User:Hashim100

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[edit] Lesson 1

[edit] Part 1

Fraction (means broken)

Example of fractions:

$\frac{3}{4},\frac{4}{3},2\frac{4}{2}$

[edit] General form of fraction

$\frac{m}{n}=\frac{m...is...called...numerator}{n...is...called...denumerator}= \frac{chosen...parts}{broken...parts}$

[edit] Explanation

When the numerator is less than the denumerator (Long hand written)

numerator < denumerator (short hand written)

This symbol < means less than

Note: In maths they use symbol to shorten the written work

$\frac{3}{4}$

it is called proper fraction.

And when the numerator is greater than the denumerator (Long hand written)

numerator > denumerator (short hand written)

This symbol > means greater than

$\frac{4}{3}$

it is called an Improper fraction

The combination of a whole number and proper fraction

$2\frac{1}{3}=2+\frac{1}{3}$

it is called mixed fraction

[edit] Extra note

Mixed fraction is the result of an Improper fraction

$2\frac{1}{3}=2+\frac{1}{3}=\frac{2\cdot3}{1\cdot3}+\frac{1}{3}=\frac{6+1}{3}= \frac{7}{3}$

[edit] Part 2 : Addition and subtraction of fraction

An example

$1\cdots\frac{2}{5}+\frac{1}{5}$

$2\cdots\frac{2}{5}+\frac{1}{5}=\frac{2+1}{5}=\frac{3}{5}$

The rule of adding fraction:

Fraction can only be added if they all have the same denumerator

Subtracting of fraction

An example

$1\cdots\frac{2}{5}-\frac{1}{5}$

$2\cdots\frac{2}{5}-\frac{1}{5}=\frac{2-1}{5}=\frac{1}{5}$

The rule of subtracting fraction:

Fraction can only be subtracted if they all have the same denumerator

[edit] Lesson 1: Part 2: Homework

Add the following fractions

$1\cdots\frac{1}{2}+\frac{1}{3}+\frac{2}{5}$

$2\cdots\frac{5}{6}+\frac{1}{5}$

$3\cdots\frac{5}{8}+\frac{2}{4}$

Subtract the following fractions

$1\cdots\frac{1}{2}-\frac{1}{3}-\frac{2}{5}$

$2\cdots\frac{5}{6}-\frac{1}{5}$

$3\cdots\frac{5}{8}-\frac{2}{4}$

[edit] Mixed exercise

$1\cdots\frac{1}{2}+\frac{1}{3}-\frac{2}{5}$

$2\cdots\frac{5}{6}+\frac{1}{5}-\frac{4}{7}$

$3\cdots\frac{2}{8}-\frac{1}{4}+\frac{3}{9}$

Adding with algebra

$1\cdots\frac{1}{x}+\frac{1}{y}$

$2\cdots\frac{2}{a}+\frac{1}{b}$

$3\cdots\frac{3}{2(x-1)}+\frac{2}{y}$

Subtracting with algebra

$1\cdots\frac{1}{x}-\frac{1}{y}$

$2\cdots\frac{2}{a}-\frac{1}{b}$

$3\cdots\frac{3}{2(x-1)}-\frac{2}{y}$

[edit] Part 3 : Indices

${b^{x}=N}\cdots1$

x is called the index and b is called the base

[edit] Multiplying indices

$2\cdot2=2^2=4$

$2\cdot2\cdot2=2^3=8$

$2\cdot2\cdot2\cdot2=2^4=16$

$2\cdot2\cdot2\cdot2\cdot2=2^5=32$

Let examine this one :

$2\cdot2\cdot2\cdot2\cdot2=2^5$

$2^2\cdot2^3=2^{2+3}=2^5$

When a set of same bases are being multiply, with same or different index, just add the indices like the above example

Caution : You can not do this

${2^3}\cdot{3^4}={6^7}$

[edit] General formula of multiplying indices

$b^{m}{\cdot}b^{m}=b^{n+m}$

[edit] Exercise

$1\cdots2^3\cdot2^4$

$2\cdots2^3\cdot2^4\cdot2^{-2}$

$3\cdots{a}^3\cdot{a}^4\cdot{a}^{-2}$

$4\cdots5^3\cdot5^3\cdot5^{8}$

[edit] Dividing indices

$\frac{2\cdot2\cdot2\cdot2\cdot2}{2}=\frac{2^5}{2}=2^{5-1}=2^4=16$

$\frac{2\cdot2\cdot2\cdot2}{2}=\frac{2^4}{2}=2^{4-1}=2^3=8$

$\frac{2\cdot2\cdot2}{2}=\frac{2^3}{2}=2^{3-1}=2^2=4$

$\frac{2\cdot2}{2}=\frac{2^2}{2}=2^{2-1}=2^1=2$

$\frac{2}{2}=2^{1-1}=2^0=1$

When a set of same bases are being divided, with same or different index, just subtract the indices like the above examples

