Addition of Algebraic Expressions is the process of collecting like terms and adding them. Identify and add the coefficients of like terms and sum them to find the final expression of given problems. We have explained clear ways to find the Addition of Algebraic Expressions. We even provided different examples for different methods to solve the addition of algebraic expressions. Therefore, check practice questions, answers, and explanations for every problem and get the best knowledge to solve algebraic expressions.

Methods to Solve an Addition of Algebraic Expressions

Students can perform the addition of algebraic expressions using two methods. Use the best and easy method for you and find the final expression. The two methods to find the addition of algebraic expressions are given below.
1. Horizontal Method
2. Column Method

How to find the addition of algebraic expressions using the Horizontal Method?

The Horizontal Method is the simplest way to find the addition of algebraic expressions. Just by following the below step by step procedure, students can estimate the addition of algebraic expressions.

• Firstly, note down the given expressions.
• Place the given expressions in a row and separate them with the addition symbol in between them.
• Re-arrange the given terms by grouping or placing like terms together.
• Simplify the coefficients of like terms.
• Finally, write the resultant expression in standard form.

How to find the addition of algebraic expressions using the Column Method?

The Column method is also called a Vertical Method. The Addition of Algebraic Expressions can be estimated using the vertical method by writing expressions in separate rows. Before you write separate rows, you need to arrange the expressions with like terms in the correct order. Have a look at the below procedure to exactly know what is the process to find the addition of algebraic expressions using the Column Method.

• Write the given expressions.
• Place one expression below the other expression with the like terms come in the same column.
• Next, add the like terms column-wise with their coefficients.
• Finally, fins the resultant expression in the standard form.

Addition of Algebraic Expressions Solved Examples

1. Add 5a + 7b – 6c, b + 2c – 3a and 2a – 5b – 3c

Solution:Given expressions are 5a + 7b – 6c, b + 2c – 3a and 2a – 5b – 3c.

Horizontal Method:
There are three variables available.
Note down the like terms and then find the sum of the numerical coefficients of all terms.
5a + 7b – 6c + b + 2c – 3a + 2a – 5b – 3c.
Arrange the like terms together.
5a – 3a + 2a + 7b + b – 5b – 6c + 2c – 3c.
Now, find the sum of the numerical coefficients of all terms.
4a + 3b – 7c.

The required expression is 4a + 3b – 7c.

Column Method:
Arrange the given expressions in the same order and write them in rows. Note down the like terms below each other and add them in column-wise.
Rearrange the given expressions.
5a + 7b – 6c, – 3a + b + 2c, 2a – 5b – 3c
Note down the like terms below each other and add them in column-wise.
+ 5a + 7b – 6c
– 3a +  b + 2c
+ 2a – 5b – 3c
—————————-
4a + 3b – 7c

The required expression is 4a + 3b – 7c.

2. Add 7x² + 8y – 9, 3y + 2 – 3x² and 3 – y + 3x².

Solution:
Given expressions are 7x² + 8y – 9, 3y + 2 – 3x² and 3 – y + 3x².

Horizontal Method:
There are three variables available.
Note down the like terms and then find the sum of the numerical coefficients of all terms.
7x² + 8y – 9 + 3y + 2 – 3x² + 3 – y + 3x².
Arrange the like terms together.
7x² – 3x² + 3x² + 8y + 3y – y – 9 + 2 + 3.
Now, find the sum of the numerical coefficients of all terms.
7x² + 10y – 4.

The required expression is 7x² + 10y – 4.

Column Method:
Arrange the given expressions in the same order and write them in rows. Note down the like terms below each other and add them in column-wise.
Rearrange the given expressions.
7x² + 8y – 9,  – 3x² + 3y + 2 and + 3x² – y + 3
Note down the like terms below each other and add them in column-wise.
+ 7x² + 8y – 9
– 3x²  + 3y + 2
+ 3x² – y  + 3
—————————-
7x² + 10y – 4

The required expression is 7x² + 10y – 4.

3. Add 4x² – 2xy + 4y², 3xy – 5y² + 9x² and 2y² + xy – 7x².

Solution:
Given expressions are 4x² – 2xy + 4y², 3xy – 5y² + 9x² and 2y² + xy – 7x².

Horizontal Method:
There are three variables available.
Note down the like terms and then find the sum of the numerical coefficients of all terms.
4x² – 2xy + 4y² + 3xy – 5y² + 9x² + 2y² + xy – 7x².
Arrange the like terms together.
4x² + 9x² – 7x² – 2xy + 3xy + xy + 4y² – 5y² + 2y².
Now, find the sum of the numerical coefficients of all terms.
6x² + 2xy + y².

The required expression is 6x² + 2xy + y².

4. Add 10a² + 3b² – c², 2b² + 6c² – 7a² and 3a² – 8b² – 6c².

Solution:
Column Method:
Arrange the given expressions in the same order and write them in rows. Note down the like terms below each other and add them in column-wise.
Rearrange the given expressions.
10a² + 3b² – c², 2b² + 6c² – 7a² and 3a² – 8b² – 6c².
Note down the like terms below each other and add them in column-wise.
+ 10a² + 3b² – c²
– 7a² + 2b² + 6c²
+ 3a² – 8b² – 6c²
—————————-
6a² – 3b² – c²

The required expression is 6a² – 3b² – c².

5. Add the 4x + 8y and 2x + 3y.
Solution:
Given expressions are 4x + 8y and 2x + 3y.

Horizontal Method:
There are two variables available.
Note down the like terms and then find the sum of the numerical coefficients of all terms.
4x + 8y + 2x + 3y
Arrange the like terms together.
4x + 2x + 8y + 3y
Now, find the sum of the numerical coefficients of all terms.
6x + 11y

The required expression is 6x + 11y.

Column Method:
Arrange the given expressions in the same order and write them in rows. Note down the like terms below each other and add them in column-wise.
Rearrange the given expressions.
4x + 8y and 2x + 3y
Note down the like terms below each other and add them in column-wise.
4x + 8y
2x + 3y
—————————-
6x + 11y

The required expression is 6x + 11y.

6. Add 3x + 9y + 5 and 4x + 3y + 2

Solution:
Given expressions are 3x + 9y + 5 and 4x + 3y + 2.

Horizontal Method:
There are two variables available.
Note down the like terms and then find the sum of the numerical coefficients of all terms.
3x + 9y + 5 + 4x + 3y + 2
Arrange the like terms together.
3x + 4x + 9y + 3y + 5 + 2
Now, find the sum of the numerical coefficients of all terms.
7x + 12y + 7

The required expression is 7x + 12y + 7.

Column Method:
Arrange the given expressions in the same order and write them in rows. Note down the like terms below each other and add them in column-wise.
Rearrange the given expressions.
3x + 9y + 5 and 4x + 3y + 2
Note down the like terms below each other and add them in column-wise.
3x + 9y + 5
4x + 3y + 2
—————————-
7x + 12y + 7

The required expression is 7x + 12y + 7.

7. Add 12x + 4y + 21z and 32x – 2y – 16z

Solution:
Given expressions are 12x + 4y + 21z and 32x – 2y – 16z.

Horizontal Method:
There are three variables available.
Note down the like terms and then find the sum of the numerical coefficients of all terms.
12x + 4y + 21z + 32x – 2y – 16z
Arrange the like terms together.
12x + 32x + 4y – 2y + 21z – 16z
Now, find the sum of the numerical coefficients of all terms.
44x + 2y -5z

The required expression is 44x + 2y -5z.

Column Method:
Arrange the given expressions in the same order and write them in rows. Note down the like terms below each other and add them in column-wise.
Rearrange the given expressions.
12x + 4y + 21z and 32x – 2y – 16z
Note down the like terms below each other and add them in column-wise.
12x + 4y + 21z
32x – 2y – 16z
—————————-
44x + 2y -5z

The required expression is 44x + 2y -5z.

8. Add 6x³ – 4y³ and 9x³ – 5y³
Solution:
Given expressions are 6x³ – 4y³ and 9x³ – 5y³.

Horizontal Method:
There are two variables available.
Note down the like terms and then find the sum of the numerical coefficients of all terms.
6x³ – 4y³ + 9x³ – 5y³
Arrange the like terms together.
6x³ + 9x³ – 4y³ – 5y³
Now, find the sum of the numerical coefficients of all terms.
15x³ – 9y³

The required expression is 15x³ – 9y³.

Column Method:
Arrange the given expressions in the same order and write them in rows. Note down the like terms below each other and add them in column-wise.
Rearrange the given expressions.
6x³ – 4y³ and 9x³ – 5y³.
Note down the like terms below each other and add them in column-wise.
9x³ – 5y³
6x³ – 4y³
—————————-
15x³ – 9y³

The required expression is 15x³ – 9y³.

9. Add 3a² + 5b² + 7c² – 9abc and 2a² – 4b² + 6c² + 8abc
Solution:
Given expressions are 3a² + 5b² + 7c² – 9abc and 2a² – 4b² + 6c² + 8abc.

Horizontal Method:
There are three variables available.
Note down the like terms and then find the sum of the numerical coefficients of all terms.
3a² + 5b² + 7c² – 9abc + 2a² – 4b² + 6c² + 8abc
Arrange the like terms together.
3a² + 2a² + 5b² – 4b² + 7c² + 6c² – 9abc + 8abc
Now, find the sum of the numerical coefficients of all terms.
5a² + b² + 13c² – abc

The required expression is 5a² + b² + 13c² – abc.

