Tessellations in Grade VIII

Hii guys, so today we are going to study about a very new and different topic of Maths which is Tessellations. How many of you know what a Tessellations is? I believe not many but still a few might know. So let's start without wasting any time.

Tessellation is the process in which we create a two dimensional plane by repeatedly joining the geometric shapes without any overlaps and gaps. In simple language a tessellation is something in which we have some shapes and we join them together to form a regular sheet without any overlaps and gaps. We can see tessellations throughout our history, from ancient architecture to modern art.

For Example the Jigsaw puzzle we used to play when we were little. Its the best example we can get for our topic. It corresponds with the everyday term tiling which refers to applications of tessellations, often made of glazed clay. It also exist in the nature like the Honeycomb which also has a Tessellation structure.

For Example -

a tessellation of triangles

a tessellation of squares

a tessellation of hexagons

The above given figures are some examples of Tesselations, for more understanding.

Now let's study the different topics that comes under it.

When discussing about the tilings a form of tessellation, we see that they are multicolored, so we need to specify whether the colors are the part of tiling or just the part of an illustration.

Let's come to the theorem in our topic, it is known as the four color theorem.

The theorem states that every tessellation in every Euclidean plane, with a set of four available colors, each tiled when colored in one color such that no tiles of equal color meet at a curve of positive length. Note that the coloring guaranteed by the four-color theorem will not in general respect the symmetries of the tessellation. To produce a coloring which does, as many as seven colors may be needed.

Any arbitrary Quadrilateral when taken can also form a tessellation with 2-fold rotational centre at the mid point of all the four sides of the Quadrilateral. In a similar manner, we can construct a parallelogram subtended by a minimal set of translation vectors, starting from a rotational center. We can easily divide this by one diagonal, and take one half of the triangle as fundamental domain. Such a triangle formed will have the same area as of the quadrilateral and can be constructed from it by cutting and pasting.

Now we come over to the types of tessellations, we classify then in two types – the regular tessellation and irregular tessellation.

A regular tessellation is one which is highly symmetric and is made of congruent regular polygon. There are only three regular tessellations which exist. They are of Equilateral triangles, squares and regular hexagons. We even have a more accurate one which is edge-to-edge tessellation. In this type of tessellation the side of one polygon is fully shared with the sides of another polygon. There is no room for partial sharing of sides.

The most common example of the apperiodic pattern is the Penrose tilings which is formed using two different types of polygons and adds beauty to the walls. We also have the self dual tessellation and the example for it is the Honeycomb. Another example for it is shown in the figure below.

 

Tessellations is not only used in architecture but also to design the computer models. In the computer graphics, tessellation technique is used manage the datasets of polygons and divide them into polygon structure. This is generally used for rendering. The data is generally tessellated into triangles also known as triangulation.

The CAD application in our computers is generally represented in analytical 3D curves and surfaces.

The mesh of a surface is usually generated per individual faces and edges so that original limit vertices are included into mesh. The following three basic fundamentals for surface mesh generator are defined to ensure the approximation of the original surface that suits the need for the further processing

• The maximum allowed distance between the planar approximation polygon and the surface. This parameter ensures that mesh is similar enough to the original analytical surface (or the polyline is similar to the original curve).
• The maximum allowed size of the approximation polygon (for triangulations it can be maximum allowed length of triangle sides). This parameter ensures enough detail for further analysis.
• The maximum allowed angle between two adjacent approximation polygons (on the same face). This parameter ensures that even very small humps or hollows that can have significant effect to analysis will not disappear in mesh.

Some geodesic domes are designed by tessellating the sphere with triangles that are as close to equilateral triangles as possible.

So as we have read above that a honeycomb is a natural tessellation, there are some more tessellations in our nature like when the lava flows, it often displays columnar jointing, due to the contraction forces that are present, it causes the lava to break into cracks when it is cooled. These cracks often gets broken in the hexagonal columns. The tessellation is even present in the flower petals, leaves, tree bark, or a fruit. It is not easily visible but when we look through the leaf etc. with the help of a microscope then we can easily see the formation.

It seems like what we have learned till now seems to be theoretical, so let's now have something mathematical in it.

Now the topic that comes under it is, Number of Sides of a Polygon versus the number of Sides at a Vertex.

If we assume an infinite long tiling, then let a be the number of sides of a polygon and b be the average number of sides which meet at the vertex. Then ( a – 2 ) ( b – 2 ) = 4. For example, we have the combinations (3, 6), (31/3, 5), (33/4, 42/7), (4, 4), (6, 3) for the tilings.

For a tiling which repeats itself, one can take the averages over the repeating part. In the general case the averages are taken as the limits for a region expanding to the whole plane. In cases like an infinite row of tiles, or tiles getting smaller and smaller outwardly, the outside is not negligible and should also be counted as a tile while taking the limit. In extreme cases the limits may not exist, or depend on how the region is expanded to infinity.

For a finite tessellation and a polyhedra we have

( a – 2 ) ( b – 2 ) = 4 (1 – X/f ) ( 1 – X/v ).

