Post on 26-Apr-2018
2
Simplify
Simplify
y5 x y4 = y9
y10 ÷ y7 =
y3
(y5)3 = y15
4a +
3y +
2a - 2y = 6a + y
4y x 5p = 20yp
5y4a2 x 4y3a = 20y7a3
40y19t6 = 8y10t4
5y9t2
Add powers
Subtract powers
Multiply powers
4a - 2a 3y -2y
4 x 5 y x p
4 x 5 y4 x y3
a2 x a1
40 ÷ 5 y19 ÷ y10
t6 ÷ t2
Simplify
a3 × a4……………………………..…………(1)
Simplify
x4 ÷ x9............................................(1)
Simplify
(q3)4 ..............................................(1)
Simplify
3p + 2q – p + 2q…………………(2)
Simplify
3n × 2p..........................................(1)
Simplify fully
3p5q × 4p3q2..................................(2)
Simplify fully
..............................................(2)
3
54 23
q
Keep numbers
without letters
separate
Algebra Revision Notes
3
Algebra Revision Notes
Expand
A) 2(y + 5) = 2y + 10
B) 3(x + 10) + 2(3x –2) =
3x + 30 + 6x - 4 = 9x + 26
(x+4)(x+5)
X2 + 4x + 5x + 20
X2 + 9x + 20
(x+10)(x-2)
X2 +10x –2x –20
X2 +8x –20
2 x y
2 x 5
Expand Expand Simplify
3x + 6x +30 - 4
Draw a Grid
X +4
X x2 +4x
+5 +5x +20
Simplify
Draw a Grid
X +10
X x2 +10x
-2 -2x -20
Simplify
Expand
3y(y + 4)........................................(2)
4(2x + 5) + 2(3x – 2)
…………………...........................(2)
Expand and simplify
(x – 6)(x + 4)
(2)
Expand and simplify
(2x + 5)(3x – 2)
(2)
Expand - remove the brackets by multiplying
4
Algebra Revision Notes
Factorise—put the brackets back in
A) 5x + 30 = 5(x + 6)
B) X2—5x = x(x -5)
B) 8x2 + 12xy = 4x(2x + 3y)
A) X2 + 6x + 8 = (x + 4) (x + 2)
A) X2 -100 = (x + 10)(x -10)
5 is the highest factor/number that
goes into both 5 and 30.
X is the common term that appears
in both parts of the equation.
The highest factor/number that goes into 8
and 12 is 4
X is also a common term that appears in both
parts of the equation
FIRST - Identify all the
factors (in pairs) that go
into 8.
1, 8
2, 4
SECOND - identify which
pair of numbers will add
or subtract to give you
the middle number.
1, 100
10, 10
25, 4
20, 5
In this example there is no amount of ‘x’ in
the middle of the equation. Therefore you
need to identify numbers which you can add
together or take away to give the answer ‘0’
Factorise 3y – 12
..........................(1)
Factorise x2 10x
……..…………………..(Total 1 mark)
Factorise completely 5x² + 10xy
.....................................(2)
Factorise x2 + 7x + 10
.......................................................(2)
Factorise x2 + 2x 15
………………………..(Total 2 marks)
5
Algebra Revision Notes
Solve - find the value of the letter/term
A) 2x + 3 = 10
10 - 3 ÷ 2
X = 7/2 or 3.5
B) 4x + 2 = 2x + 18
4x + 2 = 2x + 18
2x + 2 = 18
18 –2 ÷ 2
X = 16/2 = 8
C) 2(5x + 3) = 3x - 22
10x + 6 = 3x –
22
7x + 6 = -22
-22 –6 ÷ 7 = -28 ÷ 7
X = -4
To solve - Work the equation
backwards and do the opposite
+3 becomes –3 x2 becomes ÷2
To solve - Work the equation
backwards and do the opposite
This is a balancing equation as it has
‘x’ on both sides. Subtract the smaller amount of
‘x’ from each side
subtract
+3 becomes –3 x2 becomes ÷2
Solve
4x + 3 = 19
Solve
5t – 4 = 3t + 6
Solve
3(x – 4) = x + 24 Expand first
Subtract the smaller amount of
‘x’ from each side
6
Algebra Revision Notes
Nth Term — number patterns
Calculate the nth term
A) 5, 7, 9, 11, 13
Write the first 5 terms
B) 5n + 1
(1) (2) (3) (4) (5)
5 10 15 20 25
6 11 16 21 26
2n +3
Goes up in 5’s
Then add 1
Find the difference between
the numbers and then add a ‘n’
to it. How do you get
from the
difference to the
first number
Calculate the nth term
3, 9, 15, 21, 27
Write the first 5 terms for 7n - 2
(1) (2) (3) (4) (5)
Substitution — change the letter into a number
x = 10
y = -4
Find the value of
4x + 3y
40 –12 = 28
4 x 10 = 40 3 x -4 = -12
x = 17
y= -2
Find the value of
3x + 6y
7
Algebra Revision Notes
Inequalities - state values which satisfy / solve
State values that satisfy the inequality
A) -1 ≤ x < 3
-1, 0, 1, 2
State values that satisfy the inequality
B) -4 < x ≤ 2
-3, -2, -1, 0, 1
C) Solve
4x + 1 ≥ 21
21 - 1 ÷ 4
X ≥ 5
Include
-1 Don’t Include 3
Don’t
Include –4
Include 2
To solve - Work the equation
backwards and do the opposite
+1 becomes –1 x4 becomes ÷4
Remember to include the inequality
symbol back into your final answer.
