Download Algebra Revision Sheet – Questions 2 and 3 of Paper 1

Algebra Revision Sheet –
Questions 2 and 3 of Paper 1
Simple Equations –
Step 1 Get rid of brackets or fractions
Step 2 Take the x’s to one side of the equals sign and the numbers to the other (remember
to change the sign when crossing the ‘=’)
Step 3 Divide across by the number next to the x.
Example –
Solve for x,
3(2x - 1) = 4x
6x –3 = 4x
6x –4x = +3
2x = 3
3
x
2
Example –
6(x - 7) = 2(x + 3)
6x – 42 = 2x + 6
6x –2x = 6 + 42
4x = 48
48
x
 12
4
Get rid of brackets by multiplying
x’s to one side, numbers to the other
Divide across by 2
Solve for x,
x7 x3

2
6
Cross multiply to get rid of the fractions
Get rid of brackets by multiplying
x’s to one side, numbers to the other
Divide across by 4
Substitution –
Write out the question again substituting numbers for letters.
Example – Find the value of x 2  5 xy when x = 3 and y = -2
x 2  5 xy
= (3) 2  5(3)(2)
= 9 + 30
= 39
Write out expression
Substitute in numbers for x and y
Evaluate
1
Simultaneous Equations –
2 equations we use to find values for x and y
One equation will be linear (just x and y parts)
One equation will be quadratic ( x 2 or y 2 parts)
Example
x + 2y = 3
x  y 2  24
2
x + 2y = 3

 x = 3 – 2y
x 2  y 2  24

 (3  2 y) 2  y 2  24

 3(3 – 2y) –2y(3 – 2y) - y 2 = 24

 9 – 6y –6y + 4 y 2 - y 2 = 24

 3 y 2  12 y  15  0

 y 2  4 y  5  0

 ( y  5)( y  1)  0

 y – 5 = 0 and y + 1 = 0

 y = 5 and y = -1
Equation 1
Equation 2
Rearrange Equation 1 to get x alone
by bringing 2y over the other side.
Equation 2
Insert the above value of x into Equation 2
Open up (3  2 y) 2
Remove brackets by multiplying
Simplify to get a quadratic equation
Divide across by 3
Factorise (put into brackets)
Let each bracket equal zero
y’s to one side, numbers to the other
We must now use these two values of y to get two x values.
We do this by substituting the y values into Equation 1.
y = 5 then
x + 2y = 3
Equation 1
x + 2(5) = 3
x + 10 = 3
x = 3 – 10
x = -7
Substitute y = 5 into Equation 1
Remove bracket by multiplying
x’s to one side, numbers to the other
Evaluate
So x = -7 and y = 5
Answer (-7, 5)
y = -1 then
x + 2(-1) = 3
x –2 = 3
x=3+2
x=5
So x = 5 and y = -1
Answer (5, -1)
Write them as a pair, x value first
Substitute y = 5 into Equation 1
Remove bracket by multiplying
x’s to one side, numbers to the other
Evaluate
Write them as a pair, x value first
2
Algebraic Fractions –
These questions involve solving equations that have fractions and require you to get rid
of the fractions before you can solve the equation.
Step 1 – Find a lowest common denominator (usually all the bottom terms multiplied)
Step 2 – Multiply each top term by any terms NOT underneath it
Step 3 – Remove the common denominator
Step 4 – Bring all the terms to the left of the ‘=’ and simplify
Step 5 – Solve the quadratic equation by factorising (if possible) and letting each bracket
equal 0. If it cannot be factorised you must use the formula (see next page).
Example
Solve
1
2

3
x 1 x  3
1
2
3
Common denominator is (x+1)(x-3)


x 1 x  3 1
1( x  3)(1)  2( x  1)(1)  3( x  1)( x  3)


Top terms times all terms NOT below them
( x  1)( x  3)

 1(x – 3) + 2(x +1) = 3{(x +1)(x – 3)} Remove denominator (bottom section)

 x – 3 + 2x + 2 = 3{x(x+1) +1(x – 3)} Multiply out, open (x +1)(x – 3)
Simplify

 x – 3 + 2x + 2 = 3{ x 2  x  x  3 }
2

 x – 3 + 2x + 2 = 3{ x  2 x  3 }
Simplify
2
Simplify

 3x - 1 = 3x  6 x  9
2
Everything to one side.

