Intermediate Tier - Algebra revision

Contents :

Collecting like termsMultiplying terms togetherIndicesExpanding single bracketsExpanding double bracketsSubstitutionSolving equationsFinding nth term of a sequenceSimultaneous equationsInequalities

Factorising – common factorsFactorising – quadraticsOther algebraic manipulationsSolving quadratic equationsRearranging formulaeCurved graphsGraphs of y = mx + cGraphing inequalitiesGraphing simultaneous equationsSolving other equations graphically

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

Multiplying terms together Remember your negative numbers rules for multiplying and dividing

Signs same +ve answerSigns 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

30ac12s2

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

4st35y-6a2

81k2

Indices

(F2)4

t2 t2

b1

a2 x a3 c0 a4

x7 x4

2e7 x

3ef2

4xy3 2xy

5p5qr x 6p2q6r

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 - 1212c + 30-2d - 2gcd + 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

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 + 126a2 – 5a – 415a2 + a – 28

7p2 + 11p – 6

5. (c + 7)2 =6. (4g – 1)2 =

c2 + 14c + 4916g2 – 8g + 1

Substitution If a = 5 , b = 6 and c = 2 find the value of :

3a ac

4bca

(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

Solving equationsSolve the following equation to find the value of x : 4x + 17 = 7x – 1

17 = 7x – 4x – 117 = 3x – 1

17 + 1 = 3x18 = 3x18 = x 3

6 = xx = 6

Take 4x from both sides

Add 1 to both sides

Divide both sides by 3

Now solve these: 1. 2x + 5 = 172. 5 – x = 23. 3x + 7 = x + 154. 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 = 10x = 12

(10 x 10)+(3 x 10) = 130 Too lowToo high(12 x 12)+(3 x 12) = 208

etc.

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

1500y

2y

Find the size of each angle

Angles are:300,600,1200,1500

2v + 394v + 5

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

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 – 12n + 45n – 26n + 147n

And these sequences

n2

n2 + 2-2n + 22-3n + 4320n - 14

Simultaneous equations

4a + 3b = 176a 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 = 3418a 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 = 173b = 17 - 83b = 9b = 3

So the solutions are:a = 2 and b = 3

Now solve: 5p + 4q = 242p + 5q = 13

InequalitiesInequalities can be solved in exactly the same

way as equations

14 2x – 814 + 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 > 42. 5x – 3 123. 4x + 7 < x + 134. -6 2x + 2 < 10

Find the range of solutions for these inequalities :

X > 1 X = 2, 3, 4, 5, 6 ……orX 3 X = 3, 2, 1, 0, -1 ……orX < 2 X = 1, 0, -1, -2, ……or-2 X < 4 X = -2, -1, 0, 1, 2, 3or

Factorising – common factorsFactorising 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)

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

xx

Factor pairs

of - 22:-1, 22- 22, 1- 2, 11- 11, 2

Find the pair which

give - 9x2

-22

xx

-11-11x

2x2

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)

Fully factorise this expression:4x2 – 25

Look for 2 square numbers separated by a minus. SimplyUse 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 manipulationsCancelling 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)

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 solutionsx = - 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

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 + h4.

E = u - 4vd

5.

6. Q = 4cp - st

Answers: 1. a = P – 5 42. e = Ar b

3. g = D – c

4. h = (B – e)2

5. u = d(E + 4v)

6. p = Q + st 4c

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

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

13

4m

c

y

x

Graphs of y = mx + c Write down the equations of these lines:

Answers:y = xy = x + 2y = - x + 1y = - 2x + 2y = 3x + 1y = 4y = - 3

y

x

x = -2

x - 2

y = x

y < x

y = 3y > 3

Graphing inequalities

Find the region that

is not covered by these 3 regionsx - 2y xy > 3

Graphing simultaneous equations Solve these simultaneous equations using a graphical method : 2y + 6x = 12

y = 2x + 1Finding co-ordinates for 2y + 6x = 12using 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 + 1x = 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

x2

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

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

x2-3

y = x2 + x – 6

There are two solutions tox2 + 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.