mathematics revision guide algebra -...

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Mathematics

Revision Guide

Algebra

Grade C—B

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

qq

Keep numbers

without letters

separate

Algebra Revision Notes

3


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


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


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



This is a balancing equation as it has

‘x’ on both sides. Subtract the smaller amount of

‘x’ from each side

subtract


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

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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

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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




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)

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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

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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’

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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


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


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

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