  3x  6 x  9  3x  1 = 0
2

  3x  3x  8 = 0
If the equation can be factorised do so and
let each bracket equal 0. If not use the
FORMULA.
Turn over to see how the FORMULA can be used to find our x values. The formula
MUST be learned off by heart.
3
In this case we need to use the FORMULA
 b  b 2  4ac
2a
This formula will give us the two roots (x values) for any quadratic equation. Where
possible however it is easier to factorise and let each bracket equal 0. This formula can
be used in any question on the two papers to solve a quadratic and not just Q2 or Q3.
x=
 3x 2  3x  8 = 0
a  3 b  3 c  8
A quadratic that cannot be factorised.
The values of a, b and c for the formula.
 b  b 2  4ac
2a
The formula
 (3)  (3) 2  4(3)(8)
2(3)
Substitute in the a, b and c values.
 3  9  96
6
3  104
6
3  10.2
6
3  10.2
3  10.2
x=
and x =
6
6
13.2
 7.2
x=
and x =
6
6
x = - 2.2 and x = 1.2
Simplify
Simplify
Remove square root
Split into the + and – parts
Simplify
The roots of the equation.
4
Indices –
With questions involving indices we must break down the numbers on each side of the ‘=’ to the same base
number. We can then let the indices (powers) of each equal each other and solve the simple equation.
To break them down we need to learn the laws of indices below.
(a) 33.34  33.4  37
1
2
(b)
37
 3 7 3  3 4
3
3
1
3
(c) (32 ) 3  32.3  36
2
3
(e) 9  9  3
(f) 64 = 64  4
1
1
(h) 5  3 5 and 3  4  4
3
3
Example
Solve for x in the equation

1
44
1
3 3
 = 2 5 x
1
2
(21 )(2 )  2 5 x
2
8
1
3

4
1
4
 = 2 5 x
The Equation
1
( 2 ) ( 2 2 ) 4 = 2 5 x
1
1
2
64  4
Divide across by 3
Example
1
3
Solve for x in the equation
25 x  56 x
Change 25 into 5 2
Remove bracket to leave both sides in base 5
Let the indices (powers) equal each other
x’s to one side, numbers to the other
(5 2 ) x  56 x
5 2 x  5 6 x
2x = 6-x
2x + x = 6
3x = 6
6
x 2
3
 83  
(g) 8 = 3 8 2 =
3
(d) 30  1
= 2 5 x
1
1  5 x
2
3
 5 x
2
3 = 2(5 – x)
3 = 10 – 2x
2x = 10 – 3
2x = 7
7
x
2
Turn everything into base 2
Multiply the powers
1
2
1
(2 )(2 )  2
1
1
2
Let the powers equal each other
Remove mixed fraction
Multiply across by 2
Multiply to remove bracket
x’s to one side, numbers to the other
Divide across by 2
5
Surds –
Surds are irrational numbers in the form
Some important points:
ab = a . b and a . b = ab

Therefore



24 can be broken down into
4. 6=2 6
6
a
6
a
6
therefore



4
b
2
4
b
Terms with the same surd part can be added and subtracted
Therefore 5 6 +2 6 - 3 6 = 4 6
Any surd squared is equal to the term under the root sign
2
x x
2
x 1  x 1
a a
Therefore
5 5
2
Equations involving surds can be solved by squaring both sides of the ‘=’. This gets rid of
the surd part to leave you with a simple or quadratic equation.
Example