In this expression f represents the number of faces of a polygon and v represents the number of vertex and X is the euler characterstic.

The formula follows observing that the number of sides of a face, summed over all faces, gives twice the total number of sides in the entire tessellation, which can be expressed in terms of the number of faces and the number of vertices. Similarly the number of sides at a vertex, summed over all vertices, also gives twice the total number of sides. From the two results the formula readily follows.

In most cases the number of sides of a face is the same as the number of vertices of a face, and the number of sides meeting at a vertex is the same as the number of faces meeting at a vertex. However, in a case like two square faces touching at a corner, the number of sides of the outer face is 8, so if the number of vertices is counted the common corner has to be counted twice. Similarly the number of sides meeting at that corner is 4, so if the number of faces at that corner is counted the face meeting the corner twice has to be counted twice.

A tile with a hole, filled with one or more tiles is not permissible, because the network of all sides inside and outside is disconnected. However it is allowed with a cut so that the tile with the hole touches itself. For counting the number of sides of this tile, the cut should be counted twice.

The another topic that we are going to study is the Tessellation of other spaces. As well as tessellating the 2-dimensional Euclidean plane, it is also possible to tessellate other n-dimensional spaces by filling them with n-dimensional polytopes. Tessellations of other spaces are often referred to as honeycombs.

So this is all we have for the topic tessellations and I hope that you would have understood what a tessellation is. This topic we have learned is totally a theoretical one, but it do have some problems to do, and I expect that you will be able to understand how to solve them by using the only formula which we have learned so far in this topic.

Quadratic Equation in VIII Grade

Hello friends, in today's class we are going to learn what a quadratic equation is and how can we solve it.

 

A Quadratic equation is a second order univariate polynomial equation. A general quadratic equation can be written as a x² + b x + c, here x is the variable and a,b and c are the constants with a ≠ 0. There are two solutions of the quadratic equations, either both are real or both are complex.

There are various methods of solving quadratic equations but the ones that we have to learn are as follows-

1. factoring
2. completing the square
3. by using the quadratic formula.

Let's start solving the quadratic equations by using the above methods one by one.

 

We start by using the factoring method, let's understand by using the general formula.

The general equation is ax² + bx + c. We follow some general steps to solve the quadratc equation.

1.multiply the constants a and c without forgetting their signs.

2.Write down b, with its sign.

3.Write all the possible factors of the products of ac and find which pairs add up to give b.

4.Now rewrite the x terms as the sum of two terms with these selected numbers as the coefficients.

So by this we can find the solution of any quadratic equation.

For Example –

1. Solve x² + 5 x + 6.

Answer – the possible combination of the middle term is 1x+4x and 2x+3x.

The product of a and c is 6 x 1= 6. So the second combination is the suitable option for this quadratic expression. So now it can be written as x² + 3 x + 2 x + 6.

Now we take x common from the first two terms and 2 from the last two terms.

Now the quadratic equation becomes x ( x + 3 ) + 2 ( x + 3 ).

This time we take (x + 3) common, so the equation now becomes (x + 2 ) ( x+ 3 ).

the value of x = -3, -2.

The solution for the quadratic equation x² + 5 x + 6 is x = -2, -3.

2. Solve x² – 3 = 2 x.

We bring 2x on the left hand side. So now the equation is x² – 2 x – 3.

now breaking the middle term we see that the possible combination is -1x -1x and -3x + x.

The product of a and c is -3. So the second combination is the suitable one for this quadratic expression.

Now the equation becomes x² – 3 x + 1 x – 3.

now by taking x common from the first two terms and 1 from the last two terms we get,

x ( x – 3 ) + 1 ( x – 3 ). Now taking x - 3 common from both the terms. We get

(x + 1 ) ( x – 3 ).

x = -1, 3.

so the solution to the quadratic equation x² - 2 x – 3 is x = -1, 3.

3. Solve x² - 9.

Answer- we can see that it is in a² - b² form, so we can break it as ( a + b )( a – b ).

So our equation becomes ( x + 3 ) ( x – 3 ).

now x = -3 , 3.

So the solution to the quadratic equation x² - 9 is x = -3, 3.

 

2. By completing the squares.

For the general equation a x² + b x + c.

In this method we follow these steps.

1. take the constant to one side and make the coefficient of x² = 1.
   

2. add the square of the half of coefficient of x on both the sides in the expression as shown above.
Now we can easily find out the value of x.
Let's clear the doubts with the help of an example.
1. Solve x² - 5 x + 6.
we follow the same general steps.
= x² - 5 x = -6.
add the square of the half of coefficient of x on both sides.
= x² - 5 x + 25/4 = 25/4 – 6.
so the left hand side becomes
= (x - 5/2)² = 25/4 – 6.
= (x – 5/2) = ± ¼.
Now value of x = 5/2 + ¼ or 5/2 – ¼.
x = 3, 2.
2. Solve x² – 3 = 2 x.
= x² - 2 x = 3.
= x² - 2 x + 1 = 3+1.
= (x -1)² = 4.
= (x -1) = ± 2.
= x = 1+ 2 or 1-2.
= x = 3, -1.
3. solve x² - 9.
In this we can see that the coefficient of x = 0, so we bring the constant to the right hand side.
= x² = 9.
= x = ± 3
= x = -3, +3.
Let's now use the third method to find the solution to the quadratic expressions that is by using the quadratic formula.


