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.
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