NUMBERS and ALGEBRA
APPROXIMATION
Round the number 14.874 to:
- the whole number ≈ 15
- 2 decimal places ≈ 14.87 (2 d.p.)
- 3 significant figures ≈ 14.9 (3 s.f.)
POWERS OF NUMBERS
LOGARITHMS
ab is called a power with base a and exponent b.
a x = y ⇒ x = loga y
Ex: 24 = 16, then log2(16) = 4
x = log10 y = log(y) = lg(y) ⇒ 10 x = y
Ex: 103 = 1000, then log10(1000) = 3
(*e ≈ 2.718)
x = loge y = ln(y) ⇒ e x = y
Ex: e2 = 7.389, then ln(7.389) = 2
loga a = 1
Ex: log33 = 1
loga1 = 0
Ex: log191 = 0
loga x + loga y = loga x y
Ex: loga5 + loga4 = loga5·4 = loga20
x
loga x − loga y = loga( )
y
Ex: loga9 - loga3 = loga(9/3) = loga3
2
22 = 4
23 = 8
24 = 16
25 = 32
BOUND
26 = 64
7
If a measuring device gives a reading of x and 2 = 128
has a level of precision p, then the bond is of the 28 = 256
122 = 144
132 = 169
142 = 196
152 = 225
162 = 256
172 = 289
form xmin ≤ x < xmax,
where xmin = x - p/2 and xmax = x + p/2.
PERCENTAGE ERROR of the
MEASUREMENTS
ve − va where ve is the exact value
% error =
of the quantity and va is the
ve
approximate value obtained through measurement.
Ex: For the height h of the person, the bound on
the measurement is 165.5 cm ≤ h < 166.5, and
ve = 165.5 cm, va = 166 cm, so
% errorh =
165.5 − 166
≈ − 0.302 %
165.5
ORDER OF OPERATIONS
1. Parenthesis.
2. Exponents.
3. Multiplication and Division from left to right.
4. Addition and Subtraction from left to right.
FRACTIONS
33 = 27
34 = 81
43 = 64
44 = 256
53 = 125
11 = 121
EXPONENTS
Rule:
Example:
x a x b = x a+b
22 ⋅ 23 = 22+3 = 25
(x a)b = x ab
(32)3 = 32⋅3 = 36
i f z = x y then
63 = 2 ⋅ 33 = 23 ⋅ 33
a
a
a b
z = xy = x x
1
1
= 10−1
= x −1
10
x
1
1
−a
= 2−5
=
x
5
2
xa
2
23
x a xa
a −a ( )3 =
= 23 ⋅ 3−3
( ) = a =x y
3
3
3
y
y
x=x
n
1
2
6=6
1
a
x a = (x a) n = x n
3
loga x k = kloga x
Ex: loga54 = 4loga5
1
loga = − loga x
x
Ex: loga(1/10) = -loga10
logb x
loga x =
logb a
Ex: log74 =
1
2
1
6
26 = (26) 3 = 2 3 = 22
log10 4
= 0.7124
log107
ADDING / SUBTRACTING
LOGARITHMS
x2
Adding Fractions
3
−3
3−3
0
x0 = 1 i f x ≠ 0
2
⋅
2
=
2
=
2
=
1
ln
+ ln x y + ln8 =
Ex: Simplify
a c
a⋅d c⋅b
a⋅d+c⋅b
4y
+ =
+
=
1
MULTIPLYING / DIVIDING
b d
b⋅d d⋅b
b⋅d
ln x 2 + ln
+ ln x + lny + ln8 =
Ex:
4y
EXPONENTIALS
2 3
2⋅4 3⋅3
2⋅4+3⋅3
17 1.Convert fractions into exponents.
2ln x + ln(4y)−1 + ln x + lny + ln8 =
+ =
+
=
=
3 4
3⋅4 4⋅3
3⋅4
12 2.Recognise numbers that are powers or squares of
other numbers. Rewrite these numbers in exponential 2ln x − ln 4y + ln x + lny + ln8 =
Subtracting Fractions
form.
2ln x − (ln 4 + lny) + ln x + lny + ln8 =
a c
a⋅d c⋅b
a⋅d−c⋅b
3.Break original bases into common bases.
− =
−
=
4.Combine common bases using exponential rules. 2ln x − ln 4 − lny + ln x + lny + ln8 =
b d
b⋅d d⋅b
b⋅d
Ex:
Ex: 104 ⋅ 2−12
⋅ 128 = 104 ⋅ 2−12 ⋅ 5−5 ⋅ 128 = 2ln x + ln x − lny + lny + ln8 − ln 4 =
1+y x−1
x(1 + y) y(x − 1)
−
=
−
=
55
−1
y
x
xy
yx
(2 ⋅ 5)4 ⋅ 2−12 ⋅ 5−5 ⋅ 27 = 24 ⋅ 54 ⋅ 2−12 ⋅ 5−5 ⋅ 27 = 3ln x + ln8 − ln 4 = 3ln x + ln8 + ln(4) =
x(1 + y) − y(x − 1)
x + x y − yx + y
x +y
1
8
4−12+7
4−5
−1
−1
=
=
3ln x + ln8 ⋅ 4−1 = 3ln x + ln = 3ln x + ln2
2
⋅
5
=
2
⋅
5
=
xy
xy
xy
4
10
a
c
a
⋅
c
Multiplying Fractions
⋅ =
SURDS
MULTIPLYING / DIVIDING SURDS
b d
b⋅d
Cancel any common factors from the deno- Surds are sometimes the only way to give an Ex: Simplify
minators and the numerators and than multiply Exact Answer.
4 3
3
5
4 1 3
6
3 ) 5 ]2
( a 3 ) 2 ( a a 2 b)4
[(a
( a 5b)4
1⋅1
1
Ex: 3 5
Rules:
Examples:
=
⋅ 8
=
⋅
⋅
=
=
4
5 4 3
2n
4
2
6
3
4
6 10
2⋅2
4
5
4
2n
(
a
b)
(
a
)
(a )
( a b)6
5 = |5|
x = |x|
a
c
a
⋅
d
Dividing Fractions
÷ =
2
10 2
6
6
2n + 1
5
b d
b⋅c
2n+1
5
a 3 b3
a 5 ( a 5 a b )4
x
=
x
3
=
3
( 25 − 12
)
5 ⋅
=
⋅
= a
Ex: 5 ÷ 3 = 5 ⋅ 4 = 5
3 3
12
2 )6
1 )6
a2b4
(a 8 (b 8
a5
2n + 1
2n + 1
3
3
8 4
8⋅3
6
−x = −
x
−8 = − 8 = − 2
11
3
2 3
1
Multiplying/Dividing Fractions with Decimals
−2 ( 10
3 − 2 ) b ( 3 − 4 ) = a 6 b − 12
a
a
4
Convert decimals into fractions, and then n k x = nk x
81 = 81 = 3
calculate.
Rationalizing the Denominator
2
33 ⋅ 2 ⋅ 10
1
÷ 0.6 =
=
Ex: 0.33 ⋅
11
100 ⋅ 11 ⋅ 6
10
EVEN & ODD
Even ± Even = Even
Odd ± Odd = Even
Even*Anything = Even
Even ± Odd = Odd
Odd *Odd = Odd
n
x=
nk
n
n
xk
2=
x
x
= n
y
y
n
xy =
n
6
49
49
=
81
xny
x n y = n x ny
3
8y =
2 y=
23 =
81
3
6
=
8
7
9
1
2
=
1⋅
2
2⋅
2
=
2
22
=
2
2
SCIENTIFIC NOTATION
a× 10k, where 1 ≤ a < 10, and k is a positive or
8 3 y = 2 3 y negative integer.