3. For the general equation a x² + b x + c.

The value of


The two values are using the positive and the negative signs.
But how did we got this formula?
We got this by using completing the squares method, which we have just learned.
Let's derive it for our general equation,
1.


2.


3.



4.






So this is how we got our general formula. Let's now solve some questions using this quadratic formula.
1. Solve – .




Here a = 2, b = -3 and c = 1/2. now we put these values in the formula and find the values of x.




So we get two values of x.
2. Solve-

for x by using the quadratic formula. Here a = 1, b = 2 and c = 10 .

so we put these values in the formula.

So the value of x becomes,

 

Now as we have solved the quadratic equations by using the different methods, so let's now solve some word problems on Quadratic equations.

By using the same general formulas we can solve the word problems on quadratic equations, What we have to do is just form the quadratic expression from the given problem and the rest could be found using the above three formulas.

Let's solve some of them for more understanding.

 

1. Question - The length of a rectangle is 6 inches more than its width. The area of the rectangle is 91 square inches. Find the dimensions of the rectangle.

Answer - The formula for the area of a rectangle is length x breadth.

So if width is assumed as x than the other side i.e. the breadth will be x+6 as per the given condition.

So the area is x ( x + 6 ), and the area is given as 91.

So x ( x + 6 ) = 91.

now the quadratic equation is x² + 6 x – 91 = 0.

so now by solving this equation we will get x = 7, -13.

since the length of the triangle cannot be negative so the breadth of the triangle is 7 and its length is 7 + 6 = 13.

2. Question - The product of two consecutive odd integers is 1 less than four times their sum. Find the two integers?

Answer – let us assume the first number to be n. Since the second number is the consecutive odd one so the value of second number will be n + 2.

now we implement the given problem.

The product of two numbers will be n(n+2), this is equal to 1 less than 4 times its sum,

the sum is 2 n + 2, 4 time of it 4 x (2 n + 2 ), and it is 1 less = 4 ( 2 n + 2 ) -1.

so the final quadratic equation is n² + 2 n = 8 n + 8 – 1.

= n² - 6 n – 7.

By solving the above quadratic equation we get the values of x as 7, 9 and -1, 1.

So these are the two possible combinations possible for the numbers.

3. Question-The hypotenuse of a right triangle is 6 more than the shorter leg. The longer leg is three more than the shorter leg. Find the length of the shorter leg.

Answer – for solving these equations we should know about pythagoras theorem.

According to pythagoras theorem (side)² + (side)² = (hypotenuse)²

Take one side as x, than the other side is x + 3 and the hypotenuse is x + 6.

now by using the pythagoras theorem.

x² + ( x + 3 )² = ( x + 6 )²

so we get, x² - 6 x – 27.

now by solving this we will get x = -3, 9. since the side cannot be negative so the value of will be 9.

Variables in VIII Grade

Hello friends in today's session we are going to learn about variables, expressions, equations and inequalities. Let's just take them one by one. We will start with variables.

A variable in mathematics is a quantity which may change itself according to the given conditions. It is generally represented by English alphabets x, y z, a, b, c etc. If the given condition has no variables than it is said to have constants.

A constant is a value which remains unchanged like numbers 3, 4, 5 etc.

For Example- x + 2 = 6, where x is variable and 2,6 are constants.

Variables are further categorized as dependent and independent variables.

The variables whose value depend on other terms in the condition are known as the dependent variables whereas the ones which take different values freely are known as the independent variables.

When variables come together with some constants then they form an expression. Like y = f(x) is a general expression. This equation says that y is a function of x, so the variable x is an independent variable and the variable y is dependent on x.

y = f(a, b, c), here also y is a function of a,b and c. So the variable y is dependent whereas a, b and c are independent variables.

Now the expression properly is a mathematical term which uses variables and the constants with different mathematical operations.

Example-

x

2

x + 2

y + 8

x + y

2x + 9z etc.

Now we move towards the equations.

An equation is an expression which is equal to some other constant or variable. Equations are used to find the values of the given variables. The word problems in the maths are generally solved by forming equations.

For Example-

2 = 2

17 = 2 + 15

x = 7

7 = x

t + 3 = 8

3 × n +12 = 100

are the equations.

Now since we know what an equation is, so we now will try solving the given equation.

This is done by two methods either by solving the given equation to bring the value of the variable or by substituting the different values of variable that is the hit and trial method.

For Example

Solve 4z + 12 ?

this we will done by solving for z that is the first method.

To make it easy we use some general steps.

Step 1- simplify the equation to bring it in the simplest form.

