Download Algebra Revision Notes

Mathematics
Revision Guide
Algebra
Grade C—B
1
Algebra Revision Notes
Simplify
Simplify
5
4
y xy =y
a3 × a4……………………………..…………(1)
y10 ÷ y7 =
y3
5 3
Simplify
Add powers
9
Subtract powers
Simplify
x4 ÷ x9............................................(1)
Multiply powers
15
(y ) = y
Simplify
(q3)4 ..............................................(1)
4a - 2a
Keep numbers
without letters
separate
3y -2y
4a +
3y +
Simplify
3p + 2q – p + 2q…………………(2)
2a - 2y = 6a + y
Simplify
4x5
3n × 2p..........................................(1)
yxp
4y x 5p = 20yp
a2 x a1
4x5
Simplify fully
3p5q × 4p3q2..................................(2)
y4 x y3
5y4a2 x 4y3a = 20y7a3
Simplify fully
t6 ÷ t 2
40 ÷ 5
19 6
40y t
5y9t2
3q 4  2q 5
q3
y19 ÷ y10
..............................................(2)
10 4
= 8y t
2
Algebra Revision Notes
Expand - remove the brackets by multiplying
Expand
Expand
2xy
A) 2(y + 5) = 2y + 10
3y(y + 4)........................................(2)
2x5
B) 3(x + 10) + 2(3x –2) =
3x + 6x
3x + 30 + 6x - 4 =
9x +
Expand
Expand
(x+4)(x+5)
+30 - 4
26
Simplify
Draw a Grid
4(2x + 5) + 2(3x – 2)
…………………...........................(2)
Expand and simplify
(x – 6)(x + 4)
X
+4
X
x2
+4x
+5
+5x
+20
2
X + 4x + 5x + 20
Simplify
X2 + 9x + 20
(2)
Expand and simplify
Draw a Grid
(x+10)(x-2)
X
+10
X
x2
+10x
-2
-2x
-20
(2x + 5)(3x – 2)
X2 +10x –2x –20
Simplify
X2 +8x –20
(2)
3
Algebra Revision Notes
Factorise—put the brackets back in
A)
5x + 30 = 5(x + 6)
Factorise
3y – 12
..........................(1)
5 is the highest factor/number that
goes into both 5 and 30.
X2—5x = x(x -5)
B)
Factorise
……..…………………..(Total 1 mark)
X is the common term that appears
in both parts of the equation.
8x2 + 12xy =
B)
4x(2x + 3y)
x2  10x
Factorise completely
The highest factor/number that goes into 8
and 12 is 4
5x² + 10xy
.....................................(2)
X is also a common term that appears in both
parts of the equation
A)
X2 + 6x +
SECOND - identify which
pair of numbers will add
or subtract to give you
the middle number.
8 = (x + 4) (x + 2)
Factorise
x2 + 7x + 10
FIRST - Identify all the
factors (in pairs) that go
into 8.
1, 8
.......................................................(2)
2, 4
A)
X2 -100 = (x + 10)(x -10)
Factorise
x2 + 2x  15
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’
………………………..(Total 2 marks)
4
Algebra Revision Notes
Solve - find the value of the letter/term
A)
2x + 3 = 10
10 - 3 ÷
+3 becomes –3
To solve - Work the equation
backwards and do the opposite
Solve
4x + 3 = 19
2
x2 becomes ÷2
X = 7/2 or 3.5
B)
4x + 2 = 2x + 18
Subtract the smaller amount of
‘x’ from each side
This is a balancing equation as it has
‘x’ on both sides.
Solve
5t – 4 = 3t + 6
subtract
4x + 2 = 2x + 18
2x + 2 = 18
To solve - Work the equation
backwards and do the opposite
18 –2 ÷ 2
+3 becomes –3
x2 becomes ÷2
X = 16/2 = 8
C)
2(5x + 3) = 3x - 22
Solve
3(x – 4) = x + 24
Expand first
10x + 6 = 3x –
Subtract the smaller amount of
‘x’ from each side
22
7x + 6 = -22
-22 –6 ÷ 7 = -28 ÷ 7
X = -4
5
Algebra Revision Notes
Nth Term — number patterns
Calculate the nth term
A)
Calculate the nth term
5, 7, 9, 11, 13
+3
3, 9, 15, 21, 27
Find the difference between
the numbers and then add a ‘n’
to it.
2n
How do you get
from the
difference to the
first number
Write the first 5 terms
Then add 1
B)
5n + 1
Write the first 5 terms for 7n - 2
(1)
(2)
(3)
(4) (5)
Goes up in 5’s
(1)
(2)
(3)
(4) (5)
5
10
15
20
25
6
11
16
21
26
Substitution — change the letter into a number
x = 10
x = 17
y = -4
y= -2
Find the value of
Find the value of
3x + 6y
4x + 3y
4 x 10 = 40
3 x -4 = -12
40 –12 = 28
6
Algebra Revision Notes
Inequalities - state values which satisfy / solve
State values that satisfy the inequality
–2 ≤ x < 3
A)
x is an integer.
-1 ≤ x < 3
Include
-1
Don’t Include 3
-1, 0, 1, 2
Write down all the possible values of x
...........................................................
(Total 2 marks)
State values that satisfy the inequality
B)
-4 < x ≤ 2
Don’t
Include –4
Include 2
Solve the inequality
5x < 2x – 6
-3, -2, -1, 0, 1
C) Solve
.........................................(2)
4x + 1 ≥ 21
21 - 1 ÷
To solve - Work the equation
backwards and do the opposite
4
+1 becomes –1
x4 becomes ÷4
Solve the inequality
5x + 12 > 2
X ≥ 5
Remember to include the inequality
symbol back into your final answer.
Use the same one that was in the
question.
……………………………(2)
7
Algebra Revision Notes
Algebraic Graphs—straight line
Draw a graph for
Draw a table with ‘x’ values form –2 to 2
x+y=4
X
-2
-1
0
1
2
y
+6
+5
+4
+3
+2
Start at ‘0’
0 +4 = +4
2+2 = +4
8
Algebra Revision Notes
Algebraic Graphs—straight line
Draw a graph for
y= 2x +1
x+2
Forming and Solving Equations
The shape has a perimeter of 53cm.
x+3
STEP 1:
Calculate the value of x to show that this is correct.
Remember that the sides of the shape are the same.
x + 2 + x + 2 + x + 3 +x + 3
Step 1: collect all the terms together
4x + 10
Step 2: solve the equation to find out the value of ‘x’
STEP 2: Solve the equation
4x + 10 = 53
53 - 10 ÷ 4 = 40 ÷ 4
x = 10
9
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
Both coefficients of ‘y’ are the same. +3 and –3. Therefore 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.
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
5a + 3b = 9
2a – 3b = 12
a =.....................................
b = .....................................
(Total 3 marks)
10
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
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.
(b) 5x + 6y = 12
(a) x 3
9x - 12y = 33
(b) X 2
10x + 12y = 24
The symbols are not the same, therefore we can add the equations together to get
to ‘0y’
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
Solve the simultaneous equations
(a) 2x + 3y = –3
(b) 3x – 2y = 28
6y = 12-15
6y = -3
y = -3 ÷ 6
y = -0.5
x = …………………
y = …………………
(Total 4 marks)
11
Algebra Revision Notes
Trial and Improvement
x3 + 2x = 110
x3 – x = 30
The solution is between 4 and 5 to
1 decimal place. Use trial and
improvement to find a solution.
x
x3+2x
(4.5)3 + (2 x 4.5)
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
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.
Changing the Subject—rearranging the equation
y = 2x + t
t = ax
Make ‘x’ the subject
5
Make ‘x’ the subject
x → x2 → +t →
=y
y → -t → ÷ 2 → x
y-t
Work the equation forward
Work the equation
backwards and do the
opposite
=x
2
12