Column Method:
Arrange the given expressions in the same order and write them in rows. Note down the like terms below each other and add them in column-wise.
Rearrange the given expressions.
3a² + 5b² + 7c² – 9abc and 2a² – 4b² + 6c² + 8abc
Note down the like terms below each other and add them in column-wise.
3a² + 5b² + 7c² – 9abc
2a² – 4b² + 6c² + 8abc
—————————-
5a² + b² + 13c² – abc

The required expression is 5a² + b² + 13c² – abc.

10. Add 2xy² + 6x²y – 9x²y – 4xy² + 5 and 3x²y + 2xy²

Solution:
Given expressions are 2xy² + 6x²y – 9x²y – 4xy² + 5 and 3x²y + 2xy².

Horizontal Method:
There are three variables available.
Note down the like terms and then find the sum of the numerical coefficients of all terms.
2xy² + 6x²y – 9x²y – 4xy² + 5 + 3x²y + 2xy²
Arrange the like terms together.
2xy² + 2xy² + 6x²y + 3x²y – 9x²y – 4xy² + 5
Now, find the sum of the numerical coefficients of all terms.
4xy² + 9x²y – 9x²y – 4xy² + 5

The required expression is 4xy² + 9x²y – 9x²y – 4xy² + 5.

Column Method:
Arrange the given expressions in the same order and write them in rows. Note down the like terms below each other and add them in column-wise.
Rearrange the given expressions.
2xy² + 6x²y – 9x²y – 4xy² + 5 and 3x²y + 2xy²
Note down the like terms below each other and add them in column-wise.
2xy² + 6x²y – 9x²y – 4xy² + 5
2xy² + 3x²y +0    + 0    + 0
—————————-
4xy² + 9x²y – 9x²y – 4xy² + 5

The required expression is 4xy² + 9x²y – 9x²y – 4xy² + 5.

11. Add 2x² + 3y – 4z², 5y + 3x², 4x² + 9z² – 8y and 3y – 3x².

Solution:
Given expressions are 2x² + 3y – 4z², 5y + 3x², 4x² + 9z² – 8y and 3y – 3x².

Horizontal Method:
There are three variables available.
Note down the like terms and then find the sum of the numerical coefficients of all terms.
2x² + 3y – 4z² + 5y + 3x² + 4x² + 9z² – 8y + 3y – 3x²
Arrange the like terms together.
2x² + 3x² + 4x² – 3x² + 3y + 5y – 8y + 3y – 4z² + 9z²
Now, find the sum of the numerical coefficients of all terms.
6x² + 3y + 5z²

The required expression is 6x² + 3y + 5z².

Column Method:
Arrange the given expressions in the same order and write them in rows. Note down the like terms below each other and add them in column-wise.
Rearrange the given expressions.
2x² + 3y – 4z², 3x² + 5y, 4x² – 8y + 9z² and – 3x² + 3y
Note down the like terms below each other and add them in column-wise.
2x² + 3y – 4z²
3x² + 5y + 0
4x² – 8y + 9z²
– 3x² + 3y + 0
—————————-
6x² + 3y + 5z²

The required expression is 6x² + 3y + 5z².

The post Addition of Algebraic Expressions | Adding Algebraic Expressions, Solved Examples appeared first on Learn CBSE.

Subtraction of algebraic expressions is subtracting one expression from another expression. A detailed explanation is given about the Subtraction of algebraic expressions in this article. Go through each step of solving problems and understand how simply a Subtraction of algebraic expressions problem can be solved. Students can get complete knowledge of Algebraic Expressions by referring to this page.

How to Find Subtraction of Algebraic Expressions?

Follow the below-listed steps to Subtract Algebraic Expressions and arrive at the solution easily. They are along the lines

• Write the given expressions in standard form.
• After that arrange one expression under another expression with the like terms come in the same column.
• The main part of the subtraction of algebraic expressions is changing the sign of every individual term of the second expression to get the inverse of the expression.
• Lastly, add the like terms and get the final expression.

Subtraction of Algebraic Expressions Solved Examples

1. Subtract 3a + 4b – 2c from 5a – 2b + 2c

Solution:
Note down both given expressions and rearrange them if required.
3a + 4b – 2c = 3a + 4b – 2c
5a – 2b + 2c = 5a – 2b + 2c
Write both expressions one below another such that the expressions will subtract each other with the like terms come in the same column.
5a – 2b + 2c
3a + 4b – 2c
Change the signs of each term available in the second row and fins the inverse of the second expression.
3a + 4b – 2c = -(3a + 4b – 2c) = – 3a – 4b + 2c
Add the like terms by arranging two expressions in columns to get the final expression.
5a – 2b + 2c
– 3a – 4b + 2c
—————————-
2a – 6b + 4c

The required expression is 2a – 6b + 4c

2. Subtract 4x² – 7x – 5 from 6 + 2x – 3x².

Solution:
Note down both given expressions and rearrange them if required.
6 + 2x – 3x² = – 3x² + 2x + 6
4x² – 7x – 5 = 4x² – 7x – 5
Write both expressions one below another such that the expressions will subtract each other with the like terms come in the same column.
– 3x² + 2x + 6
4x² – 7x – 5
Change the signs of each term available in the second row and fins the inverse of the second expression.
4x² – 7x – 5 = -(4x² – 7x – 5) = – 4x² + 7x + 5
Add the like terms by arranging two expressions in columns to get the final expression.
– 3x² + 2x + 6
– 4x² + 7x + 5
—————————-
– 7x² + 9x + 11

The required expression is – 7x² + 9x + 11

3. Subtract 4x + 2y – 4z from 10x – 6y + 2z

Solution:
Note down both given expressions and rearrange them if required.
10x – 6y + 2z = 10x – 6y + 2z
4x + 2y – 4z = 4x + 2y – 4z
Write both expressions one below another such that the expressions will subtract each other with the like terms come in the same column.
10x – 6y + 2z
4x + 2y – 4z
Change the signs of each term available in the second row and fins the inverse of the second expression.
4x + 2y – 4z = -(4x + 2y – 4z) = – 4x – 2y + 4z
Add the like terms by arranging two expressions in columns to get the final expression.
10x – 6y + 2z
– 4x – 2y + 4z
—————————-
6x – 8y + 6z

The required expression is 6x – 8y + 6z

4. Subtract – 5ab + 2a² from 3a² + 9ab.

Solution:
Note down both given expressions and rearrange them if required.
– 5ab + 2a² = 2a² – 5ab
3a² + 9ab = 3a² + 9ab
Write both expressions one below another such that the expressions will subtract each other with the like terms come in the same column.
3a² + 9ab
2a² – 5ab
Change the signs of each term available in the second row and fins the inverse of the second expression.
2a² – 5ab = -(2a² – 5ab) = – 2a² + 5ab
Add the like terms by arranging two expressions in columns to get the final expression.
3a² + 9ab
– 2a² + 5ab
—————————-
a² + 14ab

The required expression is a² + 14ab

5. Subtract 2x² – 5xy + 9y² – 4 from 7xy – 2x² – 4y² + 7.

Solution:
Note down both given expressions and rearrange them if required.
7xy – 2x² – 4y² + 7 = – 2x² + 7xy – 4y² + 7
2x² – 5xy + 9y² – 4 = 2x² – 5xy + 9y² – 4
Write both expressions one below another such that the expressions will subtract each other with the like terms come in the same column.
– 2x² + 7xy – 4y² + 7
2x² – 5xy + 9y² – 4
Change the signs of each term available in the second row and fins the inverse of the second expression.
2x² – 5xy + 9y² – 4 = -(2x² – 5xy + 9y² – 4) = – 2x² + 5xy – 9y² + 4
Add the like terms by arranging two expressions in columns to get the final expression.
– 2x² + 7xy – 4y² + 7
– 2x² + 5xy – 9y² + 4
—————————-
– 4x² + 12xy – 13y² + 11

The required expression is – 4x² + 12xy – 13y² + 11

6. What should be subtracted from 3a³ – 5a² + 7a – 8 to obtain 2a² – 4a + 3 ?

Solution:
Let ‘S’ denote the required expression.
Given that S subtracted from 3a³ – 5a² + 7a – 8 to get 2a² – 4a + 3.
(3a³ – 5a² + 7a – 8) – S = 2a² – 4a + 3
S = (3a³ – 5a² + 7a – 8) – (2a² – 4a + 3)
Note down both given expressions and rearrange them if required.
3a³ – 5a² + 7a – 8 = 3a³ – 5a² + 7a – 8
2a² – 4a + 3 = 2a² – 4a + 3
Write both expressions one below another such that the expressions will subtract each other with the like terms come in the same column.
3a³ – 5a² + 7a – 8
0   + 2a² – 4a + 3
Change the signs of each term available in the second row and fins the inverse of the second expression.
0   + 2a² – 4a + 3 = -(0   + 2a² – 4a + 3) = – 0  – 2a² + 4a – 3
Add the like terms by arranging two expressions in columns to get the final expression.
3a³ – 5a² + 7a – 8
– 0  – 2a² + 4a – 3
—————————-
3a³ – 7a² + 11a – 11