2
2y=
4y
Ex: 35 = 3.5×101
50 000 000 000 = 5×1010
NUMBERS and ALGEBRA (CONTINUED)
SIMPLIFYING EXPRESSIONS
CROSS MULTIPLYING TECHNIQUE
ABSOLUTE VALUES
• «Like terms» are terms that have the same If both the left hand side and the right hand side of If |x| = y, then x = y or x = -y.
variables, such as x, xy, xyz and so on.
the equation are or can be expressed as fractions, Ex: Solve for x.
• When there is no number in front of variable apply the cross multiply technique.
| 3x + 1 | = 5
(called the coefficient), that number is assumed
3x + 1 = 5 or 3x + 1 = − 5
Ex: x − 3 = 2 ⇒ 3(x − 3) = 2(x + 4) ⇒
to be 1. Ex: x = 1x; xy = 1xy.
x+4
3
4
• Like terms can be combined, dy adding or
x = or x = − 2
3
3x − 9 = 2x + 8 ⇒ x = 17
subtracting coefficients. This process is called
Self check: Plug the solution you have found back Ex: Solve for x.
«simplifying».
| 2x + 1 | = x − 1
into the original equation: Suppose x = 17 is the
Ex: 3x + 4x = 7x
2x + 1 = x + 1 or 2x + 1 = − (x + 1)
solution you found for equation:
5x2y3 + x2y3 = 6x2y3
2
x−3
2
17 − 3
14
2
MULTIPLYING MONOMIALS
x = or x = 0
= ⇒
=
=
3
x+4
3
17 + 4
21
3
Multiply the coefficients and multiply similar
If |x| < a, then x < a or x > -a.
The solution is correct.
variables.
Ex: Solve for x. | 2 − 5x | < 3
!Don’t forget that if a variable has no exponent
EXPRESSION SUBSTITUTIONS
2 − 5x < 3 or 2 − 5x > − 3
shown, then the exponent is 1. Ex: x = x1
1
When an expression cannot be determined, try to
Ex: (n3)(2n6) = 2n9
x > − or x < 1
5
look for a substitution for that entire expression.
(4x2yz)(2x2y3z4) = 8x4y4z
27
6
If |x| > a, then x > a or x < -a.
MULTIPLYING POLYNOMIALS
Ex: If, y = − 3 , what is 3xy2?
x
1
27
Ex: Solve for x. | 4x + 3 | > 5
6
3 6
2 3
Multiply each term in the polynomial by the y = − 3 ⇒ x y = − 27 ⇒ (x y ) = − 27 ⇒
4x + 3 > 5 or 2 − 5x > − 3
x
2
monomial in front.
2
2
x y = − 3 Substituting -3 for xy in 3xy yields
1
x > or x < − 2
!The process of multiplying each term in the 3xy2 = 3(−3) = −9
2
parenthesis by the term in front is called
Remember x 2 = | x |
«expanding».
MULTIPLYING two BINOMIALS
2
• One by one, multiply a term from one Ex: Solve for x. x = 2 ⇒ | x | = 2
x = 2 or x = − 2
binomial with a term from the other binomial.
SIMULTANEOUS LINEAR EQUATIONS • Multiply terms in the following order Front,
LINEAR INEQUALITIES
Outside, Inside, Last (F.O.I.L.)
(a + b)(c + d ) = ac + a d + bc + bd
< - less than
≤ - less than or equal to
Solving by Substitution
> - greater than ≥ - greater than or equal to
1. From any of the two equations, write y in terms Ex: (2x + y)(3x − 2y) =
2x ⋅ 3x + 2x ⋅ (−2y) + y ⋅ 3x + y ⋅ (−2y) = If ax > b and a > 0, then
of x.
6x 2 − 4x y + 3x y − 2y 2 = 6x 2 − x y − 2y 2
x > b/a (or x ∈ (b/a; +∞)
2. Plug the expression for y into the other
If ax ≥ b and a < 0, then
equation.
