Proper Subset

As well known, a subset of A, say, set B, will have some or all the elements contained in the set A. But if the latter option is removed, then the subset is a proper set. Therefore the difference between a subset - proper subset is that the latter has at least one element less than the number of elements of the main set.
We can also define proper subset as strict subset, because of the analogy to the inequalities. The symbols ‘<’ and ‘>’ mean strict inequalities whereas the symbols ‘=’ and ‘=’ mean only just inequalities as they allow a situation of equality as well.  
On the other hand if B is the subset of A, then A is called super subset of B and if B is the strict subset of A, then A is called as proper superset of B. The subset proper set differences can be identified when the relations are expressed symbolically.
B ?  A (B is subset of A)  B ?  A (B is strict subset of A)B ?  A (B is subset of A)   B ?  A (B is strict subset of A)
Let us look into some proper subsets examples. The set {1, 2, 3} is a strict subset of the set {1, 2, 3, 4} because the former does not have the number 4 which is present in the latter. This is a simple example just to understand the basic difference. Let us see where we can see the difference more effectively. We know the domain of a logarithmic function y = ln (x) is (0,8),in interval form.
Suppose we introduce a set D as{0,8}. One can immediately recognize that x is a strict subset of D because, ‘x’ cannot assume the value of 0.
Similarly, we know the range of a sine function y = sin (x) is [-1, 1], in interval form. Suppose we introduce a set R as {-1, 1}. One can immediately recognize that x is a not a strict subset of R because, ‘y’ can assume all the value of between -1 and 1, both included.
Another practical example can be given from number system. We know all natural numbers are integers. But the converse is not true. That is, the set of integers contain natural numbers and in additionalso contain 0 and negative of whole numbers.
Therefore, set of natural numbers is a strict subset of set of integers. Same way, set of irrational numbers

Prominent Variables

Variables in math are very prominent. It is a part and parcel in math, especially in the topic of algebra. Let us try to figure out what is variable and in the process let us find the variables definition. It takes place of an unknown quantity.

Suppose we say that I am 10 years older than my brother, then at any point of time my age is (x + 10). Since at any point of time my brother’s age is also not known, I assumed that as x. Therefore it can be described as letter that represents the unknown quantity and any value can be assigned to that depending on the situation and need.

It is a general practice to use the small case letters of latter halves of English letters to denote the unknown quantities. Of course, none of the terms ‘small case’, ‘latter half’ or ‘English letters’ is a strict requirement. Though it is a general convention, there are exceptions in many cases. The unknown measures of angles of triangles are denoted in capital letters, to be consistent with the symbols of the respective vertices. Again in the very same case of geometry, the first halves of English alphabets are used for unknown quantities.

For example, the prominent statement of Pythagorean Theorem is,              a2 + b2 = c2. Also, there are many Greek letters are used for the unknowns, mostly in trigonometry. For example, the measure of angles are mostly expressed as ? (in case of units in radians) and as a (in case of units in degrees).

In general two or more of these are used in expressions or functions. As a simple example let us take the case of a linear function y = f(x), where f(x) = mx + b. This is a case of input and output function where variable(s) x denotes the input quantity and the ‘y denotes the corresponding output quantity.
Now by change of variables and solving for the same on the left we can find the behavior of the inverse of the function f(x). Let us now see in what way the variables and expressions are related. Expressions are mostly parts of functions. The degrees (the powers) of these decide the type of functions.

The function behaviors can be predicted from the degree and the sign of the leading coefficient of the function. For example, a quadratic function which of degree 2 and with a single unknown will always have rising characteristics on both ends for a positive leading coefficient.

The Method to Find the Triple Product

In mathematics there are basic operations like addition, subtraction, multiplication and division. Multiplication can also be known as the product of numbers. The numbers can be natural numbers, whole numbers, integers, and real numbers and so on. So, the product of two numbers is nothing but the multiplication of two numbers.

There are different types of quantities in mathematics. There are some which have magnitude and there are some which can have both a magnitude and direction as well. Different names have been allotted to these numbers. The former ones are known as scalars and the latter ones are known as the vectors.

The difference between the two is only about the direction they possess. The scalars do not have a direction attached to them. So, they can be handled more easily than the vectors or carrier. Since they have a direction attached to them a sign is used to represent the direction. This sign shows the difference in direction of two quantities of similar nature.

There is a concept in mathematics called the triple product. From the name itself it is understood three products or three quantities are involved in the operation. Now this can be carried out between scalar quantities or the carriers. Depending on the type of the quantities present in the operation the name of the product changes.