The required expression is 3a³ – 7a² + 11a – 11

7. Subtract 5a³ – 6a² + 2a – 10 from the sum of 5a³ + 6a² + 9 and 8a² – 4?

Solution:
Given that Subtract 5a³ – 6a² + 2a – 10 from the sum of 5a³ + 6a² + 9 and 8a² – 4.
Therefore, firstly, find the sum of 5a³ + 6a² + 9 and 8a² – 4.
Note down the like terms and then find the sum of the numerical coefficients of all terms.
5a³ + 6a² + 9 + 8a² – 4
Arrange the like terms together.
5a³ + 6a² + 8a² + 9 – 4
Now, find the sum of the numerical coefficients of all terms.
5a³ + 14a² + 5
The required expression is 5a³ + 14a² + 5.
Now, subtract 5a³ – 6a² + 2a – 10 from 5a³ + 14a² + 5.
Note down both given expressions and rearrange them if required.
5a³ + 14a² + 5 = 5a³ + 14a² + 5
5a³ – 6a² + 2a – 10 = 5a³ – 6a² + 2a – 10
Write both expressions one below another such that the expressions will subtract each other with the like terms come in the same column.
5a³ + 14a² + 0.a + 5
5a³ – 6a² + 2a – 10
Change the signs of each term available in the second row and fins the inverse of the second expression.
5a³ – 6a² + 2a – 10 = -(5a³ – 6a² + 2a – 10) = – 5a³ + 6a² – 2a + 10
Add the like terms by arranging two expressions in columns to get the final expression.
5a³ + 14a² + 0.a + 5
– 5a³ + 6a² – 2a + 10
—————————-
0 + 20a² – 2a + 15

The required expression is 20a² – 2a + 15

8. What should be subtracted from 10a³ – 15a² + 9a – 21 to obtain 6a² – 14a + 36 ?
Solution:
Let ‘S’ denote the required expression.
Given that S subtracted from 10a³ – 15a² + 9a – 21 to get 6a² – 14a + 36.
(10a³ – 15a² + 9a – 21) – S = 6a² – 14a + 36
S = (10a³ – 15a² + 9a – 21) – (6a² – 14a + 36)
Note down both given expressions and rearrange them if required.
10a³ – 15a² + 9a – 21 = 10a³ – 15a² + 9a – 21
6a² – 14a + 36 = 6a² – 14a + 36
Write both expressions one below another such that the expressions will subtract each other with the like terms come in the same column.
10a³ – 15a² + 9a – 21
0   + 6a² – 14a + 36
Change the signs of each term available in the second row and fins the inverse of the second expression.
0 + 6a² – 14a + 36 = -(0 + 6a² – 14a + 36) = – 0  – 6a² + 14a – 36
Add the like terms by arranging two expressions in columns to get the final expression.
10a³ – 15a² + 9a – 21
– 0  – 6a² + 14a – 36
—————————-
10a³ – 21a² + 23a – 57

The required expression is 10a³ – 21a² + 23a – 57

9. Subtract 21a³ – 3a² + 4a – 6 from the sum of 6a³ + 7a² + 15 and 7a² – 16?

Solution:
Given that Subtract Subtract 21a³ – 3a² + 4a – 6 from the sum of 6a³ + 7a² + 15 and 7a² – 16.
Therefore, firstly, find the sum of 6a³ + 7a² + 15 and 7a² – 16.
Note down the like terms and then find the sum of the numerical coefficients of all terms.
6a³ + 7a² + 15 + 7a² – 16
Arrange the like terms together.
6a³ + 7a² + 7a² + 15 – 16
Now, find the sum of the numerical coefficients of all terms.
6a³ + 14a² – 1
The required expression is 6a³ + 14a² – 1.
Now, subtract 21a³ – 3a² + 4a – 6 from 6a³ + 14a² – 1.
Note down both given expressions and rearrange them if required.
6a³ + 14a² – 1 = 6a³ + 14a² – 1
21a³ – 3a² + 4a – 6 = 21a³ – 3a² + 4a – 6
Write both expressions one below another such that the expressions will subtract each other with the like terms come in the same column.
6a³ + 14a² + 0.a – 1
21a³ – 3a² + 4a – 6
Change the signs of each term available in the second row and fins the inverse of the second expression.
21a³ – 3a² + 4a – 6 = -(21a³ – 3a² + 4a – 6) = – 21a³ + 3a² – 4a + 6
Add the like terms by arranging two expressions in columns to get the final expression.
6a³ + 14a² + 0.a – 1
– 21a³ + 3a² – 4a + 6
—————————-
– 15a³ + 17a² – 4a + 5

The required expression is – 15a³ + 17a² – 4a + 5

The post Subtraction of Algebraic Expressions appeared first on Learn CBSE.

The pie chart represents data in a circular graph. The pieces of the chart or graph show the data size. The various components are represented by the pieces of a circle and the whole circle is the sum of the values of all components. One can get the pie chart definition, formula, use, solved examples, and steps to create a pie chart in the below sections.

Pie Chart Definition

A pie chart is also a graph and is a type of pictorial representation of data. It divides the circle into various sectors in order to explain the numeric values. Each section is a proportionate part of the whole circle. We use a pie chart to find the composition of something. The pie chart is also known as the circle chart.

A pie chart is used for data representation. The total of all data in a pie is equal to 360° and the total value of a pie is always 100%.

The central angle for a component formula is given as

Central angle for a component = (Value of the component / Sum of the values of all components) * 360°

Steps to Make a Pie Chart

Follow the below-mentioned steps to create a pie chart/ circle chart for the given data.

• Enter the given data into a table to make the process easy for you.
• Find the sum of values in the table.
• Divide each value in the table by sum and multiply the result with 100 to get the percent.
• To get the degrees of each value, substitute the values in the formula i.e (Value of each component/sum) * 360A pie chart is used for the data representation. The total of all data in a pie is equal to 360° and the total value of a pie is always 100%.
• The central angle for a component formula is given as Central angle for a component = (Value of the component / Sum of the values of all components) * 360°.
• Draw a circle and use a protractor to measure the degree of each sector.

Advantages and Disadvantages of Pie Chart

Advantages

• It is simple and easy to understand
• Data can be represented visually as a fractional part of the whole.
• It helps in providing an effective communication tool for the uninformed audience.

Disadvantages

• It can represent only one set of data. So, you need a series to compare multiple sets.
• If you have too many pieces of data, and even you add labels and numbers may not help here, they themselves may become crowded and hard to read.

Example Questions & Answers

Example 1.

The following table shows the expenditure in percentage incurred on the construction of a house in a city:

Draw a pie chart for the above data.

Solution:

Total percentage = 100

Central angle for a component = (Value of the component / Sum of the values of all components) * 360°

Calculation of central angles:

Construction for creating a pie chart

Steps of construction:

1. Draw a circle of any convenient radius.

2. Draw a horizontal radius of the circle.

3. Draw sectors starting from the horizontal radius with central angles of 64.8°, 108°, 36°, 43.2°, and 108° respectively.

4. Shade the sectors differently using different colors and label them.

Thus, we obtain the required pie chart, shown in the above figure.

Example 2.

The marks scored by a student in his examination are shown below:

Draw a pie chart for the above data.

Solution:

The given table contains the marks of a student in an exam.

The total marks scored by a student = 411

Central angle for a component = (Each subject marks / total marks) * 360°

Calculation of central angles:

Construction for creating a pie chart

Steps of construction:

1. Draw a circle.

2. Draw sectors starting from the horizontal radius with central angles of 61.31°, 74.45°, 66.56°, 77.08°, 80.58° respectively.

3. Shade the sectors differently using different colors and label them.

Thus, we obtain the required pie chart, shown in the above figure.

Example 3.

The data on the mode of transport used by 1000 students are given below:

Represent the data on a pie chart.

Solution:

The given table shows the mode of transport used by 1000 students to reach the school.

Total Number of Students = 1000

Central angle for a component = (Value of the component / Sum of the values of all components) * 360°

Calculation of central angles:

Construction for creating a pie chart

1. Draw a circle of any convenient radius.

2. Draw a horizontal radius of the circle.

3. Draw sectors starting from the horizontal radius with central angles of 54°, 18°, 72°, 36°, and 180°.

4. Shade the sectors differently using different colors and label them.

Thus, we obtain the required pie chart, shown in the above figure.

FAQs on Pie Chart

1. What are the uses of a pie chart?

A pie chart is used to represent categorical data, to compare areas of growth in a business like turnover profit and exposure.

2. What are the pie chart examples?

The real-life examples of the pie charts are a representation of kinds of cars sold in a month, types of food items liked by people in a room, marks scored by students in a class, etc

3. What is the formula to calculate the percentage of each component of a pie chart?

The pie chart formula to calculate the central angle for a component is as follows:

Central angle for a component = (Value of the component / Sum of the values of all components) * 360°

The post Pie Chart Definition, Formula, Examples | How to Make a Pie Chart? appeared first on Learn CBSE.

A bar graph is a graph with rectangular bars. It is used to compare different categories. The most usual type of bar graph is vertical. One who is willing to learn more about the bar graph can refer to the following sections. On this page, we are providing the definition of a bar graph or column graph, solved example questions, and types of the bar graph.

Bar Graph Definition

A bar graph is a pictorial representation of grouped data in the form of horizontal or vertical rectangular bars, here the length of rectangular bars is equal to the measure of the data. The height of a rectangular bar totally depends on the numeric value it represents. The frequency distribution tables can be represented using bar charts which simplify the calculations and understanding of data.

Types of Bar Graphs or Column Graphs

In general, bar graphs can be horizontal or vertical rectangular bars. The main feature of a bar chart is its length or height. If the height of any bar graph is more, then the values are greater than the given data. In any bar graph, bars represent frequencies of distinctive values of a variable. The number of values on the x-axis of a bar graph and the y-axis of a column graph is known as the scale. The two types of bar graphs are the vertical bar graph and the horizontal bar graph. Apart from vertical, horizontal bar graphs, the other two types are grouped bar graphs, stacked bar graphs.

• Vertical Bar Graph: If the given data is represented vertically in a graph with the help of bars, where the bars denote the measure of data such graphs are called the vertical bar graphs.
• Horizontal Bar Graph: If the grouped data is represented horizontally in a graph with the help of bars, then those graphs are called the horizontal bar graphs.
• Grouped Bar Graph: It is used to represent the discrete value for more than one object. Here, the number of instances are combined together into a single bar. It is also known as the clustered graph.
• Stacked Bar Graph: It divides an aggregate into different parts. A stacked bar graph is also known as the composite bar graph.

How to draw Bar Graph or Column Graph?

• On a graph paper, draw two lines that represent x-axis (OX), y-axis (OY).
• Mark the points at equal intervals along the x-axis, also mention the names of data items whose values are plotted.
• Select a suitable scale. Find the heights of the bar for the given values.
• Mark off these heights or lengths parallel to the y-axis from the points taken in Step 2.
• Draw the bars of equal width for the heights marks in the above step along the x-axis. Those bars represent the numerical data.

Column Graph Solved example Questions

Example 1.

The annual earnings of a firm during the 8 consecutive years are given below:

Draw a bar graph representing the above data.

Solution:

Given that,

The annual earnings of a firm over the 8 consecutive years.

On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.

Take years along the x-axis at equal gaps.

Choose 1 small division = 10 dollars

Then, the heights of the bars are:

Annual income in 2010-11 = (1/10 * 150) = 15 small divisions

Annual income in 2011-12 = (1/10 * 120) = 12 small divisions

Annual income in 2012-13 = (1/10 * 170) = 17 small divisions

Annual income in 2013-14 = (1/10 * 180) = 18 small divisions

Annual income in 2014-15 = (1/10 * 160) = 16 small divisions

Annual income in 2015-16 = (1/10 * 190) = 19 small divisions

Annual income in 2016-17 = (1/10 * 200) = 20 small divisions

Annual income in 2017-18 = (1/10 * 205) = 20.5 small divisions

Bar graph showing firm annual income during eight consecutive years

At the points marked in Step 2, draw bars of equal width and of heights calculated.

Example 2.

In a firm of 500 employees, the percentage of monthly salary saved by each employee is given in the following table. Represent it using a bar graph.

Solution:

Given that,

The savings percentage of 500 employees in a firm.

Draw two lines OX, OY on graph paper.

Take savings along the x-axis, number of employees along the y-axis.

Choose 1 small division = 10 units on y-axis.

Then, the heights of the bars are:

The number of employees who are saving 20% of salary is (1/10 * 100) = 10 small divisions

The number of employees who are saving 30% of salary is (1/10 * 50) = 5 small divisions

The number of employees who are saving 50% of salary is (1/10 * 73) = 7.3 small divisions

The number of employees who are saving 60% of salary is (1/10 * 150) = 15 small divisions

The number of employees who are saving 70% of salary is (1/10 * 60) = 6 small divisions

The number of employees who are saving 80% of salary is (1/10 * 67) = 6.7 small divisions

Bar graph showing the percentage of monthly salary saved by each employee in a firm.

At the points marked in Step 2, draw bars of equal width and of heights calculated.

Example 3.

70 adults and 70 children were surveyed to find out how many servings of fruit and vegetables they eat per week. The results are shown in the bar graph below:

(i) How many adults and children combine eat between 6 and 10 servings of fruit and vegetables per week?

(ii) What is the mode for the number of servings of fruit and vegetables eaten per week by adults?

(iii) What is the difference between the number of children who eat over 2020 servings of fruit and vegetables per week and between 66 to 1010 servings of fruit and vegetables per week?

(iv) What basic trend do we notice from the data presented?

Solution:

Given graph shows how many fruits and vegetables adults and children eat per week.

(i) Here, we need to take readings of the two bars showing 6-10 servings per week category. The number of adults who are between 6 and 10 servings per week is a total of 12. The number of children who are between 6 and 10 servings per week is a total of 25.

Therefore, the number of adults and children combined who have between 6 and 10 servings per week is 12 + 25 = 37.

(ii) Simply, we need to locate the bar that represents children eating over 20 servings and the bar that represents children eating between 6 – 10 servings, find the difference between them.

The number of children who eat over 20 servings of fruit and vegetables is 4. The number of children who eat between 6 and 10 servings of fruit and vegetables is 25. Therefore, the difference between the two categories is 25 – 4 = 21

(iii) The mode is the most frequently occurring value. Since this question is about the adults, we do not need to consider the bars for children. All we need to do is see which bar is highest for the adults.

The 16 – 20 bar has the highest length i.e 22. Therefore the mode is 16 – 20 servings.

(iv) The most basic trend we can obtain from the given information is that adults eat more portions of fruit and vegetables per week than children.

FAQs on Bar Graph or Column Graph

1. How do you define a bar graph?

The bar graph represents the categorical data using rectangular bars. It shows a comparison between discrete categories.

2. What are the different types of bar graphs?

The different types of bar graphs are vertical bar graphs, grouped bar graphs, horizontal bar graphs, and stacked bar graphs.

3. Is a Histogram the same as a Bar Graph?

Bar graphs are used when your data is in categories. But histogram is used when the data is continuous.

4. When a bar graph is used?

The bar graph is used to compare the items between different groups over time. These are used to measure the changes over a period of time. When the changes are larger, a bar graph is the best option to represent the data.

The post Bar Graph or Column Graph | Difference between Bar Graph and Column Graph appeared first on Learn CBSE.

Simple Interest is a quick and easy method for calculating the Interest Charged on a Loan or Principal Amount. The Concept of Simple Interest is quite famous and is used in many sectors such as finance, automobile, and banking. Go through the further modules to know about What is Simple Interest, Formula to Calculate Simple Interest, Solved Examples on How to Calculate the Simple Interest.

What is Simple Interest?

Simple Interest is the method for calculating the Interest Amount on a Certain Amount of Sum. SI is obtained by multiplying the interest rate, principal, time duration that elapses between payments. In general, it is calculated on a daily, monthly or annual basis.

In real times, money is not borrowed or lent for free. While repaying the amount you borrowed you need to pay a certain amount of interest along with the amount you have taken. In fact, the amount of interest you repay depends on the loan amount as well as the time for which you borrow and the rate of interest.

Simple Interest Formula

You need to be familiar with the Simple Interest Formula in order to understand the concept of Finances. The formula for Simple Interest helps you find the interest amount if principal, rate of interest, and time duration are given.

Formula to Calculate Simple Interest is SI = (P × R ×T) / 100

Where, P = Principal

R = Rate of Interest (in percentage)

T = Time Duration (in years)

However, the formula to find the Amount is given by Amount (A) = Principal (P) + Interest (I).

The amount is the total money you pay back at the end of the time you borrowed.

Simple Interest Formula for Months

The Formula to find Simple Interest for Months varies slightly compared to a yearly basis. Let us consider the Principal Amount be P and Rate of Interest per Annum be R and n be the time duration in months then the formula to Calculate SI is as such

Simple Interest for n months = (P × n × R)/ (12 ×100)

Solved Examples on Simple Interest

1. Raju takes a loan of Rs 20,000 from a bank for a period of 2 years. The rate of interest is 5% per annum. Find the interest and the amount he has to pay by the end of 2 years?

Solution:

Principal or Loan Sum = 20, 000

Time Duration T = 2 Yrs

Rate of Interest R = 5%

Formula to Calculate the Simple Interest = (P × R ×T) / 100

= (20,000*5*2)/100

= 200000/100

= 2000

Amount to be repeated by the end of the year = Principal + Interest

= 20, 000+2000

= 22, 000

Therefore, Raju needs to repay a total of Rs. 22, 000/- after the end of 2 years.

2. Mohan pays Rs 13000 as an amount on the sum of Rs 10000 that he had borrowed for 3 years. Find the rate of interest?

Solution:

Amount = 13, 000

Principal = 10,000

SI = Amount – Principal

= 13, 000 – 10, 000

= 3, 000

Time Duration = 3 yrs

Rate of Interest R = ?

SI = (P × R ×T) / 100

3, 000 =(10, 000*R*3)/100

R = (3000*100)/(10,000*3)

= 300000/30000

= 10%

Therefore, Rate of Interest is 10%.

3. Neela borrowed Rs 40,000 for 2 years at a rate of 4% per annum. Find the interest accumulated at the end of 2 years?

Solution:

P = Rs 40,000

R = 4%

T = 2 years

SI = (P × R ×T) / 100

= (40, 000*4*2)/100

= 3200

Interest accumulated at the end of 2 years is Rs. 3,200/-

The post Simple Interest – Definition, Formula, Solved Examples appeared first on Learn CBSE.

Learn all about What is Simple Interest and How to Calculate Simple Interest from here. Simple Interest is paid or received over a certain period for a fixed percentage of the principal amount. Get to see Solved Examples on finding the Simplest Interest and the terms involved in it with detailed explanation. Refer to Simple Interest Formula and learn about it in detail.

When we borrow or lend money for a certain period we need to pay back along with extra money for availing the facility.

Formula to Calculate SI = PTR/100

where P is the Principal

T is the Time duration

R is the Rate of Interest in Percentage

Formula to Calculate Amount = Principal + Interest

Examples on Simple Interest

1. Find simple interest on $4000 at 10% per annum for 2 years. Also, find the amount?

Solution:

From given data P = $4000

R = 10%

T = 2 years

Substitute the given data in the formula for Simple Interest

SI = PTR/100

= (4000*2*10)/100

= 80000/100

= 800

SI = $800

Amount = Principal + Interest

= $4000+$800

= $4800

2. Calculate the simple interest on $ 8400 at 5 % p.a. for 8 months?

Solution:

Principal = $ 8400

Rate of Interest = 5%

Time Duration n = 8 months

We know the formula to calculate Simple Interest for Months is given by

SI = (P × n × R)/ (12 ×100)

Substitute the given data in the above formula

SI = (8400*8*5)/(12*100)

= $280

3. John took a loan of $10000 from a bank on 8th February 2009 at the rate of 5% p.a. and paid back the same on 2nd July 2009. Find the total amount to be paid by John?

