Online math tutoring - Learn math with live examples

Learn Math in a virtual classroom from the best online tutors. With the help of a whiteboard and animated live examples, students can easily understand every difficult Math problem. Moreover, online math help provides free math worksheets in order improve students problem solving skills.

Online Math tutoring is a smart and comfortable way of learning the subject from any location. Many students struggle a lot in Math subject and score low marks. Online Math help is the ideal option to overcome the anxiety and stress, which students often face while solving Math problems. As we know, Math subject has high importance in varied field like Engineering, Science & Technology, Banking, etc. therefore it is important for every student to learn the subject thoroughly. Online tutoring gives enormous benefits and ample time to students to master the subject . This cutting-edge mode of learning has gained immense appreciation across the globe due to personalized and interactive learning sessions.

Online Math tutoring sessions come with a wide variety of teaching and learning tools like whiteboard, attached chat options, recorder, dashboard, etc. With the help of these tools, a student can select a topic, start a session with preferred tutor at convenient time. Along with this, students can communicate through chat option and clear their doubts from an online tutors who are available round the clock in a virtual classroom. Moreover, every tutoring session can be recorded and replayed by the students to revise the topic as many times as need and want.

Many websites make Math learning more informative for students with live examples. Every topic is well explained with the help of graphs or animations, which keep students involved more in a online learning session. Right from understanding the basic concept of Math to Algebra, Calculus,Geometry and Trigonometry, each topic is illustrated with animated examples to make the session more effective and influential for students. Online Math help not only enhance your problem solving skills but also keep a tab on your performance. Regular feedback from an online tutor can actually help a student to do better in the subject.

Learning Math with live example is quite enjoyable and beneficial for those students who face difficulty in understanding the subject. Every tricky sum is being solved by highly experienced online Math tutor so as to give a thorough understanding of the subject. In addition to this, some websites provide free worksheets and math quiz to make the Math subject more interesting for students. Online Math tutoring provides different ways of learning with the help of live examples, which helps students to understand the logic behind every math problem. It is a great assistance  for students who find Math subject boring and difficult.

System of Linear Equations

System of Linear Equations is a collection of linear equation  Systems of Linear Equations that involve two equations in two variables are simplest to deal.
Suppose there are two linear equation in x and y, then each equation will represent a line in x-y plane. A solution to these equations will be the point where these lines intersect. Thus the solution will be unique value of x and y. If the equations represent parallel lines then there will be no solutions to this system. If the Linear System of Equations contains same coinciding lines then the solutions will be infinite in number.

Any system of linear equation can have following conclusions: no solution, unique solution or infinitely many solutions. A linear system is consistent if it has at least 1 solution and is said to be inconsistent if it has no solution.
Suppose a linear equation is 2x+y=0. Then there will be infinite points satisfying this equation. Like (x, y) ={(1,-2),(0,0),(2,-4)… and many more}. Now suppose there is another line x-y=0, then solutions to this will be (x, y)={(0,0),(1,1),(2,2) and so on}. A common solution to these equations is x=0, y=0. This is hence solution of this system of equations.
We can search solution of more than two equations also by drawing graph of the equations also.

Method of Solving Systems of Linear Equations:
By substitution:
Let two equations are a_1x+b_1y=c_1 and a_2x+b_2y=c_2. Solve first equation for x:
 a_1x+b_1y=c_1  
a_1x=c_1–b_1y or x=(c_1–b_1y)a_1
Substitute this value of x in second equation to get:
a_2((c_1–b_1y)a_1)+b_2y= c_2
Now you get an equation in y. Solve for y. now put the value of y in any of the two equations to solve for x.
You can substitute value of y also from an equation and then substitute it in other equation.