The given equation is already in its simplest form.

Step 2- equate the give equation to 0 by bringing all the terms to one side .

4z + 12 = 0.

Step 3 – bring the constants on one side and leave the variables on the other side.

4z= - 12.

Step 4 – make the coefficient of the variable equal to 1.

in this case we will divide the equation by 4.

4z/4 = -12/ 4 , so our equation becomes

z = -3 .

so the result after solving 4z +12 = 0 is z = - 3.

This method though is very easy for small problems but becomes very difficult for complex problems.

So we use the hit and trial method for complex problems

For Example-

Solve 4z + 12 = 0 ?

we substitute different values of z and see if it becomes equal to 0.

so put the value of z = 0

= 12= 0

which is not possible.

So we put 1

4+12 = 0

16 = 0 which is also not possible.

Now for z = -1

- 4 + 12 = 0

8 = 0

which is also not possible.

Now we put z = - 3

so

4 x – 3 + 12 = 0

-12 + 12 = 0

0=0

so this the answer for the above gives equation.

Now you will be having one question in your mind that how will you know that which value it will satisfy and how many times we will have to perform this method.

So that is the reason we generally used the first method, and avoid the hit and trial method. The method is only used when the equations formed by the first method becomes very complex.

Let's solve some word problems.

1. The sum of my age and 20 equals 40

to solve this problem, we will first form an equation. Let my age be x, so according to the problem

x + 20, that is the sum of my age plus 20, and this is equal to 40

so the equation is x + 20 = 40 .

so now by using the steps for solving a given equation. We see that the value of x = 20.

2. The difference between my age and my younger sister's age, who is 11 years old, is 5 years.

So in this problem let my age by y, my sister's age is 11 and the difference between our age is 5.

y – 11 = 5 is the equation.

Now by solving the equation by the general steps.

We get the value of y = 6.

so my age is 16.

What we have learned about the equations till now is all about the linear equations that is the equations with exponent as 1.

Now we move to the quadratic equations, those equations which have 2 as their variable powers are known as the quadratic equation. We have to keep one thing in mind before solving the equation and it is that if the equation is linear then there will be one solution for the variable, if quadratic then 2 solutions, if three than 3 solutions and so on.

For Example.

is a Quadratic equation.

The quadratic equation is solved by using a general formula known as the quadratic formula.

This is the general formula that we use to solve the quadratic equations. Now the two solutions will be

Let's now solve some problems on quadratic equations.

For Example

Solve x² + 5x + 6 = 0.

we use the quadratic formula to solve the above equation.

Here a = 1, b= 5 and c = 6.

so we will put it in the formula and get two values of x.

after solving the values x comes out to -2 and -3.

Now we move to the inequalities.

Inequality is used to compare the two sides of the given equation. It is represented by using two symbols, “<” for less than and “>” for greater than. Let's understand this with the help of some notations.

x < y = this means that x is lesser than y.

x > y = this means that x is greater than y.

The other notations used are

1. ≤ for less than or equal to.
2. ≥ for greater than or equal to.
3. ≠ for not equal to.
4. <<much lesser than.
5. >> much greater than.

These notations are not used very commonly, but they do sometimes come in the questions.

There are some properties of inequalities that we need to know.

1. transitivity .

If a > b and b > c then we can say that a > c. Or we can write it as if a < b and b < c than a < c.

2. addition and subtraction property.

If a < b then, a + c < b + c or a – c < b – c.

3. for multiplication.

If a < b then, ac < bc or a/c < b/c.

4. the inverse property.

If a < b then -a > -b.

if a > b then -a < -b.

5. the multiplication inverse.

If a < b then 1/a > 1/b. And if a > b then 1/a < 1/b.

The inequalities are generally solved graphically.

For Example.

1.Solve x + 3 < 2 ?

so we solve the inequality by taking the constant on one side and variables on the other.

= x < -1

so this is the answer for the equation.

This tells us that all the region from -1 to negative infinity is the answer.

2.. solve 2 – x > 0.

for this we will use same procedure.

Which gives x < 2

So all the region from 2 to negative infinity is the answer.

3. Solve: -2y <-8 ?

so by using the reversing property, the equation becomes.

2y > 8

y > 4.

This tells that all the region above 4 is the answer.

We can plot this one also like the others.

Now moving towards much more complex examples.

4. Solve -2 < (6-2x)/3 < 4?

we can see that in this we have two inequalities at the same time.

We will solve it by taking one inequality at a time.

So now we have two inequalities that is, -6< 6 – 2 x and 6 – 2 x < 12.

Solving the fist one we get the answer for x < 6 and the second inequality gives us x > -3.

so the final answer is -3 < x < 6.

So that's all for today and I hope that you will have no problem in solving any of the questions from Variables, expressions, equations and inequalities.