Solution:

Principal = $10000

Time = 20 days+ 31 days + 30 days + 31 days + 30 days + 2 days

= 144 days

Rate of Interest = 5 %

General Formula for SI = (PTR)/100

Changing the formula in accordance with daily basis we have SI = (P*R*n)/100*365

= (10000*5*144)/100*365

= $197.26

Amount to be paid = Principal + Interest

= $10000 + $197.26

= $10197.26

4. A some amounted to $ 4000 at 2% Per Annum for the period of 5 years. Find their sum?

Solution:

From the given data A = $ 4000

R =  2 % Per Annum

T = 5 years

P = ?

Consider the principal to be found as X.

SI = PTR/100

= x*5*2/100

= 10x/100

= x/10

A = P + I
= x+x/10

= 11x/10

We know from the given data Amount = $4000

Equating them we have

11x/10 = $4000

11x= $4000*10

x = $40000/11

x = $3636.36

Therefore, Principal is $3636.36.

5. Arun takes a loan of $10,000 to buy a truck at a rate of 10 % Simple Interest. Calculate the Annual Interest to be paid for the loan amount?

Solution:

From the given data Principal P = $10,000

T = 1 Year

R = 10%

SI  PTR/100

= (10, 000*1*10)/100

= $ 1000

Amount = Principal + Interest

= $10,000 + $ 1000

= $11, 000

6. At what rate percent per annum a sum of Rs. 900 will become Rs. 3,600 in 10 years?

Solution:

P = Rs. 900, A = Rs. 3,600
T = 10 Years
I = A – P = 3600 – 900 = Rs. 2700

I = PTR/100

2700 = 900*10*R/100

R = (2700*100)/(900*10)

= 270000/9000

= 30%

At 30% Rate of Interest the Sum of Rs. 900 will become Rs.3600 in 10 years.

7. In how many times will a sum of money becomes 5 times itself as 30% per annum SI?

Solution:

Let the Amount be ‘x’.
After time ‘t’ money will be ‘5x’ then
Interest for time ‘t’ is
SI=5x−x=4x
Given Rate = 30% Per Annum
SI = PTR/100

4x = (x*t*30)/100

4*100/30 = t

t = 400/30

= 13.33 years

The Sum x will become 5x times itself at a 30% interest rate after a span of 13.33 years.

The post What is Simple Interest? | Solved Problems to Calculate Simple Interest appeared first on Learn CBSE.

MCQ Questions for Class 10 Social Science with Answers PDF Free Download is very important for students who want to score good marks in their CBSE board examination. Students who can Practice CBSE Class 10 Social Science Multiple Choice Questions with Answers to improve your score in Board Exams.

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MCQ Questions for CBSE Class 10 Social Science History with Answers : India and the Contemporary World – II

1. The Rise of Nationalism in Europe MCQ Questions
2. The Nationalist Movement in Indo-China MCQ Questions
3. Nationalism in India MCQ Questions
4. The Making of Global World MCQ Questions
5. The Age of Industrialisation MCQ Questions
6. Work, Life, and Leisure MCQ Questions
7. Print Culture and the Modern World MCQ Questions
8. Novels, Society and History MCQ Questions

MCQ Questions for Class 10 Social Science with Answers: Geography

MCQ Questions for CBSE Class 10 Social Science Geography with Answers : Contemporary India – II

1. Resource and Development MCQ Questions
2. Forest and Wildlife Resources MCQ Questions
3. Water Resources MCQ Questions
4. Agriculture MCQ Questions
5. Minerals and Energy Resources MCQ Questions
6. Manufacturing Industries MCQ Questions
7. Lifelines of National Economy MCQ Questions

MCQ Questions for Class 10 Social Science with Answers: Civics

MCQ Questions for CBSE Class 10 Social Science Civics (Political Science) with Answers : Democratic Politics – II

1. Power Sharing MCQ Questions
2. Federalism MCQ Questions
3. Democracy and Diversity MCQ Questions
4. Gender Religion and Caste MCQ Questions
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MCQ Questions for Class 10 Social Science with Answers: Economics

MCQ Questions for CBSE Class 10 Social Science Economics with Answers : Understanding Economic Development – II

1. Development MCQ Questions
2. Sectors of Indian Economy MCQ Questions
3. Money and Credit MCQ Questions
4. Globalisation and the Indian Economy MCQ Questions
5. Consumer Rights MCQ Questions

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MCQ Questions for Class 10 Maths with Answers PDF Free Download is very important for students who want to score good marks in their CBSE board examination. Students who can Practice CBSE Maths Multiple Choice Questions with Answers for Class 10 Pdf to improve your score in Board Exams.

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1. Real Numbers MCQ Questions
2. Polynomials MCQ Questions
3. Pair of Linear Equations in Two Variables MCQ Questions
4. Quadratic Equations MCQ Questions
5. Arithmetic Progressions MCQ Questions
6. Triangles MCQ Questions
7. Coordinate Geometry MCQ Questions
8. Introduction to Trigonometry MCQ Questions
9. Some Applications of Trigonometry MCQ Questions
10. Circles MCQ Questions
11. Constructions MCQ Questions
12. Areas Related to Circles MCQ Questions
13. Surface Areas and Volumes MCQ Questions
14. Statistics MCQ Questions
15. Probability MCQ Questions

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MCQ Questions for Class 10 Science with Answers PDF Free Download is very important for students who want to score good marks in their CBSE board examination. Students who can Practice Chapter-wise CBSE Class 10 Science Multiple Choice Questions with Answers to improve your score in Board Exams.

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MCQ Questions for Class 10 Science Chemistry with Answers

• Chapter 1 Chemical Reactions and Equations MCQ Questions
• Chapter 2 Acids, Bases and Salts MCQ Questions
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MCQ Questions for Class 10 Science Biology with Answers

• Chapter 6 Life Processes MCQ Questions
• Chapter 7 Control and Coordination MCQ Questions
• Chapter 8 How do Organisms Reproduce? MCQ Questions
• Chapter 9 Heredity and Evolution MCQ Questions
• Chapter 15 Our Environment MCQ Questions
• Chapter 16 Management of Natural Resources MCQ Questions

MCQ Questions for Class 10 Science Physics with Answers

• Chapter 10 Light Reflection and Refraction MCQ Questions
• Chapter 11 Human Eye and Colourful World MCQ Questions
• Chapter 12 Electricity MCQ Questions
• Chapter 13 Magnetic Effects of Electric Current MCQ Questions
• Chapter 14 Sources of Energy MCQ Questions

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On this page, you will learn completely about the rule of separation of division of algebraic fractions. Get to see the solved example questions in the coming sections of this article. Solved Examples on Separation of Division of Algebraic Fractions will make you familiar with the concept in a better way and you can solve related problems easily.

Rule of Separation of Division

• (x + y) / z = x/z + y/z
• (x – y) / z = x/z – y/z
• z / (x + y) ≠ z/x + z/y

From the above three expressions, we can observe that the denominator of the fractions should be the same to perform addition or subtraction operator between them. The rule of separation of division says that to calculate sum or difference between two or more fractions, you need to make a common denominator for them and add or subtract numerators.

Examples on Separation of Division of Algebraic Fractions

Example 1.

Find the difference of fractions by taking the common denominator: x / bc – y / ab?

Solution:

We observe the two denominators are bc and ab and their L.C.M. is abc. So, abc is the least quantity which is divisible by ab and bc. To subtract those fractions, you must make a common denominator i.e abc. To make denominator as abc for x / bc multiply it with a, and multiply y / ab with c.

Therefore, we can write

(x * a) / abc – (y * c) / abc

= (ax + cy) / abc

Example 2.

Find the sum of fractions by taking the common denominator: a / xy + b / xz + c / yz.

Solution:

There are three denominators xy, xz, and yz, and their L.C.M. is xyz. To make the fractions with the common denominator, the numerator and denominator of these are to be multiplied by xyz ÷ xy = z in case of a/xy, xyz ÷ yz = x in case of c/yz, xyz ÷ xz = y in case of b/xz.

Therefore, we can write

a / xy + b / xz + c / yz

= (a.z) / xyz + (b.y) / xyz + (c.x) / xyz

= (az + by + cx) / xyz

Example 3.

Solve fractions p/qr + q/pr – r/pq by taking the common denominator.

Solution:

We can observe that three fractions have denominators as qr, pr, and pq their L.C.M is pqr. To make a common denominator for three fractions, their numerators should be multiplied by pqr ÷ qr = p for p/qr, pqr ÷ pr = q for q/pr, and pqr ÷ pq = r for r/pq.

Therefore, we can write

p/qr + q/pr – r/pq

= (p.p) / pqr + (q.q) / pqr – (r.r) / pqr

= (p² + q² – r²) / pqr.

The post Rule of Separation of Division of Algebraic Fractions | Dividing Algebraic Fractions appeared first on Learn CBSE.

Learn about how to perform sum and difference of two or more algebraic fractions on this page. We are giving a detailed step by step procedure that helps to calculate the addition, subtraction of algebraic fractions easily. Have a look at some example questions and answers for a better understanding of the concept.

How to Add and Subtract Algebraic Fractions?

You may feel that performing addition or subtraction of algebraic fractions is a bit difficult. To help you out in solving those questions, we are providing the step by step procedure in the below sections of this page. Follow these steps while solving the questions.

• If the denominator of fractions is the same, then just add or subtract the numerators and keep the denominator as it is.
• If the denominator of the algebraic fractions is different, then find the lowest common multiple of those denominators.
• Express all fractions in terms of the lowest common denominator.
• Perform the required operation among the numerators to obtain the result.