Systems of Linear Equations Word Problems
Q.1) Cost of 2 chairs and 1 table is 1000 while cost of 1 chair and 3 tables is 1500. Find cost of each.
Solution) let cost of chair=x and of table=y.
2x+y=1000………(1)
x+3y=1500………(2)
From second equation: x=1500-3y
Substituting x in first equation = 2(1500-3y)+y=1000
3000-6y+y=1000
3000-1000-5y=0
2000=5y
y=(2000/5)=400
Putting y=400 in first equation we get: 2x+400=1000
2x=(1000-400)=600
x=600/2=300
Linear equation can also be solved by equating coefficients:
Equate coefficients of x by multiplying equation (2) by 2:
2x+2(3y)=2(1500)
Subtract this equation from (1):
   2x+y=1000
-(2x+6y=3000)
     0-5y= -2000
5y=2000
Y=400
Now put value of y in any equation.
You can equate coefficients of y also and then subtract the two equations to get value of x first.

Divison of two numbers is given by the following relation

Divison of two numbers is given by the following relation:
m/n=q
Here m is being divided by n and the result of the divison is q which is known as quotient. Let see how to do Division by 2 Digit Numbers which means dividing any dividend by 2 Digit Divisor

Step 1) Put the two digit divisor before the divison braces and put the dividend no. below the divison bar.
Step 2) Check the first digit of the dividend. If it is smaller than the divisor then take the first two digits of the dividend. Now determine how many times of divisor produces those two digits of dividend or produces a number which is just less than the dividend digits. Let x times of divisor give above result.
Step 3) Now multiply the no. x by divisor, let the result is y. Put y under the first two digits of dividend.
Step 4) Subtract no. ‘y’ from first two digits of dividend. Let the result is z. number z will be less than the divisor. So, bring down the third digit of dividend beside z. Now again follow the same steps from step 2.
Step 5) Continue following these steps till no more digits are left in the dividend and you get a remainder which is less than divisor.

Let’s use method of Long Division Two Digit Divisors through some examples:
Example 1) Divide number 7139 by 16.

Step 1) Firstly, check the first digit of the dividend. It is 7 and is smaller than divisor 16. So we will consider two digits of dividend i.e. 71. Now the largest multiple of 16 which is smaller than 71 is 64. As 16*4 = 64 so, write 4 (quotient) on right hand side of dividends and 64 below 71. Now, subtract 64 from 71.  
16)7139(4                                                                
       64                                                                                                                                                                                                                                   .       7
Step 2) Now consider number 7. As this is smaller than divisor 16, so we will bring digit 3 down with number 7. Now we will repeat the above step again. Find the largest multiple of 16 which is smaller than 73. This number is 64 which is equal to 16*4 . Write 4 at the place of quotients and subtract 64 from 73.
 16)7139(44                                                                
      64
----------
      73
-     64
----------
      99
-     96
----------
-      3
Step 3) Repeat the above steps again. This is done below:
16)7139(446
     64
--------
     73
-    64
--------
     99
-    96
--------
      3

3 is remainder.
This is how a Division 2 Digit Divisor is done. You can practice Divisibility by 2 digit divisors: 234/34, 5678/89, 7553/123etc.

What is Ratio

When we say one banana for every three apples, the relationship between the banana and the apple is shown by a term called Ratio. It is used in comparing and showing the relationship between two entities. It is denoted using the symbol colon (:) between the two values.

In the above example the proportion between banana to apple would be banana: apple read as ‘banana to apple’ the value of which would be 1:3 read as ‘one is to three’.

Hence we can say ratios tell the relationship between two values that is how one number is related to the other. It may be denoted as a fraction also, for instance the two values which are to be compared are X and Y then the proportion between them can be shown either as X:Y or X/Y or just X to Y. In the above example the proportion shows that apples are three times bananas.

One important point to remember while writing the balance is that the order should not be changed that is the respective numbers should not be interchanged.

If for instance there are 3 pencils for every 5 pens, the balance when considered as pencils to pens should also be written in the same order pencils:pens, 3:5 and not 5:3 which would mean pens to pencils

Let us now determine the value of Y, if X=6 and the balance of X to Y is 3:4. To find the value of Y first we need to determine how many times X is divisible by the corresponding part of the balance (3:4) which can be calculated by dividing 6 with 3 which gives 2.

Now we just need to multiply this 3 with the corresponding balance part of Y which gives 2x4=8. When the proportion is 3:4 and the value of X=6 then the value of Y=8.  Ratio definition can be given as comparison between two things which tells the relationship between the two. Let us now take a glance at the various ratio problems which help to understand the concept.

