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.

Vector Subtraction

Vector Subtraction is a special case of vctor addition. 
Let there are two Vectors A and B then subtraction of Vector B from A is A – B which can also be written as A + (-B).
So, A – B = A + (-B)
Here (–B) is a Vector that is parallel to Vector B but is in reverse or opposite direction of Vector B.
According to parallelogram law of addition, if there are two Vectors such that they form the two consecutive sides of a parallelogram then the sum of the Vectors is represented by the diagonal of the parallelogram passing through the point at which the tails of two Vectors meet. The geometrical representation of A-B is given in following diagram.

The above method is known as head to tail method.
You can find Subtraction of Vectors by using perpendicular components method. In this method you need to split each Vector into perpendicular components. We normally split the Vectors in x and y directions.
Suppose a Vector of magnitude 5 makes an angle of 30o with horizontal then its vertical component will be 5sin(30)j and horizontal component will be 5cos(30)i. i and j are unit Vectors along the x and y directions respectively. If a component is in downward or left direction assign a negative sign to it. Similarly Split all the given Vectors in such a way. Now add all the horizontal components together and add all the vertical components separately. Subtract a component if its magnitude is negative. Now calculate the resultant magnitude of the Vectors using Pythagoras theorem R2 = A2 + B2 . Here A is sum of horizontal components, B is sum of vertical components and R is resultant.
The angle that the resultant Vector makes with the horizontal is given as: Ө = tan-1|B|/|A|  . Here|A| and |B| are magnitudes of the Vectors.
Subtracting Vectors is similar to addition of Vector. The difference is just of the negative sign.
Let us see some Vector Subtraction Examples now:
Example 1) Subtract Vector B = 9i+7j+2k from A =3i + 5j + k.
Solution) To subtract B from A we write them as A-B = (3i + 5j + k) – (9i - 7j + 2k)
Now add the Vectors in the direction i, j, k separately. As the given Vectors are three dimensional we are considering z direction also. i, j, k refers to unit Vectors in x, y and z direction.
(3-9)i = -6i
(5- (-7))j= 12j
And (1-2)k = -k
A-B=(-6i+12j-k)
|A-B| = √(〖(-6)〗^2+〖12〗^2+〖(-1)〗^2 ) =√(36+144+1) = √181

Venn diagram

An illustrative representation of relationship between and among the given sets that have something in common are called Venn Diagrams. These diagrams are represented using circles, the most commonly used ones are the two circles and three circle Venn-diagrams. It is used mainly to depict the intersections of sets which mean to show the objects or things common to all the sets. The Picture of Venn diagram with two circles or three circles consists of a rectangle which has in it two circles or three circles respectively. Intersection of sets is shown as overlapping circles, if no common objects then the circles are shown separately.

The total number of elements in the set is shown at the right corner of the rectangle using a symbol ‘µ’ or ‘E’. The object or elements which do not belong to any of the sets is shown outside the circle within the rectangle. A glance at the Venn diagram help to understand the complete picture of the relationship among the sets given and also help to find the unknowns values using  the diagram.

Example of Venn Diagram is as follows: set X={1,2,3,5,7,9} and set Y={7,9,11,13}, the intersection of these sets given by X∩Y={7,9}.
The Venn-diagram showing the relationship between these two sets is as given below, the pink region shows the intersection of the sets, X∩Y.It also shows the elements only in the set X {1,3,5} in green color and the elements only in the set Y {11,13} in yellow color.

Venn diagram compare and contrast the relationship between the elements of the sets, they are basically used to visualize the relationship between two or three sets. They are also used to compare and contrast the characteristics of the elements of the sets.  An outline about the sets can be easily created using the Venn-diagrams. Venn Diagram with three circles is a type of diagram which involves three sets.  Let us learn more about this using a simple example: A survey was done on a group of children as how many like to play baseball, football and ice hockey shown by the Venn-diagram below,

The above Venn-diagram gives the following information about the three sets, F, B and I.

• Total number of children (E) would be sum of all the numbers on the three circles, 5+14+8+2+16+5+20=70, so E=70.
• n(F)=29, n(B)=47, n(I)=31
• only n(F)=5, only n(B)=20, only n(I)=16
• n(F∩B∩I)=8, n(F∩B)=22, n(B∩I)=13, n(F∩I)=10
• only n(F∩B)=14, only n(B∩I)=5, only n(F∩I)=2

Greatest common divisor

The numbers when multiplied together to get another number are called the factors of the number. For example consider the number fourteen, 2 when multiplied with 7 gives 14 and hence 2 and 7 are the factors of the number 14 which includes 1 and also 14. The factors of 6 are 1, 2, 3 and 6.
The factors of 9 are 1,3 and 9. The common factors of 6 and 9 are 1 and 3. So, we can define common factors as the factors which are common to all the numbers considered.

