Computing Fractions

Introduction:
          The whole part is divided in to different parts. Each parts is called fraction of the whole thing. The fraction is shown as a/b, where a is referred as numerator and b is referred as denominator. Those denominator and numerators are involed in fraction. Fractions are differentiated by their values of numerator and denominator. They are computed and described below.
The example of fraction is `2/6`

Common Factor for Computing Fractions:

This common factor is used for solving and reducing fractions:

Example:

16 =4*4

20 =4*5

       Take all common numbers.

       The product of common numbers and remaining numbers is called common factor.

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Description for Computing Fractions with Examples:

Computing fractions for addition:

Example: `2/5` + `3/5`

Here denominators are same.

So numerators are added  `(2 + 3)/5` .

Therefore the resultant fraction is `5/5`

The result for the addtion fraction is 1

Example: `2/5 ` + `3/4`

Here denominators are different. So we have to find least common divisor.

The lease common divisor for 4 and 5 is 20.

In `2/ 5` , the denominator 5 is 4 times in the least common divisor.

So we have to multiply the numeratory 2 by 4. 2 * 4 = 8.

In `3/4,` the donminator 4 is 5 times in the least common divisor. so we have to multiply the numerator 3 by 4. 3 * 5 = 15

Now, we can add the denominator. so we will get `(8 +15)/20` = `23/20` 

Computing fractions for subtraction: `8/5 - 3/5`

Here denominators are same. So we are doing like below.

Numerators are subtracted. `(8 - 3)/5` .

So we get `5/5` =1

Example: `2/5 - 3/4`

Here denominators are different. So we have to find least common divisor.

The lease common divisor for 4 and 5 is 20.

In `2/ 5` , the denominator 5 is 4 times in the least common divisor.

So we have to multiply the numeratory 2 by 4. 2 * 4 = 8.

In `3/4` , the donminator 4 is 5 times in the least common divisor.

So we have to multiply the numerator 3 by 4. 3 * 5 = 15

Now, we can subtract the denominator. so we will get `(8 -15)/20 = -7/20`

Computing fractions for multiplication:

`3/5 xx 5/7`

Here 3 and 5 are numerators and  5 and 7 are denominators

Multiplying the numerator by numerator:

3 * 5 = 15

Multiplying the denominator by deminator

5 * 7 = 35

Therefore the fraction is  `15/35`

therefore the result for the multiplication fractorion is `3/7` 

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Computing fractions for division:

`2/5-:3/5`

Firest we have to find  reciprocal for the second fraction is 5/3.

Then we have to multiply this reciprocal with dividend fraction like below

`2/5 xx 5/3`

So we will get `10/15` .

Therefore the result is `2/3`

Math Problems for Pre Algebra

Introduction :
Pre algebra is a branch of mathematics that is applied to make the mathematical model of the real-world situations and to handle a problems that we might not solve the problems using the simple arithmetic. Using words, algebra utilize a symbol to make statements. Algebra consist of real numbers, complex numbers, linear equations, vectors etc. In algebra, we are frequently using the letters that stands for the numbers in mathematics.  Algebra uses symbols of the arithmetic operations for adding, subtracting, dividing and multiplying. In this article we shall discuss about solve math problems for pre algebra.

Sample Problem for Solve Math Problems for Pre Algebra:

Solve math problems for pre algebra problem 1:

Solve the given math problem for pre algebra equation and find out x value of the equation 3(3x - 2) + 5 = 2(4x + 5) – 10.

Solution:

We are going to find the x value of the given pre algebra equation.

In the first we are going to multiply the term – 5 in both sides of the equation. We get

3(3x - 2) + 5 – 5 = 2(4x + 5) – 10 -5

3(3x - 2) = 2(4x + 5) - 15

In the next step grouping the same values in the above equation, we get

9x -8x = 6 + 10 – 15

x = 1

The value of the x in a given equation is 1.

Solve math problems for pre algebra problem 2:

Solve the given math problem for pre algebraic equation step by step and find the x value (5x – 6) + 8 = (6 x – 8) + 8.

Solution:

We are going to find the x value of the given pre algebra equation.

In the first we are going to multiply the value – 8 in both sides of the equation. We get

(5x – 6) + 8 – 8 = (6 x – 8) + 8 – 8

(5x – 6) = (6 x – 8)

In the next step grouping the same values in the above equation, we get

5x -6x = 6 – 8

-x = -2

x = 2

The solution of the x in a given equation is 2.

If we have any doubt in the solution means we check the answer of given problem. Substituting x = 2 in the given equation, we get

5(2) – 6 + 8 = 6(2) – 8 + 8

10 – 6 + 8 = 12 – 8 + 8

12 = 12

The left hand side and right hand side of the answers are same. So the solution is correct.

Practice Problem for Solve Math Problems for Pre Algebra:

Solve the given algebraic equation 1 + 2(1 + 4)² - 4

Answer: 47

Solve the given algebraic expression 2 + 3(1 + 4)² - 20

Answer: 57