There are 8 children, 3 are boys and 5 girls. What is the ratio of boys to girls, girls to boys, the total children to boys and total children to girls? Given the total number of children=8, boys=3 and girls=5. So, the proportion of boys to girls is 3:5; the proportion of girls to boys is 5:3; the proportion of total number of children to boys is 8:3 and the proportion of total number of children to girls is 8:5.

Change of Base Formula for Logarithms

Logarithm is a means of expressing a number using exponents. Example log101000 is equal to 3 as 1000 is a cube of ten and can be written as log_10 10^3. Hence the value is 3.
The common base for logarithms is base ten and the other base is the natural logarithm base –e. At times while calculating logarithms we come across base other than 10 and the base e, in such cases the base change can be done using a special formula.

Logarithm change of base formula can be given as, log x to base a = log x to base b/log a to base b.  To understand how to arrive to this base change formula let us go through the following steps:
Consider y=log_a x, we get x = a^y
Taking log_b on both sides would result in log_b x = log_b a^y
Applying the power rule to the above equation gives, log_b x = y log_b a
Now dividing on both sides with log_b a gives, log_b x/ log_b a= y log_b a/ log_b a
So, we get, y = log_b x/ log_b a

Let us now consider a simple example, the value of log 27. This can be written as log_10 7/log_10 2

The value can be calculated as log 7=0.845 and log2= 0.3010. When these values are divided the final answer would be 2.80730…; thus using loga x= log_bx/log_b a, the change of base formula logarithms value of the given logarithmic expression can be found easily. Using Log base change formula it becomes easy to evaluate logarithms with different base. Here the logarithm is written as a fraction with the logarithm of the number as the numerator and the logarithm of the base as the denominator, such as log_a x = log x/log a.

Then each of the logarithms is evaluated using the log table or a scientific calculator, the final value is got by dividing these values. The evaluation of other logarithms with base different from natural logarithm base or the common logarithm base can be done using the base change formula, log_a x = log_b x/log_b a. Let us now evaluate the logarithm log_5 9. This problem can be solved by either using natural logarithm or the common logarithm. Using the natural logarithm that is base-e it would be, log_5 9 = ln9/ln5 = 2.1972/1.6094 which would be approximately equal to 1.3652… Now using the common logarithm that is the base ten it would be, log_5 9 = log 9/log5= 0.9542/0.6989 = 1.3652… Using either of the logarithms we arrive at the same result.

Simple interest

Definition:
Consider a house that one would have rented. The tenant has to pay some amount of money to the owner of the house as rent for using the property. Similarly if a person borrows money from another person, he has to pay some amount of money as rent for using the borrowed money. This charge paid for use of funds is called interest. Therefore the amount charged on a fixed amount of principal, that is lent by a lender for a specific period of time is called simple interest. In simple interest the principle amount over the period of loan remains constant and is not reduced or increased.
Formula for simple interest:
Some important terms related to simple interest:
(1) Principal (P): The money borrowed or lent.
(2) Interest (I): The additional amount paid to the lender, for the use of the money borrowed.
(3) Rate( R ): Interest for one year per 100 units of currency.
(4) Time (T): The time period for which the money is borrowed.
(5) Simple interest or (S.I.): When the interest is paid to the lender regularly every year or every half year, we call the interest simple interest.
(6) Amount (A): Principal + Interest = amount at the end of the term of T years.

Formula used for calculating simple interest is like this:
S.I. = P x R x T
100
A = P + S.I.

When we calculate simple interest, the following points need to be noted:
(1) Rate of 4% per annum means $ 4 for every $ 100 per year. Similarly a rate of 1.5% per month means $ 1.5 for every $ 100 per month = $ 1.5 * 12 = $ 18 for every $ 100 per  year = 18% per annum.
(2) When time is given in days, we convert it to years by dividing by 365. When time is given in  months, we convert it to years by dividing by 12. When dates are given, the day on which the sum is borrowed is not included but the day on which the money is returned is included, while counting the number of days.