Solved Example Questions on Sum & Difference of Algebraic Fractions

Example 1.

Find the sum of a / (a – b) + b / (a² – b²)?

Solution:

We can observe that the denominators of the fractions are different. Those are (a-b) and (a² – b²).

The factors of denominators are (a – b), and (a + b) (a – b).

L.C.M of (a-b), (a² – b²) is (a – b) (a + b)

To make the two fractions having common denominator both the numerator and denominator of these are to be multiplied by (a * (a + b)) / ((a + b) (a – b)) in case of a / (a – b), (b * 1) / ((a – b) (a + b)) in case of b / (a² – b²).

Therefore, a / (a – b) + b / (a² – b²)

= (a * (a + b)) / ((a + b) (a – b)) + b / ((a + b) (a – b))

= (a (a + b) + b) / ((a + b) (a – b))

= (a² + ab + b) / (a² – b²).

Example 2:

Find the difference of (x² + 5x + 6) / (7x + 7y) – (y² – 8y + 16) / (x² – xy)?

Solution:

We can observe that the denominators of the fractions are different. Those are (7x + 7y), (x² – xy).

The factors of denominators are 7 (x + y), x (x – y).

Least common multiple of denominators (7x + 7y), (x² – xy) is 7x (x + y) (x – y)

To make the two fractions having common denominator both the numerator and denominator of these are to be multiplied by [(x² + 5x + 6) * x (x – y)] / [7x (x + y) (x – y)] in case of (x² + 5x + 6) / (7x + 7y), [(y² – 8y + 16) * 7 (x + y)] / [7x (x + y) (x – y)] in case of (y² – 8y + 16) / (x² – xy).

Therefore, (x² + 5x + 6) / (7x + 7y) – (y² – 8y + 16) / (x² – xy)

= [(x² + 5x + 6) * x (x – y)] / [7x (x + y) (x – y)] – [(y² – 8y + 16) * 7 (x + y)] / [7x (x + y) (x – y)]

= [x (x – y) (x² + 3x + 2x + 6)] / [7x (x + y) (x – y)] – [7 (x + y) (y² – 4y – 4y + 16)] / [7x (x + y) (x – y)]

= [x (x – y) (x (x + 3) + 2 (x + 3))] / [7x (x + y) (x – y)] – [7 (x + y) (y (y – 4) – 4 (y – 4))] / [7x (x + y) (x – y)]

= [x (x – y) (x + 3) (x + 2)] / [7x (x + y) (x – y)] – [7 (x + y) (y – 4) (y – 4)] / [7x (x + y) (x – y)]

= [x (x – y) (x + 3) (x + 2) – 7 (x + y) (y – 4) (y – 4)] / [7x (x + y) (x – y)]

= [x (x – y) (x + 3) (x + 2) – 7 (x + y) (y – 4)²] / [7x (x + y) (x – y)].

Example 3.

Simplify the algebraic fractions 1 / (m – n) – 1 / (m + n) + 2n / (m² – n²)?

Solution:

We can say that all the denominators are different, those are (m – n), (m + n), (m² – n²)

The factors of denominators are (m – n), (m + n), (m + n) (m – n)

L.C.M of denominators is (m + n) (m – n)

To make the fractions having common denominator both the numerator and denominator of these are to be multiplied by (m + n) / [(m + n) (m – n)] in case of 1 / (m – n), (m – n) / [(m + n) (m – n)] in case of 1 / (m + n), 2n / [(m + n) (m – n)] in case of 2n / (m² – n²).

Therefore, 1 / (m – n) – 1 / (m + n) + 2n / (m² – n²)

= (m + n) / [(m + n) (m – n)] – (m – n) / [(m + n) (m – n)] + 2n / [(m + n) (m – n)]

= [m + n – (m – n) + 2n] / [(m + n) (m – n)]

= [m + n – m + n + 2n] / [(m + n) (m – n)]

= 4n / [(m + n) (m – n)].

The post Sum and Difference of Algebraic Fractions | Addition and Subtraction of Algebraic Fractions appeared first on Learn CBSE.

In this article, you will learn about simplifying algebraic fractions, reducing the fraction to its lowest term, performing arithmetical operations on algebraic fractions. Get the solved example questions on algebraic fractions to understand the concept better. Each and Every Problem is explained with Step by Step Solutions so that you can learn the Procedure on how to solve related problems easily.

Solved Examples on Algebraic Fractions

Example 1.

Reduce the algebraic fractions to their lowest terms?

(i) (3x² – 6y²) / (6x – 12y)

(ii) (5x² – 5y²) / (25x² + 50xy + 25y²)

(iii) (3ab – 3a²) / (3a² – 6ab + 3b²)

Solution:

(i) (6x² – 6y²) / (12x – 12y)

Factorizing the numerator and denominator separately and cancel the common factors we get,

= (6 (x² – y²)) / ((12 (x – y))

= (x² – y²) / (2 (x – y))

= ((x – y) (x + y)) / (2 (x – y))

= (x + y) / 2

(ii) (5x² – 5y²) / (25x² + 50xy + 25y²)

Factorizing the numerator and denominator separately and cancel the common factors we get,

= (5 (x² – y²)) / (25 (x² + 2xy + y²))

= (x² – y²) / (5 (x² + 2xy + y²))

= [(x + y) (x – y)] / [5 (x² + xy + xy + y²)]

= [(x + y) (x – y)] / [5 (x( x + y) + y (x + y)]

= [(x + y) (x – y)] / [5 ((x + y) (x + y)]

= (x – y) / [5 (x + y)]

(iii) (3ab – 3a²) / (3a² – 6ab + 3b²)

Factorizing the numerator and denominator separately and cancel the common factors we get,

= [3a (b – a)] / [(3a² – 3ab – 3ab + 3b²)]

= [3a (b – a)] / [(3a(a – b) – 3b(a – b))]

= [3a (b – a)] / [(3a – 3b) (a – b)]

= [-3a (a – b)] / [(3a – 3b) (a – b)]

= -3a / 3(a – b)

= -a / (a – b)

Example 2.

Simplify the algebraic fractions?

(i) [1/x + 1/y] / [1/x² – 1/y²]

(ii) [(u + v) / 2u – 2v) + (v – u) / (2v + 2u) + 2v² / (u² – v²)] [1/v – 1/u]

(iii) [(a³ – ab² + b³) / (a – b)³ – b / (a – b)] [(a² – 2ab + 2b²) / (a² – ab + b²) – b/a]

Solution:

(i) [1/x + 1/y] / [1/x² – 1/y²]

Factorize the numerator

1/x + 1/y = (y + x) / (xy)

Factorize the denominator

1/x² – 1/y² = (y² – x²) / (x²y²)

= (y + x) (y – x) / x²y²

Simplification of the given expression after factorizing the numerator and the denominator:

[(y + x) / (xy)] / [(y + x) (y – x) / x²y²]

= [(y + x) * x²y²] / [(y + x) (y – x) * xy]

= xy / (y – x)

(ii) [(u + v) / 2u – 2v) + (v – u) / (2v + 2u) + 2v² / (u² – v²)] * [1/v – 1/u]

Factorize the denominators

2(u – v), 2 (u + v), (u +v) (u – v)

L.C.M of first expression is 2 (u + v) (u – v), second expression is uv

= [((u + v) * (u + v) / 2(u – v) (u + v)) + ((v – u) * (v – u) / 2 (u + v) (u – v)) + 2v² * 2 / 2 (u + v) (u – v)] * [u – v / vu]

= [(u + v)² / 2(u – v) (u + v) + (v – u)² / 2(u – v) (u + v) + 4v² / 2(u – v) (u + v)] * [u – v / vu]

= [((u + v)² + (v – u)² + 4v²) / 2(u – v) (u + v)] * [u – v / vu]

= [(u² + v² + 2uv + v² + u² – 2uv + 4v²) / 2(u – v) (u + v)] * [u – v / vu]

= [(2u² + 6v²) / 2(u – v) (u + v)] * [u – v / vu]

= [2u³ -2u²v + 6v²u + 6v³] / [2(u – v) (u + v)vu]

= 2u² (u – v) + 6v² (u + v) / [2(u – v) (u + v)vu]

= 2[u² (u – v) + 3v² (u + v)] / [2(u – v) (u + v)vu]

= [u² (u – v) + 3v² (u + v)] / [(u – v) (u + v)vu]

(iii) [(a³ – ab² + b³) / (a – b)³ – b / (a – b)] [(a² – 2ab + 2b²) / (a² – ab + b²) – b/a]

Factorize the denominators

(a – b)³, (a – b) and a² – ab + b², a

L.C.M of (a – b)³, (a – b) is (a – b)³, L.C.M of a² – ab + b², a is a (a² – ab + b²)

Express all fractions in terms of the lowest common denominator.

= [(a³ – ab² + b³) / (a – b)³ – (b (a – b)²) / (a – b)³] [(a (a² – 2ab + 2b²)) / a (a² – ab + b²) – b (a² – ab + b²) / a (a² – ab + b²)]

= [(a³ – ab² + b³) / (a – b)³ – (b (a² – 2ab + b²) / (a – b)³] * [(a³ – 2a²b + 2ab²)) / a (a² – ab + b²) – (a²b – ab² + b³) / a (a² – ab + b²)]

= [(a³ – ab² + b³) / (a – b)³ – (a²b – 2ab² + b³) / (a – b)³] * [(a³ – 2a²b + 2ab²) / a (a² – ab + b²) – (a²b – ab² + b³) / a (a² – ab + b²)]

= [(a³ – ab² + b³ – a²b + 2ab² – b³) / (a – b)³] * [(a³ – 2a²b + 2ab² – a²b + ab² – b³) / a (a² – ab + b²)]

= [(a³ + ab² – a²b) / (a – b)³] * [(a³ – 3a²b + 3ab² – b³) / a (a² – ab + b²)]

= [(a³ + ab² – a²b) / (a – b)³] * [(a – b)³ / a (a² – ab + b²)]

= [(a³ + ab² – a²b) (a – b)³] / [(a – b)³ a (a² – ab + b²)]

= [a (a² – ab + b²)] / [a (a² – ab + b²)]

= 1

Example 3.