VIII Grade Terms and Sequences

Hello friends, earlier we have studied about Proportional and non-proportional linear relationships. Now today we all are going to learn about the next topic that is terms in arithmetic sequences. It is one of the important areas of study which not just play an important role in eighth grade but also in our higher grades. Before proceeding further let's talk about the basic concept behind Sequence. In simplest mathematics manner it is a systematic list of objects. In mathematics the sequence list can be numbers, algebraic expressions, fractions etc. If the objects in the list are numbers then it is called as numeric sequence. Moving forward we will discuss numeric sequence topic in detail. Hence the sequence with number we mean the numeric sequence. Now each of the number used to form a sequence is called a term. In a sequence there can be finite number of terms or it can have infinite number of terms as well. Finite number of sequences can be stated as finite sequence and a sequence with infinite number of terms can be stated as infinite sequence. If we try to add a series of sequences then we get a series. So the total of the terms used in the sequences is called a series.


Let's take an example to understand the basic concept behind numeric sequences. A series : 1,3,5,7,9 is a finite number sequence with 5 terms. The another expression that is 3 + 6 + 9 + 12 + 15 + 18 is also a finite numeric sequence with six terms.
Similarly we can easily identify the infinite sequences like 3,1,6,4,5,4,...... etc.


Now moving further we are going to see the different types of sequences. There are various kinds of sequences in mathematics, but in eighth standard we all are going to discuss about three types of sequences.

1. Arithmetic Sequence :
   

In a simplest manner we can define it as the sequence of expressions or numbers in which the difference between a term and its previous term is a constant
For example : 2,4,6,8,10 ….... , 5,7,9,......................, etc.


In the above two sequences, each term is different from the previous term by a constant number and this statement is always true for any of the two consecutive terms. This kind of difference is known as the common difference. In simple words any of the term used in an arithmetic sequence is obtained by adding a constant number to the previously used term.
The most common form of an arithmetic sequence with the first term is “b” and the common difference d is given by
b, b + d, b + 2d,…………………………….b + (n – 1) d,………..


It is noticeable that first term is “b”, second term in the sequence is b + d and the third term in the above mentioned sequence is b + 2d etc.... So in this manner we proceed further and the nth term will be
b + (n – 1)d. In short we can say that an arithmetic sequence can be defined by the first term and the common difference.

2. Geometric Sequence
   

The next type of sequence is Geometric sequence which is basically a sequence in which the ratio of any term is a constant. Let's take an example to understand it better:
,100,200,400,800……………………….., 7,21,63,189……………………
In both of the above examples each term used in the sequence maintains a definite or we can say a constant ratio with the previous term. This can be stated in a more defined manner as if we take any term in the sequence and try to divide it by the previously formulated term we get the same number. This type of numbers in a sequence is called that common ratio of the geometric sequence.
In normal way we can understand it in a manner that any term of a geometric sequence is obtained by multiplying a constant number to the previous term
The common form of an arithmetic sequence with first term “a” and common ratio that is “r” is given by
a, ar,ar2,ar3,ar4………..ar(n-1)………….


Here it is also noticeable that the first term used is “a”, the second term in use are ar, and the third term in the sequence is ar2. So the nth term used in the sequence will be ar(n-1)
In short we can say that a geometric sequence is defined by the first term and the common ratio

3. Harmonic sequence :
   

The next kind of sequence used is harmonic sequence as it has its reciprocals in arithmetic sequence. This shows that if we are going to take any of the arithmetic sequence, the reciprocals of the terms used in the sequence are said to be in a harmonic sequence.
Let's take an example to understand it better : ½,1/3, ¼, 1/5, 1/6 …...... 1/5, 1/10, 1/15.... etc.
The most common form of the harmonic sequence can be expressed from the general form of the arithmetic sequence used by taking the reciprocal of the terms.
Let's take some example to illustrate the above mentioned terms and methodology:
Example 1: Here the problem is that the first term of an arithmetic sequence is 4 and the fifth term of the sequence is 20. Find the sequence
Solution: Let the first term b = 4
By using above mentioned sequence or general form we can find the fifth term from the formula for nth term b+ (n-1) d
Now the fifth term is b+(4-1)d= 20
b+3d= 20
4+4d=20
4d = 20-4=16
d= 16/4=4
So b=4 and d=4 the sequence is
4,8,12,16,20.........
Example 2: The next question is that the first term used in a geometric sequence is 6 and the fourth term used is 48. We need to find out the basic sequence form :
Solution : Let the first term is “a” = 6
Now the fourth term that can be calculated from the formula for nth term ar(n-1)
The fourth term which is to be used is ar(4-1) = ar3= 48
ar3= 48
6r3=48
r3= 48/6=8 so we can get r=2
So a=6 and r=2 the sequence we get after formulating this problem is :
6,12,18,24,..........


The other topic which we are going to understand consist of following terms that are Variables, expressions, equations and inequalities. Now take individual topic at a time :
Inequality: An inequality tells that two values are not equal. For example a ≠ b shows that a is not equal to b. Slope formula plays an important role in graphing linear inequalities. So we need to know what slope formula means. Slope of a line describes the steepness, incline or grade of the straight line.