 Let us now learn about Greatest common factor which is also called Highest Common Factor and at times Greatest Common Divisor. It can be defined as the greatest integer which divides the number evenly that is without any remainder. In short it is called GCD.

For example, GCD of 12 and 18 is 6 as 6 is the greatest integer which divides 12 and 18 evenly. Let us now find the Greatest Common Divisor of the numbers 36 and 72. The first step to find the GCD would be to write all the factors of each of the numbers, factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36; factors of 72 are 1, 2, 3, 4, 6, 9, 12, 18, 36 and 72.

The next step is to identify the greatest integer which is common to both the lists, 36 is the largest integer common to both the lists and hence GCD of 36 and 72 is 36.

There are different methods to find the GCD of two or more numbers the simplest being the prime factorization method where prime factors of each numbers are listed and then the greatest prime factor common is determined.

The other method is Euclid’s algorithm which involves a process of repeated division given by the Greatest Common Divisor formula, gcd(a,b)=gcd(b, a mod b) where [a mod b] is got by { a – b[a/b]}, mod here means modulus. There are variations in the formula according to the values of a and b.

Let us now learn about properties of GCD which can be given as:

• For any integers ‘a’ and ‘b’ there exist integers ‘p’ and ‘q’ such that [p.a+q.b]=gcd(a,b)
• If ‘a’ and ‘b’ are relatively prime integers then there exist an integer ‘p’ such that pa=1(mod b)
• If ‘a’, ‘b’ and ‘c’ are relatively prime integers and if a divides bc then a divides c

Estimation and Rounding

Estimation and rounding is a important concept in basic mathematics. Number system is the concept that defines everything about numbers and estimation and rounding is a concept covered within number system.

Let’s have a look at both the concepts in this post.
The process of rounding a number estimating its nearest value is called estimation and rounding. For example: After she started breathing exercises during pregnancy , her medical issues lessen to about 59.8%. Here, we can also say the same thing in another way: After she started breathing exercises during pregnancy , her medical issues lessen to about 60%. This is rounding a number estimating to its nearest value. This is an approximate value of the given number.

There are certain rules to be followed while incorporating estimation and rounding, two are mentioned as below:
Rule 1: If the number that needs to be rounded is followed by a number which is greater or equal to 5, it can be rounded to its nearest greater number.

• Example 1: The hospital bag checklist of her NGO includes about 68.8% of things for baby and child care.
The hospital bag checklist of her NGO includes about 70% of things for baby and child care.

• Example 2: These are my most comfortable shoes ; my feet stays almost 95% relaxed wearing the same.
These are my most comfortable shoes; my feet stays almost 96% relaxed wearing the same.
Rule 2: If the number that needs to be rounded is followed by a number which is less than 5, it can be rounded to its nearest smaller number.

• Example 1: She bought apples for 61 rupees per kg for her daughter.
She bought apples for 60 rupees per kg for her daughter.

• Example 2: I have gained almost 6.3 kg of weight after getting married.
I have gained almost 6 kg of weight after getting married.
These are the basics about estimation and rounding.

Finding Determinant Value of 3x3 Matrices

Matrices, like any other expressions,are also subjected to basic algebraic operations. Multiplication is one of those. In addition to normal restrictions on matrix operations, multiplication of two matrices has additional restrictions and the method is a bit strange.

Suppose A and B are two matrices, their product is defined only if the number of columns of A and the number of rows of B must be same. Else, the multiplication is not defined. The algorithm for multiplying 3x3 matrices is described below.

The first element of first row and first column of the product C of matrix A and matrix B is the sum of the products of the elements of the first row with the corresponding elements of the first column of B.

For the second element in the same row, the products are now first row of Avs second column of B. Like this the entire elements of C are found.

With the restrictions and strange algorithm of the process, it can clearly be concluded that multiplication of matrices is not commutative.

A matrix just conveys the details of the data and it does not give any solution to any question on the relations between the items. On the other hand if the same arrangement of data indicates an operation for a solution, then that particular arrangement is called the determinant of the same matrix.

A matrix is denoted by containing the items of a data within a set of elongated square brackets. In a determinant the same are enclosed by two thin vertical bars (similar to absolute value symbol). The following is an example.

The determinant of the matrix indicates the solution by the operation (a1 *b2- a2*b1).
The determinant of a matrix with 3 rows and 3 columns is called a determinant 3x3 matrix or just as determinant 3x3.
To find determinant of 3x3 matrix, first rewrite the elements in the same order but with the symbol of a determinant. To find the value of determinant of a 3x3, the following is the procedure.
Consider a11, the first element in the first row and column. By deleting the row and column that includes a11, you are left with a 2 x 2 determinant which is called as the minor of the element a11and denoted as, M11.
Same way, for the second and third elements in the first row a12 and a13 , the minors are M12 and M13 respectively. Now the formula for the value of the determinant D is,
D = (a11*M11  -a12*M12 + a13*M13)