Simplify the sum and difference of algebraic fractions?

(i) (2x – 3y) / x + (4x² – 5y²) / xy

(ii) x / ac – x / bc + x / ab

Solution:

(i) (2x – 3y) / x + (4x² – 5y²) / xy

L.C.M of denominators is xy.

Express all fractions in terms of the lowest common denominator.

= y(2x – 3y) / xy + (4x² – 5y²) / xy

= (2xy – 3y² + 4x² – 5y²) / xy

= (4x² + 2xy – 8y²) / xy

= 2(2x² + xy – 4y²) / xy

(ii) x / ac – x / bc + x / ab

L.C.M of all denominators is abc

Express all fractions in terms of the lowest common denominator.

= bx / abc – ax / abc + cx / abc

= (bx – ax + cx) / abc

= x(b – a + c) / abc

Example 4.

Simplify the product and quotient of algebraic fractions

(i) (3x² – 3y²) / (12x – 12y)

(ii) (x – y) : (1/x + 1/y)

Solution:

(i) (3x² – 3y²) / (12x – 12y)

Factorize numerators and denominators

= 3(x² – y²) / 12 (x – y)

= (x + y) (x – y) / 4 (x – y)

= (x + y) / 4

(ii) (x – y) : (1/x + 1/y)

= (x – y) : (y + x) / xy)

= xy (x – y) / (x + y)

The post Problems on Algebraic Fractions | Simplifying Algebraic Fractions appeared first on Learn CBSE.

One of the main operations performed for Algebraic Expression is Division. Deeply understand how division operations are performed on Algebraic Expressions and learn the easy way to solve Algebraic Expression Division Problems. Solving Division of Algebraic Expression Problems is the opposite process of Multiplication of Algebraic Expression.

Rule to Find Division of Algebraic Expression

If x is a variable and a, b are positive integers such that a > b then (x^a ÷ x^b) = x^(a − b).

Types of Algebraic Expression Division

There are different types of Division when it comes to Dividing Algebraic Expressions. They are as such

1. Division of a Monomial by a Monomial
2. Division of a Polynomial by a Monomial
3. Division of a Polynomial by a Polynomial

How to Find Division of a Monomial by a Monomial?

1. The coefficients of the quotient of two monomials are equal to the quotient of their numerical coefficients which are multiplied by the quotient of their coefficients.
2. The variable part of the quotient of two monomials is equal to the quotient of the variables in the given monomials.

Rule:

Quotient of two monomials = (quotient of their numerical coefficients) x (the quotient of their variables)

Solved Examples

1. Divide 6x²y³ by -4xy

Solution:

Given that Divide 6x²y³ by -4xy
Use quotient law to solve the given problem.
x^m ÷ xⁿ = x^{m – n}
x²/x = x^(2-1) = x
y³/y = y^(3-1) = y²
-(6/4) = -3/2
-3/2 x y²

The required expression is -3/2xy²

2. Divide 48x³yz² by -8xyz

Solution:

Given that Divide 48x³yz² by -8xyz
Use quotient law to solve the given problem.
x^m ÷ xⁿ = x^{m – n}
x³/x = x^(3-1) = x²
y/y = 1
z²/z = z^(2-1) = z
(48/-8) = -6
-6x²z

The required expression is -6x²z

3. Divide -24x³yz³ by -6xyz²

Solution:

Given that Divide -24x³yz³ by -6xyz²
Use quotient law to solve the given problem.
x^m ÷ xⁿ = x^{m – n}
x³/x = x^(3-1) = x²
y/y = y^(1-1) = y^0 = 1
z³/z² = z^(3-2) = z
(-24/-6) = 4
4x²z

The required expression is 4x²z

How to Find Division of a Polynomial by a Monomial?

Division of a Polynomial by a Monomial can be calculated by dividing each term of the polynomial by the monomial.

1. Note down the polynomial and the monomial.
2. Consider polynomial as Dividend and monomial as the divisor.
3. Separate the terms of the polynomial (dividend) in the descending order of their exponents.
4. Finally, divide every term of the polynomial (dividend) by monomial.
5. Simplify and write the required expression.

Solved Examples

1. Divide 8x⁵ + 20x⁴ – 16x² by 4x²

Solution:

Given that Divide 8x⁵ + 20x⁴ – 16x² by 4x²
Here the dividend is 8x⁵ + 20x⁴ – 16x² and the divisor is 4x²
(8x⁵ + 20x⁴ – 16x²) ÷ 4x²
Divide every term of the polynomial (dividend) by monomial.
(8x⁵ ÷ 4x² ) +( 20x⁴ ÷ 4x² ) – ( 16x² ÷ 4x²
(8/4)(x^5/x^2) + (20/4)(x^4/x^2) – (16/4)(x^2/x^2)
2(x^5-2) + 5(x^4-2) – 4(x^2-2)
2x^3 + 5x^2 – 4

The required expression is 2x³ + 5x² – 4

2. Divide 21x³y + 12x²y² – 9xy by 3xy

Solution:

Given that Divide 21x³y + 12x²y² – 9xy by 3xy
Here the dividend is 21x³y + 12x²y² – 9xy and the divisor is 3xy
21x³y + 12x²y² – 9xy ÷ 3xy
Divide every term of the polynomial (dividend) by monomial.
(21x³y ÷ 3xy) + (12x²y² ÷ 3xy) – (9xy ÷ 3xy)
(21/3)(x³/x)(y/y) + (12/3)(x²/x)(y²/y) – (9/3)(xy/xy)
7(x^3-1)(y^1-1) + 4 (x^2-1)(y^2-1) – 3
7x^2 + 4xy – 3

The required expression is 7x² + 4xy – 3

3. 10x + 20x² + 30x⁴ – 50x⁵ ÷ by 5x

Solution:

Given that Divide 10x + 20x² + 30x⁴ – 50x⁵ by 5x
Here the dividend is 10x + 20x² + 30x⁴ – 50x⁵  and the divisor is 5x
Arrange the dividend in the descending order of their exponents.
( – 50x⁵ + 30x⁴ + 20x² + 10x ) ÷ 5x
Divide every term of the polynomial (dividend) by monomial.
(-50/5)(x⁵/x) + (30/5)(x⁴/x) + (20/5)(x²/x) + (10/5)(x/x)
(-50/5)(x^5-1) + (30/5)(x^4-1) + (20/5)(x^2-1) + (10/5)(x^1-1)
-10x⁴ + 6x³ + 4x + 2

The required expression is -10x⁴ + 6x³ + 4x + 2

How to Find Division of a Polynomial by a Polynomial?

Get to know the Procedure on How to Divide Polynomial by a Polynomial in the below modules. They are along the lines

1. Note down the polynomial and the monomial.
2. Consider polynomial as Dividend and monomial as the divisor.
3. Separate the terms of the polynomial (dividend) in the descending order of their exponents.
4. Divide the first term of the polynomial (dividend) by the first term of the polynomial (divisor) and get the first term of the quotient.
5. Multiply every term of the divisor by the first term of the quotient and then subtract the result from the dividend.
6. In the next step, take the remainder if you have any and consider it as a new dividend then go with the above process.
7. Repeat the same process till you get the remainder as 0 or a polynomial of degree less than that of the divisor.

Solved Examples

1. Divide 18 – 12a² – 8a by (3 + 2a)

Solution:

Given that Divide 18 – 12a² – 8a by (3 + 2a)
Here the dividend is – 12a² – 8a + 18 and the divisor is 2a + 3
Multiply 6a with (2a + 3) then subtract the resultant expression from – 12a² – 8a + 18
-6a (2a + 3) = – 12a² – 18a
(– 12a² – 8a + 18) + (- 12a² – 18a) = 10a + 18
Now, multiply 5 with (2a + 3) then subtract the resultant expression from 10a + 18
5(2a + 3) = 10a + 15
10a + 18 – 10a – 15 = 3
Quotient = -6a + 5
Remainder = 3

Verification:
Dividend = divisor × quotient + remainder
– 12a² – 8a + 18 = (3 + 2a) × (-6a + 5) + 3
= -18a + 15 – 12a² + 10a + 3
= – 12a² – 8a + 18
– 12a² – 8a + 18 = – 12a² – 8a + 18

The required answer is -6a + 5

2. Divide 8x² + 12x + 4 by (4x + 4).

Solution:

Given that Divide 8x² + 12x + 4 by (4x + 4)
Here the dividend is 8x² + 12x + 4 and the divisor is 4x + 4.
Multiply 2x with (4x + 4) then subtract the resultant expression from 8x² + 12x + 4
2x(4x + 4) = 8x² + 8x
(8x² + 12x + 4) – (8x² + 8x) = 4x + 4
Multiply 1 with (4x + 4) then subtract the resultant expression from 4x + 4
1(4x + 4) = 4x + 4
(4x + 4) – (4x + 4) = 0
Quotient = 2x + 1
Remainder = 0

Verification:
Dividend = divisor × quotient + remainder
8x² + 12x + 4 = (4x + 4) × (2x + 1) + 0
= 8x² + 4x + 8x + 4 + 0
= 8x² + 12x + 4
8x² + 12x + 4 = 8x² + 12x + 4