[edit] Special case

$2^0=1\cdots1$

This is a very important case

[edit] General formula of dividing indices

$\frac{b^m}{b^m}=b^{n-m}$

[edit] Exercise

$1\cdots\frac{3^5}{3^3}$

$2\cdots\frac{6^{-2}}{6^3}$

$3\cdots\frac{a^5}{a^3}$

[edit] The power of indices

$2^2\cdot2^2\cdot2^2=2^{2+2+2}=\left(2^2\right)^3=2^{2\cdot3}=2^6$

Examples

$\left(a^5\right)^4=a^{5\cdot4}=a^{20}$

$\left(3^2\right)^5=2^{2\cdot5}=a^{15}$

[edit] General power of indices

$\left(b^n\right)^m=b^{nm}$

[edit] Exercise

$1\cdots\left(2^2\right)^5$

$2\cdots\left(3^3\right)^{-3}$

$3\cdots\frac{\left(4^4\right)^6}{\left(4^4\right)^3}$

[edit] The end of lesson 1

[edit] Lesson 2

[edit] Linear equation

Example of linear equation

$3 x + 4 = 7$

3x, where x is called the variable and the 3 is called the coefficient of x

4 and 7 are just the constant

[edit] Solving linear equation

Solve for x

$4x-4=16\cdots(1)$

Solution

$4x-4=16\cdots(1)$

$4x-4+4=16+4\cdots(2)$

$4x=20\cdots(3)$

$\frac{4x}{4}=\frac{20}{4}\cdots(4)$

$x=5\cdots(5)$

Check the answer

Solve for x

$4x-4=16\cdots(1)$

$4(5)-4=16\cdots(2)$

$20-4=16\cdots(3)$

$16=16\cdots(4)$

[edit] Exercise

$5x+5=30\cdots(1)$

$-5x-5=5\cdots(2)$

$ax+b=c\cdots(3)$

$\frac{3}{x}+\frac{1}{2}=\frac{5}{4}\cdots(4)$

$3y-4=4y+5\cdots(5)$

$\frac{3y}{3}+\frac{1}{2}=\frac{5y}{4}-3\cdots(6)$

Question number 7

Sam gave to his brother 3 box of sweet, 15 are missing the remaining left in the box 21

Find how many was in the box before 15 was missing

[edit] Quadratic equation

Example of quadratic equations

$x^2-4=0\cdots(1)$

$x^2+4x=0\cdots(2)$

$x^2+4x+4=0\cdots(3)$

[edit] Ways of solving quadratic equation

There 4 ways of solving this kind of equation, they are:

By formula

$ax^2+bx+c=0\cdots(1)$

$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\cdots(2)$

$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}\cdots(3)$

Quadratic equation has two solutions (x1 and x2)

By factorisation

By Completing the square

By graph

[edit] The differences between Linear and quadratic equations

$ax-b=0\cdots(1)$

$ax^2+bx+c=0\cdots(2)$

In linear equation the variable x is raised to the power of

$x^1\cdots(1)$

Where as in the quadratic equation the variable x is raised to the power of

$x^2\cdots(2)$

[edit] Solving the quadratic by formula

An example

$x^2+5x+6=0\cdots(1)$

Step 1:

Where a = 1, b = 5 and c = 6

Step 2:

Formula

$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\cdots(2)$

$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}\cdots(3)$

Step 3:

$x_1=\frac{-5+\sqrt{5^2-4\cdot1\cdot6}}{2\cdot1}\cdots(1)$

$x_1=\frac{-5+\sqrt{25-24}}{2}\cdots(2)$

$x_1=\frac{-5+\sqrt{1}}{2}\cdots(3)$

$x_1=\frac{-5+1}{2}\cdots(4)$

$x_1=\frac{-4}{2}\cdots(5)$

$x_1=-2\cdots(6)$

step 4:

$x_2=\frac{-5-\sqrt{5^2-4\cdot1\cdot6}}{2\cdot1}\cdots(1)$

$x_2=\frac{-5-\sqrt{25-24}}{2}\cdots(2)$

$x_2=\frac{-5-\sqrt{1}}{2}\cdots(3)$

$x_2=\frac{-5-1}{2}\cdots(4)$

$x_2=\frac{-6}{2}\cdots(5)$

$x_2=-3\cdots(6)$

Steps 5:

The solution of:

$x^2+5x+6=0\cdots(1)$

is

$x_1=-2\cdots(2)$

$x_2=-3\cdots(3)$

Check the answer

$x^2+5x+6=0\cdots(1)$

$(-2)^2+5(-2)+6=0\cdots(2)$

$4-10+6=0\cdots(3)$

$-6+6=0\cdots(4)$

$0=0\cdots(5)$

$x^2+5x+6=0\cdots(1)$

$(-3)^2+5(-3)+6=0\cdots(2)$

$9-15+6=0\cdots(3)$

$-6+6=0\cdots(4)$

$0=0\cdots(5)$

[edit] Exercise

$x^2+x-6=0\cdots(1)$

$x^2+9x+20=0\cdots(2)$

$2x^2-5x-12=0\cdots(3)$

[edit] Basic Factorising of Linear and quadratic equations

Factorise the following expression

Example 1:

$y=3x+6\cdots(1)$

Step 1:

$y=3\cdot{x}+3\cdot{2}\cdots(2)$

Step 2:

Common 3 and not (x+2)

Step 3:

$y=3\left(x+2\right)\cdots(3)$

Example 2:

$z=xy+y^2\cdots(1)$

Step 1:

$z=x\cdot{y}+y\cdot{y}\cdots(2)$

Step 2:

Common y and not (x+y)

Step 3:

$z=y\left(x+y\right)\cdots(3)$

Example 3:

$T=x\left(a+b\right)+y\left(a+b\right)\cdots(1)$

Step 1:

$T=x\left(a+b\right)+y\left(a+b\right)\cdots(2)$

Step 2:

Common (a+b) and not (x+y)

Step 3:

$T=\left(a+b\right)\left(x+y\right)\cdots(3)$

[edit] Exercise

Factorise the following expression

$x=m^2+m^y\cdots(1)$

$z=10x^2y+15xy^2\cdots(2)$

$y=5x-10\cdots(3)$

[edit] Factorisation of quadratic equation

General quadratic form

$ax^2+bx+c=0\cdots(1)$

Special condition

Let a = 1

$x^2+bx+c=0\cdots(2)$

The question will be of this form

$x^2+bx+c=0\cdots(1)$

The solution will be of this form

$\left(x+A\right)\left(x+B\right)=0\cdots(2)$

To find A and B we need to solve these basic 2 equations, they are:

$A\cdot{B}=c\cdots(1)$

$A+B=b\cdots(2)$

Examples

$x^2+5x+6=0\cdots(1)$

Step 1:

Solution of this form

$\left(x+A\right)\left(x+B\right)=0\cdots(1)$

Step 2:

Solve these 2 equations

$A\cdot{B}=6\cdots(2)$

$A+B=5\cdots(3)$

Step 3:

$2\cdot{3}=6\cdots(2)$

$2+3=5\cdots(3)$

Step 4

Where A = 2 and B = 3

Step 5

$\left(x+A\right)\left(x+B\right)=0\cdots(4)$

Step 6:

$\left(x+2\right)\left(x+3\right)=0\cdots(5)$

Step 7:

Let x = -2

$\left(0\right)\left(x+3\right)=0\cdots(5)$

$0=0\cdots(6)$

Step 8:

Let x = -3

$\left(-3+3\right)\left(x+3\right)=0\cdots(7)$

$0=0\cdots(8)$

Step 9:

The solution of this quadratic equation:

$x^2+5x+6=0\cdots(1)$

is

x = -2 and x = -3

Step 10:

Check the answer

Step 1:

$\left(-2\right)^2+5\left(-2\right)+6=0\cdots(2)$

$4-10+6=0\cdots(3)$

$-6+6=0\cdots(4)$

$0=0\cdots(5)$

Step 2:

$\left(-3\right)^2+5\left(-3\right)+6=0\cdots(2)$

$9-15+6=0\cdots(3)$

$-6+6=0\cdots(4)$

$0=0\cdots(5)$

[edit] Exercise

$x^2+x-2=0\cdots(1)$

$x^2+7x+12=0\cdots(2)$

$x^2-6x+8=0\cdots(3)$

[edit] Simultaneous equation

2 by 2 simultaneous equation

Example 1:

$x+y=5\cdots(1)$

$x-y=1\cdots(2)$

Example 2:

$2x-3y=3\cdots(1)$

$5x+y=16\cdots(2)$

[edit] First method of solving simultaneous equation

The elimination method

Example

$x+y=5\cdots(1)$

$x-y=1\cdots(2)$

Step 1:

(1) + (2)

$x+x-y+y=1+5\cdots(3)$

$2x=6\cdots(4)$

$\frac{2x}{2}=\frac{6}{2}\cdots(5)$

$x=3\cdots(6)$

Step 2:

Substitute x = 3 into (1)

$x+y=5\cdots(1)$

$3+y=5\cdots(2)$

$3+y-3=5-3\cdots(3)$

$y=2\cdots(4)$

Answer

The solution of this simultaneous equation

$x+y=5\cdots(1)$

$x-y=1\cdots(2)$

is

x =3 and y = 2

Check the answer

Step 1:

$x+y=5\cdots(1)$

$x-y=1\cdots(2)$

Step 2:

$3+2=5\cdots(1)$

$3-2=1\cdots(2)$

Example 2:

$2x-3y=3\cdots(1)$

$5x+y=16\cdots(2)$

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