Ratio, Proportion and Variation - Algebra

* Ratio

            - When a number x is to y denoted as x:y where y is not equal to zero. 
            - Simply the fraction of two numbers.

            x:y = x/y

Examples: 

1. What is the ratio of 5 ft to 85 in?

If 1ft = 12 in, then 5 ft is equal to 60. Hence,

60ft/85ft = 12/17 = 12:17

2. What is the ratio of the mass of the blood to the mass of the body of a 75 kg man? If the mass of blood of the man is 5 kg.
5kg/75kg = 1/15 = 1:15

* Proportion
            - is the equality of two ratios or fraction.

a:b = c:d or  a/b = c/d 

Where:

- a and d are called extremes
- b and c are called means

We have Laws in Proportion and here are the following:

1. The product of extremes is equal to the product of means.

From a/b = c/d we will get,

ad = bc

Example:

1/4 = 2/8  

(1)(8) = (4)(2)

        8 = 8

2. The reciprocals of both ratio in a proportion are equal.
b/a = d/c

Example:

1/4 = 5/20

4 = 20/5

4 = 4

3. Interchanging the mean of the first ratio with the extreme of the other will not affect the proportion and vice versa.
a/c = b/d     or    d/c = b/a 

Example:

1/2 = 4/8 by interchanging the mean and extreme,

1/4 = 2/8

1/4 = 1/4

4. Adding the mean on the extreme of the left-side of the equation while adding the extreme on the mean of the right-side of the equation (Vise-versa) will not change the equality.  

(a+b)/b = (c+d)/d 

Example:

1/3 =  5/15

(1+3)/3 = (5+15)/15

4/3 = 20/15

4/3 = 4/3

5. Subtracting the mean on the extreme of the left-side of the equation while subtracting the extreme on the mean of the right-side of the equation (Vise-versa) will not change the equality.

 
(a - b)/b = (c - d)/d 

Example:

1/5 = 5/25

(1-5)/5 = (5-25)/25

- 4/5 = -20/25 

-4/5 = - 4/5

6. (a+b)/(a-b) = (c+d)/(c-d)

Example:
2/7 = 6/21

(2+7)/(2-7) = (6+21)/(6-21)

-9/5 = -27/15

-9/5 = -9/5 

* Variation

3 Principles of Variation:

1. If y varies directly as x then, y = kx.

2. If y varies inversely as x then, y = k/x

3. If y varies jointly as x and z then, y = kxz

Note:

 
- k is the constant of variation.

Example:

* If y varies inversely as x and directly as z and if y = 4 when x = 3 and z = 5. Find z when x = 4 and y = 6.

First, solve for k:(Substitute the first given where x,y and z are known)

y = kz/x

k = xy/z

k = (3 * 4)/5

k = 12/5 or 2.4

Now that we have the value of k, we can solve for z in the second set of values of x and y.

y = kz/x

z = xy/k

z = (4 * 6 )/ 2.4

z = 24/2.4

z = 10

Hence, the value of z is 10 when x is 4 and y is 6.

Functions - Algebra

* Operations of Functions

(f + g)(x) = f(x) + g(x)

(f - g)(x) = f(x) - g(x)

(f . g)(x) = f(x) . g(x)

                                    Note: g(x) is not equal to zero

Where: 

f(x) and g(x) are functions and read as "f of x and g of x."

Example:

Given:

f(x) = 3x + 2
g(x) = 5x - 4

Required: (f + g)(x), (f - g)(x), (f . g)(x), (f/g)(x)

Solution:
* (f + g)(x) = f(x) + g(x)

= 3x + 2 + (5x - 4)

= 3x + 2 + 5x - 4

= 8x - 2

* (f - g)(x) = f(x) - g(x)

= 3x + 2 - (5x - 4)

= 3x + 2 - 5x + 4

= -2x + 6

* (f . g)(x) = f(x) . g(x)

= (3x + 2)(5x - 4)  by using the FOIL method

= (3x)(5x) + (3x)(-4) + (2)(5x) + (2)(-4)

= 15x² - 12x + 10x - 8

= 15x² - 2x - 8

 

* Compositions of Functions

f[g(x)] = fog composition of f of g on x
g[f(x)] = gof composition of g of f on x


Examples:

Given:

f(x) = 3x - 2
g(x) = 2x + 3

Required: fog and gof

Solution: 

* fog
f[g(x)] = 3(2x + 3) - 2

= 6x + 9 - 2

= 6x + 7

* gof
g[f(x)] = 2(3x - 2) + 3

= 6x - 4 + 3

= 6x - 1

Given:

f(x) = x² - 3
g(x) = x + 2

Required: fog and gof

Solution:

* f[g(x)] = (x + 2)² – 3 by using Square of Binomials

= x² + 4x + 4 - 3

= x² + 4x + 1

* g[f(x)] = (x² - 3) + 2

= x² - 1

- it is just like substituting the value of one function inside the other.

* Inverse of a Function

To find the inverse of a given function we need to follow these 4 steps:

1. Let f(x) be equal to y
2. Interchange x and y and vise-versa.
3. Solve for y. 
4. Change y to f^-1(x).

Examples: 

Given:

f(x) = 4x - 5 

Required: f^-1(x)

Solution: 

Step 1:

f(x) = 4x - 5

y = 4x - 5

Step 2:

y = 4x - 5

x = 4y - 5

Step 3:

x = 4y - 5
4y = x + 5

Step 4:

Checking: 

f[f ^-1(x)] = x

x + 5 - 5 = x
x = x


Note:
- If we get the composition of a function and its inverse, the answer will be equal to x. This can be used for checking if the inverse that we have solved is correct.

Factoring - Algebra

Factoring is the reverse process of special product but it is harder to analyse especially if the given has big values. It is an essential part of algebra because it is used for simplifying expressions and functions.

* Types of Factoring

1. Common Monomial Factor

            - Separation of the common variable or numerical coefficient from two or more monomials in an expression.

Formula:

ax + ay = a(x + y)

Examples:

* 2x - 6y = 2(x - 3y)

* x³ + 2x² - x = x(x² + 2x - 1)

2. Difference of Two Squares

Formula:

x² - y² = (x + y)(x - y)

Examples:

* 9x² - 81 = (3x + 9)(3x - 9)

* 25a⁴ - 1 = (5a² +1)(5a² - 1)

3. Perfect Square Trinomial

Formula:

x² + 2xy + y² = (x + y)²
x² - 2xy + y² = (x - y)²

Examples: 

* 25x² - 40x + 16 = (5x - 4)²

* 100x⁴ + 100x² + 25 = (10x² + 5)²

4. Sum and Difference of Two Cubes

Formula: 

x³ + y³ = (x + y)(x² - xy + y²)
x³ - y³ = (x - y)(x² + xy + y²)

Examples:

* 8x³ - 1 = (2x)³ - 1³

= (2x - 1)(4x² + 2x + 1)

* 27x⁹ + 8 = (3x³)³ + (2)³

= (3x³ + 2)(9x⁶ - 6x³ + 4)

* Factoring in Simplifying an Expression

Examples: