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Algebra Formulas

The following table contains alternative ways of expressing algebraic functions.

General

a-x = 1 / ax

a1/x = x√a

a-1/x = 1 / x√a

(a.b)x = ax . bx

(ax)y = ax.y

aˣ⁺ʸ = aˣ . aʸ
aˣˉʸ = aˣ / aʸ

e.g.: a²˙⁵ = a² . a⁰˙⁵
e.g.: a⁰˙²⁵ = a¹ / a⁰˙⁷⁵

xⁿ/yⁿ + 1 = (xⁿ + yⁿ) / yⁿ

(x - a) / A = x/A - a/A

(x² - a²) = (x - a).(x + a)

Simplify: (x + b)/(x² - a²)
Note: (x² - a²) = (x - a).(x + a)

(x + b)/(x² - a²) = (x + b) / [(x - a).(x + a)]
(x + b) / [(x - a).(x + a)] = A / (x - a) + B / (x + a) {where A & B are unknown}
(x + b) = A.[(x - a).(x + a)] / (x - a) + B.[(x - a).(x + a)] / (x + a)
(x + b) = A.(x + a) + B.(x - a)
Find A:
set x = a
(x + b) = A.(a + a) + B.(a - a)
(x + b) = 2.a.A
A = (x + b) / 2.a
Find B:
set x = -a
(x + b) = A.(-a + a) + B.(-a - a)
(x + b) = B.(-a - a)
B = (x + b) / 2.-a
(x + b)/(x² - a²) = (x + b)/[2.a.(x - a)] + (x + b)/[2.-a.(x + a)]
cancel (x + b):
1/(x² - a²) = 1/[2.a.(x - a)] - 1/[2.a.(x + a)]

Continue in order to prove the above:
1/[(x - a).(x + a)] = 1/[2.a.(x - a)] - 1/[2.a.(x + a)]
1 = [(x - a).(x + a)]/[2.a.(x - a)] - [(x - a).(x + a)]/[2.a.(x + a)]
1 = (x + a)/(2.a) - (x - a)/(2.a)
1 = [1/(2.a)] . [(x + a) - (x - a)]
2.a = [(x + a) - (x - a)]
2.a = x + a - x + a
2.a = a + a

Factorial: e.g. 5! = 5x4x3x2x1

Binomial

(a+b)n = an + nC1.a(n-1).b + nC2.a(n-2).b2 + ..... + nCr.a(n-r).br + ..... bn
Where:
nCr = n! / (n-1)!.r!
nC1 = n
nC2 = n! / (n-2)!.2! = n.(n-1) / 2!
nC3 = n! / (n-3)!.3! = n.(n-1)(n-2) / 3!
etc.
first few terms are as follows;
(a+b)n = an + n.a(n-1).b + n(n-1).a(n-2).b²/2! + n(n-1)(n-2).a(n-3).b³/3! + ..... + bn

If; 0 = ax² + bx + c
then; x = -b ± (b² - 4.a.c)½ / 2.a

Simple and Compound Interest

The following table contains formulas for calculating simple and compound interest.

Where: P = the principal sum, p = percentage interest, n = payment term (years), q = payments per year, I = interest paid over full term, m = amount of each payment & Pn = total amount paid over 'n' years

Simple

I = P.p.n

Pn = I + P

Compound

I = Pn - P

Pn = P.(1 + p/q)n.q

m = Pn / n.q

Net-Present-Value

V = Pn / (1+p/q)n.q

Discount(simple) = Pn(simple) - V

Discount(compound) = Pn(compound) - V

Progressions

The following table contains the formulas for arithmetic and geometric progressions.

Note: r = 2nd term ÷ 1st term, d = 2nd term - 1st term, n = number of terms

Arithmetic

nth term = 1st term + d.(n-1)

Σn terms = n.[2 . 1st term.(n-1)] / 2

Geometric

nth term = 1st term . r(n-1))

Σn terms = 1st term . (1 - rn) / (1 - r)      [r < 1]

Σn terms = 1st term . (rn - 1) / (r - 1)      [r > 1]

Logarithms

The following table contains alternative ways of expressing logarithmic functions.

Note: 'base' refers to the logarithmic base, which can be any positive number
The most common bases are 10 and 2.71828182845905 (the base for natural logs, normally written thus 'ln(x)')

e = 1 + 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + ..... = 2.71828182845905

ex = 1 + x¹/1 + x²/2 + x³/3 + x⁴/4 + x⁵/5 + x⁶/6 + x⁷/7 + ....

e-x = 1 - x¹/1 + x²/2 - x³/3 + x⁴/4 - x⁵/5 + x⁶/6 - x⁷/7 + ....

y = Logbase(basey)

Log x = Logbase(x)
Anti-Log x = basex

logbase(x) = loga(x) / loga(base)
Where 'a' can be any number; 1, 2.71828182845905, 10, etc.

base(a+b+c) = base(a).base(b).base(c)
e.g.: exp(a+b+c) = exp(a).exp(b).exp(c)

Logbase(x) - Logbase(y) = log10(base).x/y

Logbase(xa) = a.Logbase(x)

Logbase(x) - Logbase(y) = z
Same as: x/y = basez

1/exp(x) = exp(-x)

eln(x) = x

Xn = Y
n = Log(Y) / Log(X)

Determinants

Determinants are a means of solving simultaneous equations, e.g.
a₁.x + b₁.y + c₁.z = 0
a₂.x + b₂.y + c₂.z = 0
a₃.x + b₃.y + c₃.z = 0

which can be written thus:
ǀa₁,b₁,c₁ǀ
ǀa₂,b₂,c₂ǀ
ǀa₃,b₃,c₃ǀ

The following table contains the procedure for solving determinants.

2ᴺᴰ Order

a₁.w + b₁.x = 0
a₂.w + b₂.x = 0

|a₁,b₁| = a₁.b₂ - a₂.b₁
|a₂,b₂|

To solve a 2ᴺᴰ Order equation you perform the calculation as shown above

3ᴿᴰ Order

a₁.w + b₁.x + c₁.y = 0
a₂.w + b₂.x + c₂.y = 0
a₃.w + b₃.x + c₃.y = 0

|a₁,b₁,c₁| = a₁.|b₂.c₂| - a₂.|b₁.c₁| + a₃.|b₁.c₁|
|a₂,b₂,c₂|         |b₃.c₃|       |b₃.c₃|         |b₂.c₂|
|a₃,b₃,c₃|

To solve a 3ᴿᴰ Order equation you convert to 2ᴺᴰ Order equations as shown above then solve 2ᴺᴰ Order equations

4ᵀᴴ Order

a₁.w + b₁.x + c₁.y + d₁.z = 0
a₂.w + b₂.x + c₂.y + d₂.z = 0
a₃.w + b₃.x + c₃.y + d₃.z = 0
a₄.w + b₄.x + c₄.y + d₄.z = 0

|a₁,b₁,c₁.d₁| = a₁.|b₂.c₂.d₂| - a₂.|b₁.c₁.d₁| + a₃.|b₁.c₁.d₁| - a₄.|b₁.c₁.d₁|
|a₂,b₂,c₂,d₂|         |b₃.c₃.d₃|       |b₃.c₃.d₃|         |b₂.c₂.d₂|        |b₂.c₂.d₂|
|a₃,b₃,c₃,d₃|         |b₄.c₄.d₄|        |b₄.c₄.d₄|         |b₄.c₄.d₄|        |b₃.c₃.d₃|
|a₄,b₄,c₄,d₄|

To solve a 4ᵀᴴ Order equation you convert to 3ᴿᴰ Order equations as shown above, then convert to 2ᴺᴰ Order equations then solve 2ᴺᴰ Order equations

The same procedure may be followed for all subsequent Order equations: 5ᵀᴴ, 6ᵀᴴ, 7ᵀᴴ, etc.

Graph with a positive slope

Fig 1. Finding x and y

Slope

A calculation method to find a point on a graph of positive or negative slope

Positive Slope

with reference to Fig 1

x = (x₂-x₁).(y-y₁)/(y₂-y₁) + x₁

y = (y₂-y₁).(x-x₁)/(x₂-x₁) + y₁

 

Negative Slope

Graph with a negative slope

Fig 2. Finding x and y

with reference to Fig 2

x = (x₂-x₁).(y-y₂)/(y₁-y₂) + x₁

y = (y₁-y₂).(x-x₁)/(x₂-x₁) + y₂

 

Further Reading

You will find further reading on this subject in reference publications(19)

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