Use the same one that was in the
question.
–2 ≤ x < 3
x is an integer.
Write down all the possible values of x
...........................................................
(Total 2 marks)
Solve the inequality
5x < 2x – 6
.........................................(2)
Solve the inequality
5x + 12 > 2
……………………………(2)
8
Algebra Revision Notes
Algebraic Graphs—straight line
Draw a graph for
x + y = 4
X -2 -1 0 1 2
y +6 +5 +4 +3 +2
Draw a table with ‘x’ values form –2 to 2
2+2 = +4 Start at ‘0’
0 +4 = +4
9
Algebra Revision Notes
Algebraic Graphs—straight line
Draw a graph for
y= 2x +1
x + 2 Forming and Solving Equations
x + 3
The shape has a perimeter of 53cm.
Calculate the value of x to show that this is correct.
STEP 1:
x + 2 + x + 2 + x + 3 +x + 3
4x + 10
STEP 2: Solve the equation
4x + 10 = 53
53 - 10 ÷ 4 = 40 ÷ 4 x = 10
Remember that the sides of the shape are the same.
Step 1: collect all the terms together
Step 2: solve the equation to find out the value of ‘x’
10
Algebra Revision Notes
Simultaneous Equations—same coefficient
When one of the letters has the same coefficient
(a) 2x + 3y = 0
(b) x—3y = 9
Add the equations together
3x = 9
Solve the equations
x = 9 ÷ 3
x = 3
Now you know the value of ‘x’ put this back into one
of the equations to calculate the value of ‘y’.
The value of ‘x’ put back into the first equation (a)
(2 x 3) + 3y = 0
6 + 3y = 0
3y = -6
y = -6 ÷ 3
y = -2
Both coefficients of ‘y’ are the same. +3 and –3. There-
fore we can add the equations together to get ‘0y’
TIP
If the symbols ‘+’ or ‘-’ are the same for the term you are trying to
eliminate then you subtract the equations from each other.
If the symbols are different for each term then you can add the
equations together.
5a + 3b = 9
2a – 3b = 12
a =.....................................
b = ..................................... (Total 3 marks)
11
Algebra Revision Notes
Simultaneous Equations—different coefficient
When the coefficients are not the same– you have to make them the same!
(a) 3x—4y = 11
(b) 5x + 6y = 12
(a) x 3 9x - 12y = 33
(b) X 2 10x + 12y = 24
Add equation (a) to equation (b)
19x = 57
Solve the equation
x = 57 ÷ 19
x = 3
Put the value of ‘x’ back into equation (b) to find the value of ‘y’
I have used equation (b) as it has a ‘+’ rather than a ‘-’.
(5 x 3) + 6y = 12
15 + 6y = 12
6y = 12-15
6y = -3
y = -3 ÷ 6
y = -0.5
As the coefficients are not the same, we need to make them the same. To do this we can
multiply the equation marked (a) by 3 and equation (b) by 2. This will give me 12y for both.
The symbols are not the same, therefore we can add the equations together to get
to ‘0y’
Solve the simultaneous equations
(a) 2x + 3y = –3
(b) 3x – 2y = 28
x = …………………
y = ………………… (Total 4 marks)
12
Trial and Improvement
Algebra Revision Notes
x3 + 2x = 110
The solution is between 4 and 5 to
1 decimal place. Use trial and
improvement to find a solution.
x x3+2x
4.5 100.125 Too small
4.6 106.536 Too small
4.7 113.223 Too big
4.65 109.844 Less than 110
so 4.7 is closer
to 1dp
(4.5)3 + (2 x 4.5)
Changing the Subject—rearranging the equation
x3 – x = 30
The solution is between 3 and 4.
Use a trial and improvement
method to find this solution.
Give your answer correct to 1
decimal place.
y = 2x + t
Make ‘x’ the subject
x → x2 → +t →
=y
y → -t → ÷ 2 → x
y - t
2
Work the equation forward
Work the equation
backwards and do the
opposite
= x
t = ax
Make ‘x’ the subject
5