackets Split the double brackets into 2
single brackets and then expand
each bracket and simplify
(3a + 4)(2a – 5)
= 3a(2a – 5) + 4(2a – 5)
= 6a2 – 15a + 8a – 20
“3a lots of 2a – 5
and 4 lots of 2a – 5”
= 6a2 – 7a – 20
If a single bracket is squared
(a + 5)2 change it into double
brackets (a + 5)(a + 5)
Expand these brackets and simplify :
1. (c + 2)(c + 6) =
2. (2a + 1)(3a – 4) =
3. (3a – 4)(5a + 7) =
4. (p + 2)(7p – 3) =
c2 + 8c + 12
6a2 – 5a – 4
15a2 + a – 28
7p2 + 11p – 6
5. (c + 7)2 =
6. (4g – 1)2 =
c2 + 14c + 49
16g2 – 8g + 1

8. Substitution If a = 5 , b = 6 and c = 2 find the value of :
3a ac
4bc
a
(3a)2
a2 –3b
c2
(5b3 – ac)2
ab – 2c c(b – a)
4b2
Now find the value of each of these expressions if
a = - 8 , b = 3.7 and c = 2/3
15 4 144 10
26 2 225
7 9.6 1 144 900

9. Solving equations
Solve the following equation to find the value of x :
4x + 17 = 7x – 1
17 = 7x – 4x – 1
17 = 3x – 1
17 + 1 = 3x
18 = 3x
18 = x
3
6 = x
x = 6
 Take 4x from both sides
 Add 1 to both sides
 Divide both sides by 3
Now solve these:
1. 2x + 5 = 17
2. 5 – x = 2
3. 3x + 7 = x + 15
4. 4(x + 3) = 20
Some equations cannot
be solved in this way and
“Trial and Improvement”
methods are required
Find x to 1 d.p. if: x2 + 3x = 200
Try Calculation Comment
x = 10
x = 12
(10 x 10)+(3 x 10) = 130 Too low
Too high
(12 x 12)+(3 x 12) = 208
etc.

10. Solving equations from angle problems
Rule involved:
Angles in a quad = 3600
4y + 2y + y + 150 = 360
7y + 150 = 360
7y = 360 – 150
7y = 210
y = 210/7
y = 300
4y
1500
y
2y
Find the size
of each angle
Angles are:
300,600,1200,1500
Find the
value of v 4v + 5 = 2v + 39
4v - 2v + 5 = 39
2v + 5 = 39
2v = 39 - 5
2v = 34
v = 34/2
v = 170
Check: (4 x 17) + 5 = 73 , (2 x 17) + 39 = 73
Rule involved:
“Z” angles are
equal

11. Finding nth term of a sequence
This sequence is
the 2 times table
shifted a little
5 , 7 , 9 , 11 , 13 , 15 ,…….……
Position number (n)
1 2 3 4 5 6
Each term is found by the position number
times 2 then add another 3. So the rule for the
sequence is nth term = 2n + 3
2 4 6 8 10 12
Find the rules of these sequences
 1, 3, 5, 7, 9,…
 6, 8, 10, 12,…….
 3, 8, 13, 18,……
 20,26,32,38,………
 7, 14, 21,28,……
 1, 4, 9, 16, 25,…
 3, 6,11,18,27…….
 20, 18, 16, 14,…
 40,37,34,31,………
 6, 26,46,66,……
100th term = 2 x 100 + 3 = 203
2n – 1
2n + 4
5n – 2
6n + 14
7n
And these sequences
n2
n2 + 2
-2n + 22
-3n + 43
20n - 14

12. Simultaneous equations
4a + 3b = 17
6a  2b = 6
1 Multiply the equations up until the second unknowns have the
same sized number in front of them
x 2
x 3
8a + 6b = 34
18a  6b = 18
2 Eliminate the second unknown
by combining the 2 equations
using either SSS or SDA
+
26a = 52
a = 52
26
a = 2
3 Find the second unknown by substituting back into
one of the equations
Put a = 2 into: 4a + 3b = 17
8 + 3b = 17
3b = 17 - 8
3b = 9
b = 3
So the solutions are:
a = 2 and b = 3
Now solve:
5p + 4q = 24
2p + 5q = 13