(x + y)2 = x 2 + 2x y + y 2
x ≤ b/a (or x ∈ (−∞; b/a)
3. Compute y by plugging the value of x into the
2
2
2
Ex: (5x + 2y) = (5x) + 2 ⋅ 5x ⋅ 2y + (2y) =
If ax > b and a = 0, then
expression found in step 2.
2
2
25x + 20x y + 4y
b < 0 (or x ∈ (−∞; +∞)
Ex: 3x − 2y = 4 (1)
If xy > 0, then x > 0, y > 0 or x < 0, y < 0
(x − y)2 = x 2 − 2x y + y 2
{2x − y = 3 (2)
If x/y > 0, then x > 0, y > 0 or x < 0, y < 0
Ex: ( 5 − 3)2 =
If xy < 0, then x > 0, y < 0 or x < 0, y < 0
Multiply equation (1) by 2 on both sides.
2
2
If x/y < 0, then x > 0, y < 0 or x < 0, y < 0
( 5) − 2 ⋅ 5 ⋅ 3 + ( 3) =
Multiply equation (2) by 3 on both sides.
Adding or subtracting the same expression to
6x − 4y = 8 (3)
5 − 2 15 + 3 = 8 − 2 15
both sides of an inequality does not change the
{6x − 3y = 9 (4)
2
2
x − y = (x + y)(x − y)
inequality. Ex: Solve for x.
Subtracting equation (4) from equation (3) gives
2
2
2x − 1 > x + 2 ⇒ 2x > x + 3 ⇒ x > 3
(3 − 5)(3 + 5) = 3 − ( 5) = 9 − 5 = 4
-y = -1 or y = 1.
Ex:
Multiplying or dividing the same positive
Cube of a Binomals
Plugging y back into eq.(1) obtains x = 2.
number to both sides of an inequality does not
(x + y)3 = x 3 + 3x 2 y + 3x y 2 + y 3
change inequality. Ex: Solve for x.
With linear systems of equations, there are three
2x − 1 ≥ 3 ⇒ 2x ≥ 4 ⇒ x ≥ 2
possible outcomes in terms of number of
(x − y)3 = x 3 − 3x 2 y + 3x y 2 − y 3
Multiplying or dividing the same negative
solutions
3
3
2
2
number to both sides of an inequality reverses
x + y = (x + y)(x − x y + y )
• One solution
the sign of the inequality. Ex: Solve for x.
Ex: 2x − 3y = 8
−3x − 1 < 5 ⇒ −3x < 6 ⇒ x > −2
x 3 − y 3 = (x − y)(x 2 + x y + y 2)
{3x − 2y = 7 The solution is x = 1, y = 2
! Don’t divide both side by variables (such as x
• Infinitely many solutions
PRIME NUMBERS
or y), because they might be negative.
Ex: x + y = 5
(1)
The multiplication of two integers yields a third If both numbers are positive, then the inequality
{2x + 2y = 10 (2)
relation between the multiplicative inverses is
The second equation is obtained from the first integer. The first two are called factors, and the third opposite of that between the original numbers.
is called the product. If an integer N is divisible by an
by multiplying by 2.
if a ≤ b, then 1/a ≥ 1/b
integer x, x is a divisor of N. Any integer N that has Ex:
• No solutions at all
exactly two distinct positive divisors, 1 and N, is said
if 0 < a ≤ b, then 1/a ≥ 1/b > 0
Ex: x + y = 3
to be a prime. Ex.: 2, 3, 5, 7, 11, 13, 17, 19, 23 and
if a ≤ b < 0, then 0 > 1/a ≥ 1/b
{2x + 2y = 8
29. Numbers that have more than two factors are
called composite numbers.
if a < 0 < b, then 1/a < 0 < 1/b
No solution because x + y = 3 and x + y = 3.5
Ex: 5(3x + 1) = 15x + 5
2x2(2x2y3 − 4xz4) = 4x4y3 − 8x3z4