Let's take an example to understand the linear inequalities: How to solve a compound inequality and graph the solutions?
Example: -6 < 2x - 4 < 12

-6 < 2x - 4 < 12
add 4 to all 3 parts
-2 < 2x <16
divide 2 from all 3 parts
-1 < x < 8

To graph the following equation, you put an open circle or we can say mark it by a dot on the point (-1,0) and then you put an open circle on the point (8,0).Then draw a line between the 2.

We need to be careful while solving inequalities, as they are harder to solve than equations and require more attention. You can multiply an equation by a positive or negative number and get an equivalent equation. But while solving Inequalities remember this: when multiplying it by a negative number, you need to change sign of the inequality.

A linear inequality describes an area of the coordinate plane that has a boundary line. In simple way in linear inequalities everything is on one side of a line on a graph.

In mathematics a linear inequality is an inequality which involves a linear function. For solving inequalities we need to learn the symbols of inequalities like the symbol < means less than and the symbol > means greater than and the symbol ≦ or ≤ less than or equal to etc.


Equation: An equation is basically explained as an assertion that two algebraic expressions are equal. Let's take an example: 3a + 1 = 4
Here we can get

a = 1 is the solution. This is an example of a linear equation. An example of a quadratic equation is a² - a - 2 = 0. This equation has the two solutions

a = -1 or a = 2.

Algebra and Variable relationship: If we talk in simple words then we will see that algebra is simply the art of replacing variables in place of numbers. In solving algebraic problems simplifying algebraic terms is important. Simplifying here refers to breaking the large expressions into smaller ones so that it becomes easy to solve.

So this is all about in our today's class. In next class we will continue with this topic. As this topic is very lengthy considering the syllabus of eighth standard.

VIII Grade Algebra

Friends today we all are going to start our eighth grade mathematics and I am going to discuss about Algebra section. Before proceeding further let's talk about the basic concepts of Algebra. Algebra is a very vast area of study of mathematics which almost covers 90 percent of mathematics. In simple mathematical manner we can say that it is a branch of mathematics which deals with the study of the rules of operations and relations. Equation term can be explained as a mathematical expressions which shows the equality of two expressions. For example b + z = 6. In most general manner we can say that any combination of literal symbols and numbers that results from algebraic operations (addition, subtraction, multiplication, division, raising to a power, and extracting a root) are called as Algebraic equations. The two most important types of such Algebraic equations are linear equations which is written in the form y = mx + b, and quadratic equations that can be represented in the form y = ax² + bx + c. In general Algebraic equations are useful for modeling real life phenomena.

Following steps are used to simplify an algebraic equation. Firstly remove all the fractions in the equation Then remove the parentheses . Combine all the like terms so that we get all the variables and terms together. Move all the variable terms by adding or subtracting on both sides of the equal sign so the variable terms are all on one side of the equal sign. And finally if there is any multiplication sign then remove it by dividing.

Let's take an example to understand it better. The problem is to find the product of the following algebraic equations:
(b + 5)(b – 3) here multiply each multinomial term to the another multinomial term like
b x b – 3 x b + 5 x b – 3 x 5 = b² + 2b – 15

Now let's take the first sub topic of eighth grade Algebra problem that is Proportional and non proportional linear relationships. First question comes in our mind is what is a proportion ? In simplest of manner we can say that forming a relationship with other parts or quantities is called as proportion. Relationship between the variables is basically a way in which the variables change. The relationship can be linear, directly proportional and non linear.

Before proceeding further let's talk about Rational expression. Rational expression is an expression which can be written as a fraction a/b. Here a is the numerator and b is a denominator. The most important thing to understand is that denominator can never be zero.
Let's take an example:
The numbers 5/3 and -6/11 are rational numbers.
The number 5 is also an expression : 5/1 = 5 and any number in decimal form also an expression for example: 3.33 is a rational number : 3.33 = 333/100.
Some theorems which tell about Rational Expressions:
First one is that any integer is a rational. For example a number n = n/1.
Second: the representation of rational number as a fraction is not unique. Like 3/4 = 6/8 = -9/-12.
Third : Every nonzero rational number or a rational that do not contain a 0 has a representation in lowest term.
Closure Property of Rational Number shown as:
Adding and Subtracting of a fraction is done by using this property: x/y + a/b = xb + ay / yb.
For Multiplying a fraction this property comes in an account: a/b x c/d = ac/bd
and for dividing a fraction we use this property: a/b divide c/d = ad/bc.

For simplifying rational expression, we must need to have  good factoring skills. It requires two steps in solving a rational expressions.
factor the numerator and denominator is the first step and the second step is divide all common factors that the numerator and denominator have.
Dividing a rational number is the most difficult part as it requires key skills. Now we are going to learn how can we divide a rational expression:
12/5 divide by9/5 then we need to take a reciprocal of 9/5 . The reciprocal of 9/5 is 5/9.
Multiply 12/5 with the reciprocal. 12/5 x 5/9 = 12/9 = 4/3.