Solve
4  2x  3  8
2x  3  8  4
Take 4 to the other side of ‘=’
2x  3  4
Simplify
2x  3   42
2x – 3 = 16
2x = 16 + 3
2x = 19
19
1
x=
=9
2
2
2
Square both side to remove surd part
x’s to one side, numbers to the other
Simplify
Divide across by 2
Example
x( x 
x x
9
x
3
x
3 x
x 33
x
Simplify
3
3
( x
)( x 
)
x
x
x
9
x
)
3
)
3 x
x
( x
x

3
x
3.3
)
Open up the first bracket
)
Remove brackets by multiplying
x x
Multiply
Simplify
6
Functions –
Questions with functions involve replacing the x in an expression with a number. They
are generally a part (c) and can be asked in many different ways. Quite often it will ask
you to put two numbers in for x and leaves you with a simultaneous equation. For
example if it asks you to find f(3) you put 3 in for x. If it asks you to find f(-5) you put –5
into the equation for x.
Below are 2 examples showing two different types of question
Example
f(x) = (2x + p)(x – p)
given that f(2) = 0 find the value of p.
f(x) = (2x + p)(x – p)
f(2) = (2(2) + p)(2 – p) = 0
(4 + p)(2 - p) = 0
4 + p = 0 and 2 - p = 0
p = -4 and p = 2
Example
write out the function
replace x in the function with 2 and let it = 0
let each bracket = 0
f(x) = ax 2  bx  8
If f(1) = -9 and f(-1) = 3 find the value of a and b
f(x) = ax 2  bx  8
f(1) = a(1) 2  b(1)  8 = -9
a + b – 8 = -9
a + b = -9 + 8
a + b = -1
write out function
put in x = 1 and let function = -9
Simplify
Simplify
This will be Equation 1 of simultaneous
f(x) = ax 2  bx  8
f(-1) = a(1) 2  b(1)  8 = 3
a–b–8=3
a–b=3+8
a – b = 11
write out function
put in x = -1 and let function = 3
Simplify
Simplify
This will be Equation 2 of simultaneous
a + b = -1
a – b = 11
2a = 10
a = 5
Equation 1
Simultaneous Equation
Equation 2
Simultaneous Equation
b’s cancel, a + a = 2a, -1 + 11 = 10
divide across by 2
a + b = -1
5 + b = -1
b = -1 –5
b = -6
Equation 1
Put a value in to get the b value
b’s to one side, numbers to the other
evaluate
a = 5 and b = -6
values for a and b
7
Factor Theorem – Long Division
In this type of question we are generally dealing with a cubic equation such as
x 3  3x 2  4 x  12 = 0
There are a number of things they can ask in this question. The bold writing indicates the different things
we may have to attempt to solve the equation.
Show that something is a factor of the equation e.g. show that x – 2 is a factor.
If x – 2 is a factor then x = 2 is a root and by putting 2 into the equation it will equal 0
Let us see if x –2 is a factor by letting x = 2
x 3  3x 2  4 x  12 = 0
23  3(2) 2  4(2)  12  0
8 + 3(4) – 8 – 12 = 0
8 + 12 – 8 –12 = 0
Write out equation
Substitute in x = 2
Simplify
This is true so x –2 is a factor
If we are NOT told the factor we must find it out through trial and error.
We start off by putting x = 1 into the equation and seeing if it equals 0
If so then x – 1 is a factor.
If not we try x = -1 and see if this equals 0. If so then x + 1 is a factor.
If not we try x = 2 and see if this equals 0. If so then x – 2 is a factor.
If not we try x = -2 and so on until we have one of the factors.
When we have a factor they may ask us to find OTHER factors. We do this through long division.
In the above example we have found out that x – 2 is a factor of x 3  3x 2  4 x  12
To find the other factors divide x – 2 into x 3  3x 2  4 x  12
x 2  5x  6
x  2 x 3  3x 2  4 x  12
Divide x 3 by x
x 3  2x 2
Multiply (x –2) by x 2
5x 2  4 x  12
5x 2  10 x
6x – 12
6x – 12
0
Subtract (change signs) and divide 5x 2 by x
Multiply (x –2) by 5x
Subtract (change signs) and divide 6x by x
Multiply (x –2) by 6
Subtract (change signs)
(x – 2)( x 2  5x  6 ) = 0
(x – 2)(x + 3)(x + 2) = 0
The factors of the equation
We can factorise the second bracket further
When we have the factors (brackets) we find the roots by letting each bracket = 0.
Finding the roots means the same as solving the equation. Basically means get the x values.
(x – 2)(x + 3)(x + 2) = 0
x–2=0
x+3=0
x=2
x = -3
x+2=0
x = -2
Factors
Let each bracket equal 0
The roots of the equation.
Remember if it the question asks us to solve then we must find what x =
8
Inequalities –
These are similar to simple equations but use the greater than, less than signs, greater than or equal to and
less than or equal to signs (  and  ).
The rules for solving are similar to solving normal equations however one important difference is that if we
decide to change all the signs we MUST change the direction of the inequality also.
Example
if –x
 4 then x  -4
Change the signs, change the direction of inequality
If asked to draw a number line we must pay close attention to whether the numbers are:
x  R - these are rational numbers, fractions, decimals etc and are illustrated on the number line with a
shaded line.
x  Z - these are integers, all positive and negative whole numbers and are illustrated on the number line
with dots.
x  N - there are natural numbers, positive whole numbers and are illustrated on the number line with
dots.
Example
Solve 5x  1  4 x  3 for x  R and illustrate on number line.
5x  1  4 x  3
x’s to one side, numbers to the other

 5x  4 x  3  1

 x  2
Evaluate
To show this on a number line
-3
-2
-1
If the question had stated that
-3
-2
-1
0
1
2
3
4
xR
x  N or x  Z we would use the following number line
0
1
2
3
4
xN , xZ
Occasionally we will be asked to find solutions for two sets in which case we need to know the meaning of
the following symbols.
 - this means what numbers would be included in both sets.
\ - this means what is included in one set without the other
for example A\B means what numbers are in set A but not in set B
A is the set 3x –2  4
xZ
1  3x
5
xZ
B is the set
2
A
B
1  3x
5
3x  4 + 2
A is everything less than or equal to 2
2
3x  6
1 – 3x  10
B is everything greater than or equal to -3
x 2
-9  3x
-3  x
 -3,-2,-1,0,1,2 (all the numbers between –3 and 2 inclusive, these
A  B therefore -3  x  2 
Example
numbers are in both set A and set B)
A\B
therefore x  -4
(all the numbers less than –4 are in set A but are
NOT in set B, they are less than 2 but not greater than –3)
9
Rearrange –
This involves using our algebra skills to rearrange equations
Step 1 Remove brackets or fractions if necessary
Step 2 Take anything with the letter we are looking for to the left of the ‘=’ and
everything else to the right of the ‘=’.
Step 3 If there is more than one term now on the left, factorise to get the letter alone.
Step 4 Divide across by the term next to the letter we want to isolate.
Example – Express x in terms of a,b and c (this means get x by itself on left of the ‘=’)
ax + b = c
ax = c – b
x
cb
a
Bring everything with an x to the left, everything
else to the right.
Divide across by a to isolate x
Example – Express b in terms of a and c (this means get b by itself on left of the ‘=’)
8a  5b
c
b
8a – 5b = bc
-5b – bc = - 8a
5b + bc = 8a
b(5 + c) = 8a
8a
b
(5  c)
Get rid of fraction by multiplying across by b
Bring everything with a b to the left, everything else
to the right.
Change all the signs.
Factorise by taking out b
Divide across by (5 + c) to isolate b
Example – Express t in terms of p and q (this means get t by itself on left of the ‘=’)
q t
p
3t
3t(p) = q - t
Get rid of fraction by multiplying across by 3t
3tp + t = q
Bring everything with a t to the left, everything
else to the right.
t(3p + 1) = q
Factorise by taking out t
q
t
Divide across by (3p + 1) to isolate t
(3 p  1)
10