The triple vector product is nothing but an operation in which two cross-products are used. There are three of these are involved. The cross products between the first one and then the second and third are taken together. The vector triple product proof can be given mathematically. Even it can be represented geometrically. The triple vector product proof is easy to understand and requires the basic understanding of addition and subtraction.

A vector triple product example will explain the concept and make it clearer. The examples always make the concept easy to understand. Even difficult examples can be made easy with the help of examples. In geometry there exists a figure known as the parallelepiped.

When the scalar triple product is found out it helps in finding the volume of this geometrical figure. There is geometrical meaning attached to this type of product. It must be carefully understood. The scalar triple product can have different values. The values have different meanings. If it is found to be zero, then the volume of parallelepiped is found to be zero

Simultaneous equations

Simultaneous Equations: If two unknown values had to be solved at the same time then these type of equations are called Simultaneous Equations. Simultaneous forms of equations are two equations with two unknowns. They are called simultaneous because they must both be solved at the same time.

There are various methods to solve Simultaneous-equations.
Simultaneous Equation solver:
How to Solve Simultaneous Equations: These equations can be solved by  by elimination method,  Solving by substation method  and by squaring method.  These equations can also be solved by graphing.

Solving Simultaneous equations.

In the  elimination   method for solving  equations, two equations are simplified by adding them or subtracting them. This eliminates one of the variables so that the other variable can be found.
To add two equations, add the left hand expressions and right hand expressions separately. Similarly, to subtract two equations, subtract the left hand expressions from each other, and subtract the right hand expressions from each other.

Take for example two equations.  x +y=10 and x-y = 2.
See  if by adding or subtracting one variable can be  eliminated. When we add these two equations  +y and –y gets cancelled.
we get 2x =12.

Hence  x = 6  Now plug in the value of x in any one equation say x + y=10
We get 6+y=10, so y = 10-6 which equals 4.  So the solution is  x =6 and  y=4

To solve by substitution , we take one of the equations and solve for x in terms of y  or solve for y in terms of x  Then this value is substituted in the unknown equation and solved for the next variable.
For example  x = 8-y and  2x – y  =  7.
Substitute 8-y  for x in the  2x – y = 7  and solve for y.
2(8-y) - y=7 . now  distribute it.
16 – 2y -y = 7  which equals 16 – 3y  = 7.
We have -3y = -9  Hence  y = 3 .

To find  x, put  value of  y  in the equation 2x – y =7.  We get  2x -3 = 7   so 2x = 10 and x = 5

To solve by graphing bring the two equations to the   y = mx + b form  where m is the slope and b is the y intercept.  Using this first mark the  y  intercept and the slope points.  Plot the points and draw  a straight line connecting  the dots .  Note the point where the two  lines intersects  one another.  From this point of intersection of the lines we can be found the values of  x and y.

Calculating Percentage Formula Using a Formula

There can always be a loss or profit in business. This concept is well explained in mathematics. A profit is obtained only when a product is sold at a greater price than its original cost. A loss is obtained in a business when the product is sold at a lower price than the cost of the product. This is the basic difference between the terms profit and loss. So, the selling price and the cost price of a product are very important in deciding the profit or loss obtained by selling a product. Various techniques are used to earn profit or avoid loss. Both these processes are same in nature. One can earn profit only by avoiding loss.

Now when a loss is obtained in a particular transaction, percent loss can be calculated from it. This helps in better understanding of the whole situation. The percentage loss can be calculated only after the loss in a transaction is found out. So, first the loss is found out. This is found by subtracting the price at which it is sold to the price it is produced. The price at which it is sold is called the selling price and the price at which it is produced is called the cost price. On using this formula in the case of profit a positive value is obtained and in the case of a loss a negative value is obtained. The negative value indicates that the value obtained is loss.

The percent loss formula can be derived once the pct is found out. To calculate percent loss formula, it is necessary to calculate the loss first in a transaction. Once the loss is calculated it is easy to calculate percentage loss formula with its help. The pct can be calculated only when the price at which it is produced is known. The cost at which a product is produced can be taken as the cost price. First the loss is calculated and then it is divided by the cost price.

For the calculation of the pct the answer obtained is always multiplied by 100. Only then the answer obtained is converted into pct. Similar is the case with the calculation of the gain pct. There can be either a gain or a loss in a transaction. Maximum emphasis is given to obtain profit. Profit is nothing but the gain. Both can be used interchangeably.

Rational expressions

Rational expressions are those involving ration of two polynomials or division of two polynomials. Rational expression: P(x)/P(y). Here P(x) and P(y) are polynomials. Here Q(x) cannot be zero.

Dividing Rational Expressions include division of two rational-expressions each of the form given above such that denominators of both the expressions are not zero.

Let us see How to Divide Rational Expressions now:

It would be easy for you to first understand division of rational numbers first before going to division of rational-expressions. To divide a rational number p/q by r/s we do multiplication of p/q and s/r. Actually, we need to reciprocate one of the numbers (second number) and then multiply it with the other number. Multiplication of rational numbers is done by simply multiplying the numerator by numerator and denominator by denominator.

Example: divide 4/3 by 9/5.
Solution: reciprocate the number 9/5 to get 5/9 and multiply it with 4/3.
4/(3 )÷9/5
 =  4/3.5/9
 =  ((4).(5))/((3).(9) )
 =  20/27

We divide rational form of expressions in the similar way as we divide rational numbers. Just reciprocate the second number and multiply it with the first number. Also check if the numerators  and denominators have any factors. If they have then write them in that form and try to cancel out the common factors of numerator and denominator.

If first rational-expression is f(x) = P(x)/Q(x) and second is g(x) = M(x)/ N(x), then f(x)/ g(x):
 = (P(x))/(Q(x))  (N(x))/(M(x)). here we have reciprocated g(x) and multiplied it with f(x).
 = (P(x)  .  N(x))/(Q(x)  .  M(x))

Let us see some examples to Divide Rational Expressions so that we are able to perform divisions of rational-expressions easily without any problem:

Example 1) divide  (y + 5)/3y by  (y+5)/(9y^2 ).
Solution 1) As shown in above explanation, just reverse the rational expression (y+5)/(9y^2 ) and multiply it with (y + 5)/3y :
=  (y + 5)/3y. (9y^2)/(y + 5)
= now here we see that y + 5 term is common in denominator and numerator so cancel it out. Also 3y is common on 3y and 9y2, so cancel one y also to get:
= 1/1. (3 y)/(y + 5)
= (3 y)/(y + 5)

Example 2) Divide (x^2+ 2x-15)/(x^2-4x-45)  by  (x^2+ x-12)/(x^2- 5x-36).
Solution 2) we can write (x^2+ 2x-15)/(x^2-4x-45)  ÷  (x^2+ x-12)/(x^2- 5x-36) as:
 (x^2+ 2x-15)/(x^2-4x-45).  (x^2- 5x-36)/(x^2+ x-12)
= ((x +5)(x - 3))/((x - 9)  (x + 5)).((x - 9)(x + 4))/((x + 4)  (x - 3))
See that all the terms are cancelling out in numerator with denominator. So we get
 (x^2+ 2x-15)/(x^2-4x-45)  ÷  (x^2+ x-12)/(x^2- 5x-36) = 1

Importance of Math Tutoring during Exam

The first and foremost benefit of online Math tutoring is that a student can take unlimited sessions at any time from the comfort of home. Online tutors not only teach students simple steps to solve difficult Math problems, but also give useful exam tips.


It has been observed that many students get worried by the thought of Math exam as it requires more concentration with logical thinking. Students actually need to use their brain while solving tough and tricky Math sums. By using correct methods, a student can easily get an accurate answer to any Math problem. Learning problem solving techniques from an online tutor can help a student to score better in exam. Having stressed and anxiety during examination is a common problem among students. But proper guidance can reduce anxiousness of students up to a certain extent. Online tutoring websites are the best option to get exam help and assignment and homework help. In an online learning session, well qualified tutors will work on your problems and also give you last minute tips before exam.

Online Math tutoring assists student to solve problems in several ways by using correct techniques. In a virtual learning session, students get to know each step for their math problem through a whiteboard. A whiteboard comes with features like math symbols, attachment option,  chat option, etc. that makes a learning session more adaptable for students. Take unlimited sessions in a secure web environment and improve  your problem solving skills. Buy a math tutoring package, just by log-in to a tutoring website and create a loin-Id and password to experience a smart and fun way of learning.

Scoring good marks in Math exam is always a matter of concern for parents as well as tutors. But performing well in exam is not that easy for the students who feel incapable doing tricky Math problems. An online tutor will not only give you enough time to understand logic behind a math problem, but also teach you some useful steps to solve a sum in fast way. During examination it is apparent that students become stressed and feel blank. Hence, it is crucial to have a proper tutoring session where students can work on their queries and clear their doubts from an experienced tutor.

An online Math tutoring program is an ideal solution for students who suffer from exam fear. With the help of a qualified tutor, students can ask as many questions as they need and want. Apart from this, regular practice of Math problems under the guidance of an online tutor can improve a student logical reasoning and make him or her confident during examination. In a web environment,  a session can be scheduled from any location and at student's convenient time so as to give them comfortable learning.