Linear relationship can be stated that the two variables always form a straight line when graphed. The most common example to understand this is the distance time graphs of the stationary object and any of the object which is moving with the constant velocity. Directly proportional can be defined as the rate of increase in one variable is similar to the rate of increase in the other variable. The example to understand: force is directly proportional to its mass. Non Linear relationship means that the rate of increase in one variable is distinct from the rate of increase in the other variable.

Let's discuss about Non Proportional linear relationship. This can be explained in the general form of linear expression that is y = mx + b, Here b is not equal to zero, m is the slope of the line or we can say that it represents the constant rate of change, b = Y intercept form. The graph of a non proportional linear relationship is a straight line which can never pass through the origin.

Let's take an example to understand the Non proportional linear relationships (y = mx + b, b not equal to 0).
The taxi provider company charges a flat fee of Rs. 50 plus Rs. 30 per mile to ride in a taxi.
Assumed that the flat fee is acquired as soon as the person enters the taxi.
What we need to do is to find out the cost of the taxi ride, multiply the total number of miles traveled by Rs. 30 and then add it with Rs. 50 (Rs. 50 is the flat fee) to the product.
If we are going to simplify the above equation then we need to follow this : If y shows the total cost of a taxi ride of x miles, then the relationship can be represented as an equation in the form of y = mx + b, here m represents the cost per mile (Rs. 30/mile) and b represents the flat fee (Rs. 50).
Total Cost = Cost per mile * Number of Miles + Flat Fee
Y = 30 * X + 50, or we can also make it y = 30x + 50


Another example to show Non-proportional Linear relationship with negative slope. The problem is that a ten inch candle burns at a constant rate of one inch per hour.
What we need to do: To find out the exact height of the candle you need to multiply the number of hours that the candle burns by 1 inch per hour, and than subtracting the product from the candle's initial height that is ten inches.
Solution : Similarly if y shows the height of the candle after x hours of burning, then the relationship can be explained in the form of an equation that is Y = mx + b, here m represents the rate at which the candle burns which is I inch per hour and b shows the initial height of the candle which is 10 inches.
Height of Candle = Rate at which it Burns * Number of Hours Burned + Initial Height
y = -1 * x + 10, or it can make it y = 10 -1x


We can explain proportional relationship by the following methodology. Let's take a general method to understand it. In any of the relationship let's take between y and z is proportional, it means that as y changes , x also changes by the same percentage. It shows that if y grows by 20 percent of y, z also grows by 20 percent of z. In an algebraic form we have already discussed it that is y = mx, where m is a constant.


Directly proportional means : A very common delusion is that two variables are directly proportional if one increases as the other increases or vice versa. Another thing is that two variables are explained to be directly proportional if and only if their ratio is a constant for all the values of each variable. Thus the most important outcome is that one variable is divided by the other , the answer is always a constant.


Now the next topic we all are going to understand in next section is Sequences: A sequence is basically a systematic listing of objects. In mathematical world the list formed from sequence can be numbers, algebraic expressions etc. If the objects in the list are numbers then it is a numeric sequence. There are many types of sequences in mathematics but the most common types of sequences are:
Arithmetic Sequence
Geometric Sequence
Harmonic Sequence
In next class we will discuss sequences in detail and also going to learn about the types of sequences in mathematics.

Eighth Grade syllabus

Hello friends, here I am going to discuss about the overall syllabus of eighth grade mathematics which we supposed to do. Students often think that the only purpose of figuring things out is to get the answer with the minimum amount of ado and effort. But there are certain things which student needs to follow to get the desired answer in the most optimum way. If student opt to choose shortest path or longest path to arrive an answer then there is a possibility of getting a wrong answer so students need to follow the optimum path to arrive an answer. A daily practice and hard work is necessary to arrive at an answer. The most important thing to remember is that there is no alternative of hard work.

Following are the topics which we all are going to study in eight standard.

1. Algebra : Analyzing and representing linear functions and solving linear equations and systems of linear equations.
2. Geometry and Measurement: Analyzing two- and three-dimensional space and figures by using distance and angle.
3. Data Analysis and Number and Operations and Algebra: Analyzing and summarizing data sets.

In more generalize manner, the topics to cover are

1. Integers
2. Perimeter
3. Area
4. Algebraic Expression
5. Equations
6. Perimeter
7. Fractions
8. Decimals

In Grade 8, student must focus on the following main categories:

(1) Understanding linear relationship along with proportional relationship and numerical relationship. Concentrating more on arithmetic sequences, variables sequences etc.....formulating and reasoning about expressions and equations, including molding a combination in bivariate data with a linear equation, and solving linear equations and systems of linear equations. Clinching the concept of a function and using functions to describe quantitative relationships.

(2) analyzing two- and three-dimensional space and figures using distance, angle, similarity, and congruence, and understanding and applying the Pythagorean Theorem. Few of the other concepts are : Geometric concepts, triangle inequality theorem, tessellations along with graphing equations (Linear equations or non-linear equations).