13. Inequalities
Inequalities can be solved in exactly the same way
as equations
14  2x – 8
14 + 8  2x
22  2x
11  x
22  x
2
x  11
Add 8 to
both sides
Divide both
sides by 2
Remember to
turn the sign
round as well
The difference is that
inequalities can be given
as a range of results
Or on a scale:
Here x can be equal to :
11, 12, 13, 14, 15, ……
8 9 10 11 12 13 14
1. 3x + 1 > 4
2. 5x – 3  12
3. 4x + 7 < x + 13
4. -6  2x + 2 < 10
Find the range of solutions for these inequalities :
X > 1 X = 2, 3, 4, 5, 6 ……
or
X  3 X = 3, 2, 1, 0, -1 ……
or
X < 2 X = 1, 0, -1, -2, ……
or
-2  X < 4 X = -2, -1, 0, 1, 2, 3
or

14. Factorising – common factors
Factorising is basically the
reverse of expanding brackets.
Instead of removing brackets
you are putting them in and
placing all the common factors
in front.
5x2 + 10xy = 5x(x + 2y)
Factorising
Expanding
Factorise the following (and check by expanding):
 15 – 3x =
 2a + 10 =
 ab – 5a =
 a2 + 6a =
 8x2 – 4x =
3(5 – x)
2(a + 5)
a(b – 5)
a(a + 6)
4x(2x – 1)
 10pq + 2p =
 20xy – 16x =
 24ab + 16a2 =
 r2 + 2 r =
 3a2 – 9a3 =
2p(5q + 1)
4x(5y - 4)
8a(3b + 2a)
r(r + 2)
3a2(1 – 3a)

15. Factorising – quadratics Here the factorising is the reverse of
expanding double brackets
x2 + 4x – 21 = (x + 7)(x – 3)
Factorising
Expanding
Factorise x2 – 9x - 22
To help use
a 2 x 2 box
x2
- 22
x
x
Factor
pairs
of - 22:
-1, 22
- 22, 1
- 2, 11
- 11, 2
Find the
pair which
give - 9
x2
-22
x
x
-11
-11x
2x
2
Answer = (x + 2)(x – 11)
Factorise the following:
 x2 + 4x + 3 =
 x2 - 3x + 2 =
 x2 + 7x - 30 =
 x2 - 4x - 12 =
 x2 + 7x + 10 =
(x + 3)(x + 1)
(x – 2)(x – 1)
(x + 10)(x – 3)
(x + 2)(x – 6)
(x + 2)(x + 5)

16. Fully factorise this expression:
4x2 – 25
Look for 2 square numbers
separated by a minus. Simply
Use the square root of each
and a “+” and a “–” to get:
(2x + 5)(2x – 5)
Fully factorise these:
(a) 81x2 – 1
(b) ¼ – t2
(c) 16y2 + 64
Answers:
(a) (9x + 1)(9x – 1)
(b) (½ + t)(½ – t)
(c) 16(y2 + 4)
Factorising a difference of
two squares
Other algebraic manipulations
Cancelling common factors in
expressions and equations
These two expressions are equal:
6v2 – 9x + 27z  2v2 – 3x + 9z
3
As are these:
2(x + 3)2  2(x + 3)
(x + 3)
The first equation can be reduced
to the second:
4x2 – 8x + 16 = 0  x2 – 2x + 4 = 0
Simplify these:
(a) 5x2 + 15x – 10 = 0
(b) 4(x + 1)(x – 2)
x + 1
Answers:
(a) x2 + 3x – 2 = 0 (b) 4(x – 2)

17. Solving quadratic equations (using factorisation)
Solve this equation:
x2 + 5x – 14 = 0
(x + 7)(x – 2) = 0
x + 7 = 0 or x – 2 = 0
 Factorise first
Now make each bracket
equal to zero separately
 2 solutions
x = - 7 or x = 2
Solve these:
 x2 + 4x + 4 =
 x2 - 7x + 10 =
 x2 + 12x + 35 =
 x2 - 5x - 6 =
 x2 + x - 6 =
(x + 2)(x + 2)
(x – 5)(x – 2)
(x + 7)(x + 5)
(x + 1)(x – 6)
(x + 3)(x – 2)
 x = -2 or x = -2
 x = 5 or x = 2
 x = -7 or x = -5
 x = -1 or x = 6
 x = -3 or x = 2

18. Rearranging formulae
a
a
V = u + at
V
V
a =
V - u
t
x t + u
 t - u
Rearrange the following formula so
that a is the subject
Now rearrange these
P = 4a + 5
1.
A = be
r
2.
D = g2 + c
3.
B = e + h
4.
E = u - 4v
d
5.
6. Q = 4cp - st
Answers:
1. a = P – 5
4
2. e = Ar
b
3. g = D – c
4. h = (B – e)2
5. u = d(E + 4v)
6. p = Q + st
4c

19. Curved graphs
y
x y
x
y
x
y = x2
Any curve starting
with x2 is “U” shaped
y = x2 3
y = x3
y = x3 + 2
Any curve starting
with x3 is this shape
y = 1/x
y = 5/x
Any curve with
a number /x
is this shape
There are three specific types of
curved graphs that you may be
asked to recognise.
If you are asked to draw an accurate
curved graph (e.g. y = x2 + 3x – 1)
simply substitute x values to find y
values and thus the co-ordinates

20. Graphs of y = mx + c
In the equation:
y = mx + c
m = the gradient
(how far up for every
one along)
c = the intercept
(where the line
crosses the y axis)
y
x
Y = 3x + 4
1
3
4
m
c

21. y
x
Graphs of y = mx + c Write down the equations of these lines:
Answers:
y = x
y = x + 2
y = - x + 1
y = - 2x + 2
y = 3x + 1
y = 4
y = - 3

22. y
x
x = -2
x  - 2
y = x
y < x
y = 3
y > 3
Graphing inequalities
Find the
region that
is not
covered
by these
3 regions
x  - 2
y  x
y > 3

23. Graphing simultaneous equations Solve these simultaneous
equations using a graphical
method : 2y + 6x = 12
y = 2x + 1
Finding co-ordinates for 2y + 6x = 12
using the “cover up” method:
y = 0  2y + 6x = 12  x = 2  (2, 0)
x = 0  2y + 6x = 12  y = 6  (0, 6)
Finding co-ordinates for y = 2x + 1
x = 0  y = (2x0) + 1  y = 1  (0, 1)
x = 1  y = (2x1) + 1  y = 3  (1, 3)
x = 2  y = (2x2) + 1  y = 5  (2, 5)
y
x
2
4
6
8
-8
-6
-4
1 2 3
-4 -3 -2 -1 4
-2
2y + 6x = 12 y = 2x + 1
The co-ordinate of the point where
the two graphs cross is (1, 3).
Therefore, the solutions to the
simultaneous equations are:
x = 1 and y = 3

24. Solving other equations graphically If an equation equals 0 then
its solutions lie at the points
where the graph of the
equation crosses the x-axis.
e.g. Solve this equation graphically:
x2 + x – 6 = 0
All you do is plot the
equation y = x2 + x – 6
and find where it
crosses the x-axis
(the line y=0)
y
x
2
-3
y = x2 + x – 6
There are two solutions to
x2 + x – 6 = 0
x = - 3 and x =2
Similarly to solve a cubic equation (e.g. x3 + 2x2 – 4 = 0) find where the
curve y = x3 + 2x2 – 4 crosses the x-axis. These points are the solutions.