Solving Complex Fractions

Complex fractions are those fractions which have a fraction in numerator or denominator or in both. For example, 1/ (¾), ½ / 3, (¾ )/ (½ ) are some of the complex fractions

Let us now learn to simplify complex fractions. While simplifying complex fraction, let us consider the following example for better understanding,

Complex Fractions

Simplify the complex fraction 4a²b/(8/ab)
Here the numerator is 4a²b and
        the denominator is 8/ab
It can be written as 4a²b  ÷ (8/ab)
to simplify, we need to flip the fraction in the denominator and multiply it with the terms in the numerator as follows,
             4a²b × ab/8 (division of fractions)
Now, we can simplify the terms
           4/8 × a²b × ab
which gives us ½ a³b²


Complex fraction solver
In solving complex fractions, we can use one more method, which is the LCD method. Let us learn how to solve  complex fractions with some example problems

1.Simplify (4/5)/(2/15)
Solution: Numerator = 4/5
Denominator = 2/15
  LCD of 5 and 15 (denominators of the two fractions) is 15
Multiply LCD with each of the fractions of the numerator and the denominator
4/5 x 15 = 4 x 3 = 12
2/15 x 15 = 2 x 1 = 2
The simplified fraction, 12/2 = 6

2.Simplify (1/a + 1/b) ÷ (1/a – 1/b)
Solution: (1/a + 1/b) = (b + a)/ab
(1/a – 1/b) = (b – a)/ab
LCD in this case would be ‘ab’
(b+a)/ab x ab = (b+a)
(b-a)/ab x ab = (b-a)
the simplified fraction is (b+a)/(b-a)

3.Simplify (x²/4)/(y/x)
Solution: x²/4 .  x/y  
= x³/4y
4.How to simplify complex fraction given below
               (a+b)/(x-y) / (a² - b²)/(x² - y²)
Solution: We re-write the given complex fraction,
          (a+b)/ (x-y)  ×  (x² - y²)/(a² - b²)
we have, (a² - b²) = (a+b)(a-b)
        (a+b)/ (x-y) × (x+y)(x-y)/(a+b)(a-b)
on simplification, we get   (x+y)/(a-b)
5.simplify the complex fraction given,
4 ¾ /3 ½
Solution: Numerator = 4 ¾ = 19/4
               Denominator = 3 ½ = 7/2
LCD of 4 and 2 is 4
19/4 x 4 = 19
7/2 x 4 = 14
the simplified fraction is 19/14

Algebraic Numbers

In today's post i will help you in learning the concept of algebraic numbers.

Algebraic number is a number that exist in a polynomial equations having integers, co efficient and so on. There are different types of algebraic numbers and these are as follows namely:

Natural numbers
Rational numbers
Irrational numbers
Complex numbers

Next time i will help you with some other concept such as algebraic manipulation.

You can also avail your help from online tutoring. Not just in algebra but in other concepts such as calculus tutoring and so on.

Do post your comments.

Sample Statistics

Let's learn about mean in sample statistics in today's session of learning.

Sample statistics is nothing but making use of numerical data in order to do some data research and data analysis. It is also the technique to summarize numerical data in statistics. Mean is used to summarize this numerical data in statistics and therefore, it is such an important concept in sample statistics.

Next time i will help you with the concept of errors in sampling in statistics. One can also connect with an online tutor for more help. Not just in statistics but one can avail to the help of algebra tutors as well.

Do post your comments.

Mixed fractions

Let's learn about mixed fractions and subtracting mixed fractions in today's learning.

Mixed fractions are those fractions which are formed of a whole number and a fraction. In order to subtract mixed fractions, we need to first convert the mixed fractions to improper fractions and then use the same method of subtract fractions. Below are the steps of subtracting mixed fractions:

Step 1) Convert mixed to improper fraction.
Step 2) If denominators are different, find the LCM and take a common denominator.
Step 3) Subtract the numerators.

For more help avail to an online tutor and get your required help. Not just fractions but you can avail to free algebra tutoring and so on as well.

Do post your comments.

probability examples

In this blog we will learn about probability examples,

Ex 1: What is the probability of getting a sum is 4 or 8 when throwing two dice simultaneously?

Solution:Now we will see how to solve probability examples

Let S = sample space, S = {(1, 1), (1, 2), (1,3)...(6, 5), (6, 6)}, n(S) = 36

A be the event of getting a sum is 4, n(A) = {(1, 3), (2, 2), (3, 1)} = 3

B be the event of getting a sum is 8, n(B) = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} = 5

P(A) = =

P(B) = =

P(A or B) = P(A) + P(B)

P(4 or 8) = + = .

In the next blog we will learn about linear definition and math dictionary algebra.Hope you like the above example of probability examples,please leave your comments if you have any doubts.