(3) Analyzing and measuring 3 dimensional figures and applying Pythagorean theorem and calculating measurements of figures using formulas for measurements. In addition to these, we are also going to learn scale drawing and proportions to convert to equivalent measurements.
(4) In the number and operation category of eight syllabus, I am going to discuss about problem involving percents along with rationalization techniques used with rational numbers and scientific notations and unit rate. Understanding and formulating the properties of real numbers, estimation of solutions, numbers, sequences etc...
(5) Finally we all are going to see the most interesting topic of eighth standard that is Probability and statistics where we formulate the dependency of events and calculating conditional probability. Some of the statistics analysis like mean, mode, median along with sampling techniques.

Now I am going to discuss about the most important topics of mathematics that are Algebra, Geometry and measurements, and Data analysis, and number and operations and algebra. So start with the first section that is Algebra.
Algebra :

In the algebra section what students going to use is linear functions, and systems of linear equations to represent and analyze the information provided and to solve a variety of problems. Students just recognize a proportion (Y/X = K, or Y = KX) as a special case of a linear equation having a form of

y = mx + b, understanding the properties that the constant of proportionality (K) is the slope and the resulting graph is a line through the origin. Students need to understand that Slope of a line is a concept which tells us how a straight line angles away from the horizontal or, we can say that it describes the steepness, incline or grade of the straight line. What students need to understand is that the slope (m) of a line is a constant rate of change.  Student recognize that tabular and graphical representations are usually only partial representation (translate among verbal, tabular, graphical, and algebraic representations of functions), and they describe how such conditions of a function as slope and Y intercept appear in distinct representations. Proceeding further students need to solve systems of two linear equations in two variables and relate the systems to pairs of lines that are parallel, or are the same lines in the plane. The overall result of the above discussion is that students need to use linear equations, systems of linear equations, linear functions, and their understanding of the slope of a line to analyze situations and solve problems.

Geometry and measurements:

In this section students deal with facts and figures. Geometry is an important area of mathematics which deals with the shape, size, relative position of figures, and the properties of space. It is all about shapes and their properties. Geometry is of two types : Plane geometry and solid geometry. Plane geometry deals with the shapes on a flat surface like lines, circles and triangles ... shapes that can be drawn on a piece of paper whereas Solid geometry is all about three dimensional objects like cubes, prisms and pyramids. Students need to use fundamental facts about distance and angles to describe and analyze figures and situations in two and three dimensional space and to solve problems, including those with multiple steps. What we are going to understand or prove is that particular configurations of lines give rise to similar triangles as the congruent angles created when a traversal cuts parallel lines. Students apply this statements for the similar triangles which helps them to solve a variety of problems. Some of the problems related to above statements are to find heights and distances. They must use the facts about the angles that are formed or formulated when a transversal cuts parallel lines to explain why the sum of the measures of the angles in a triangle is 180 degrees, and they apply this fact about triangles to find unknown measures of angles. Students also use Pythagorean theorem and explain why it is valid by using different methods. The Pythagorean Theorem states that in a right triangle the squares of the two short sides add to the square of the long side. If we call the lengths of the two short sides “a” and “b”, and the length of the long side “c” this leads to the familiar statement

c² = a² + b²

Students apply the Pythagorean theorem to calculate distances between points in the Cartesian coordinate plane to measure lengths and analyze polynomials and polyhedra.

Data analysis and number and operations and algebra :

Here we are going to discuss about probability, conditional probability, statistical analysis etc. Probability is a way of telling or expressing a knowledge that an event will occur or has occurred. The probability of an event occurring given that another event has already occurred is called a conditional probability. Statistics, on the other hand is the practice or science of collecting and analyzing numerical data in large quantities. It is basically a study of the collection, organization, analysis and interpretation of the data. So, the two things probability and statistics together play an important role in finding out measures of central value, measures of spread of different data and this helps in comparing of two data. Both probability and statistics are interrelated with each other and play an important role in analyzing the data. Students have to use descriptive statistics which comes loaded with concepts like mean, median, and range, to summarize and compare data sets. They also need to organize and display data to act and answer questions. Comparing the information comes out from the mean and the median and investigating the different effects that changes in data values have on these measures of center. Students need to understand the most important concept that a measure of center alone does not completely describe a data set because very different data set can share the same measure of center. Students have an option to select any of the two (mean or the median) as the appropriate measure of center for a given purpose.

Angles and triangles:

Here students need to use different ideas about distance and angles, and how they vary or behave under the following situations or conditions translations, rotations, reflections, and dilations, and ideas about congruence and similarity to describe and analyze two-dimensional figures and to solve problems. If we talk about concepts behind triangle then they need to show that the sum total of the angles in a triangle is the angle formed by a straight line. The various configurations of the lines play an important role to find out a similar triangles because of the angles created when a transversal cuts parallel lines. The another topic which students needs to work out is Volume by solving problems involving cones, cylinders and spheres. Now in next class we all going to practice each and every section of eighth grade mathematics syllabus. Starting from the Algebra section to the Probability section.