The required answer is 2x + 1

3. Divide 3x² + 18x + 24 by (3x + 12)

Solution:

Given that Divide 3x² + 18x + 24 by (3x + 12)
Here the dividend is 3x² + 18x + 24 and the divisor is (3x + 12)
Multiply x with (3x + 12) then subtract the resultant expression from 3x² + 18x + 24
x(3x + 12) = 3x² + 12x
(3x² + 18x + 24) – (3x² + 12x) = 6x + 24
Multiply 2 with (3x + 12) then subtract the resultant expression from 6x + 24
2(3x + 12) = 6x + 24
(6x + 24) – (6x + 24) = 0
Quotient = x + 2
Remainder = 0

Verification:
Dividend = divisor × quotient + remainder
3x² + 18x + 24 = (3x + 12) × (x + 2) + 0
= 3x² + 6x + 12x + 24 + 0
= 3x² + 18x + 24
3x² + 18x + 24 = 3x² + 18x + 24

The required answer is x + 2

4. Divide 18x – 12x² + 2x³ – 4 by (2x – 4).

Solution:

Given that Divide 18x – 12x² + 2x³ – 4 by (2x – 4)
Here the dividend is 2x³ – 12x² + 18x  – 4 and the divisor is (2x – 4)
Multiply x² with (2x – 4) then subtract the resultant expression from 2x³ – 12x² + 18x  – 4
x²  (2x – 4) = 2x³ – 4x²
(2x³ – 12x² + 18x  – 4) – (2x³ – 4x²) = – 8x² + 18x  – 4
Multiply -4x with (2x – 4) then subtract the resultant expression from – 8x² + 18x  – 4
-4x (2x – 4) = -8x² + 16x
(- 8x² + 18x  – 4) – (-8x² + 16x) = 2x – 4
Multiply 1 with (2x – 4) then subtract the resultant expression from 2x – 4
1 (2x – 4) = 2x – 4 – 2x + 4 = 0
Quotient = x² – 4x + 1
Remainder = 0

Verification:
Dividend = divisor × quotient + remainder
18x – 12x² + 2x³ – 4 = (2x – 4) × (x² – 4x + 1) + 0
= 2x³ – 8x² + 2x – 4x² + 16x – 4 + 0
= 2x³ – 12x² + 18x – 4
2x³ – 12x² + 18x – 4 = 2x³ – 12x² + 18x – 4

The required answer is x² – 4x + 1

5. Divide (87x – 18x² – 84) by (9x – 12).

Solution:

Given that Divide (87x – 18x² – 84) by (9x – 12)
Here the dividend is – 18x² + 87x  – 84 and the divisor is (9x – 12)
Multiply -2x with (9x – 12) then subtract the resultant expression from – 18x² + 87x  – 84
-2x (9x – 12) = – 18x² + 24x
(- 18x² + 87x  – 84) – (- 18x² + 24x) = 63x  – 84
Multiply 7 with (9x – 12) then subtract the resultant expression from 63x  – 84
7 (9x – 12) = 63x – 84
(63x  – 84) – (63x  – 84) = 0
Quotient = -2x + 7
Remainder = 0

Verification:
Dividend = divisor × quotient + remainder
(87x – 18x² – 84) = (9x – 12) × (-2x + 7) + 0
= – 18x² + 63x + 24x – 84
= 87x – 18x² – 84
87x – 18x² – 84 = 87x – 18x² – 84

The required answer is -2x + 7

6. Divide (20x³- 16x² + 12x + 72) by (12 – 8x + 4x²)

Solution:

Given that Divide (20x³- 16x² + 12x + 72) by (12 – 8x + 4x²)
Here the dividend is (20x³- 16x² + 12x + 72) and the divisor is (12 – 8x + 4x²)
Multiply 5x with (4x² -8x + 12) then subtract the resultant expression from 20x³- 16x² + 12x + 72
5x (4x² -8x + 12) = 20x³- 40x² + 60x
(20x³- 16x² + 12x + 72) – (20x³- 40x² + 60x) = 24x² – 48x + 72
Multiply 6 with (4x² -8x + 12) then subtract the resultant expression from 24x² – 48x + 72
6 (4x² -8x + 12) = 24x² – 48x + 72
(24x² – 48x – 72) – (24x² – 48x + 72) = 0
Quotient = 5x + 6
Remainder = 0

Verification:
Dividend = divisor × quotient + remainder
(20x³- 16x² + 12x + 72) = (4x² -8x + 12) × (5x + 6) + 0
= 20x³ + 24x² – 40x² – 48x + 60x + 72
= (20x³- 16x² + 12x + 72)
(20x³- 16x² + 12x + 72) = (20x³- 16x² + 12x + 72)

The required answer is 5x + 6

7. Using division, show that (2x – 2) is a factor of (2x³ – 2).

Solution:

Given that Using division, show that (2x – 2) is a factor of (2x³ – 2)
Divide (2x³ – 2) by (2x – 2)
Here the dividend is (2x³ – 2) and the divisor is (2x – 2)
Multiply x with (2x – 2) then subtract the resultant expression from (2x³ – 2)
x (2x – 2) = 2x³- 2x
(2x³ – 2) – (2x³- 2x) = 2x – 2
Multiply 1 with (2x – 2) then subtract the resultant expression from (2x – 2)
(2x – 2) – (2x – 2) = 0
Quotient = x + 1
Remainder = 0

(2x – 2) divides (2x³ – 2). Therfore, (2x – 2) is the factor of (2x³ – 2)

8. Find the quotient and remainder when (14 + 30x – 26x² + 10x³) is divided by (8 – 6x + 2x²)?

Solution:

Given that Divide (14 + 30x – 26x² + 10x³) by (8 – 6x + 2x²)
Here the dividend is (14 + 30x – 26x² + 10x³) and the divisor is (8 – 6x + 2x²)
Rearrange the given expressions.
divide (10x³ – 26x² + 30x + 14) by (2x² – 6x + 8)
Multiply 5x with (2x² – 6x + 8) then subtract the resultant expression from (10x³ – 26x² + 30x + 14)
5x (2x² – 6x + 8) = 10x³ – 30x² + 40x
(10x³ – 26x² + 30x + 14) – (10x³ – 30x² + 40x) = 4x² – 10x + 14
Multiply 2 with (2x² – 6x + 8) then subtract the resultant expression from 4x² – 10x + 14
2 (2x² – 6x + 8) = 4x² – 12x + 16
(4x² – 10x + 14) – (4x² – 12x + 16) = 2x – 2
Quotient = 5x + 2
Remainder = 2x – 2

The Quotient is 5x + 2 and the Remainder is 2x – 2

9. Divide (30x⁴ + 51x³ – 186x² + 90x – 9) by (6x² + 21x – 3)

Solution:

Given that Divide (30x⁴ + 51x³ – 186x² + 90x – 9) by (6x² + 21x – 3)
Here the dividend is (30x⁴ + 51x³ – 186x² + 90x – 9) and the divisor is (6x² + 21x – 3)
Multiply 5x² with (6x² + 21x – 3) then subtract the resultant expression from (30x⁴ + 51x³ – 186x² + 90x – 9)
5x² (6x² + 21x – 3) = 30x⁴ + 105x³ – 15x²
(30x⁴ + 51x³ – 186x² + 90x – 9) – (30x⁴ + 105x³ – 15x²) = -54x³ – 171x² + 90x – 9
Multiply 9x with (6x² + 21x – 3) then subtract the resultant expression from -54x³ – 171x² + 90x – 9
-9x (6x² + 21x – 3) = -54x³ – 189x² + 27x
(-54x³ – 171x² + 90x – 9) – (-54x³ – 189x² + 27x) = 18x² + 63x – 9
Multiply 3 with (6x² + 21x – 3) then subtract the resultant expression from 18x² + 63x – 9
3 (6x² + 21x – 3) = 18x² + 63x – 9
(18x² + 63x – 9) – (18x² + 63x – 9) = 0
Quotient = 5x² – 9x + 3
Remainder = 0

Verification:
Dividend = divisor × quotient + remainder
(30x⁴ + 51x³ – 186x² + 90x – 9) = (6x² + 21x – 3) × (5x² – 9x + 3) + 0
= 30x⁴ -54x³ + 18x² -105x³ – 189x² +63x – 15x² +27x – 9
= (30x⁴ + 51x³ – 186x² + 90x – 9)
(30x⁴ + 51x³ – 186x² + 90x – 9) = (30x⁴ + 51x³ – 186x² + 90x – 9)

The required answer is 5x² – 9x + 3

The post Division of Algebraic Expression | How to Divide Algebraic Expressions? appeared first on Learn CBSE.

Formula and Framing the Formula are the concepts used to explain the relationship between different variables. Let us discuss deeply every individual concept of Formula and also Framing the Formula. It is easy to find the relation between two variables by framing a formula with simple steps. Every concept is clearly explained with the solved examples in the below article. Therefore, completely read the entire article and follow every step to get complete knowledge on Formula and Framing the Formula.

List of Topics for Formula and Framing the Formula

• Change the Subject of a Formula
• Changing the Subject in an Equation or Formula
• Practice Test on Framing the Formula

Formula

A formula is a relation between different variables. The formula can be written as an equation with the help of variables and also mathematical symbols. When you look at the equation, it is clearly stating how a variable is related to another variable.

Example:

1. Let us consider a square which is of side a and the perimeter of it is p, then the formula will be p = 4a.
Here the formula shows the exact relation between the perimeter of a square and also the side of a square. It is easy to find the unknown quantity using the known quantity when the values of all the quantities are known.
2. If the perimeter of a rectangle p is twice the sum of its breadth b and length l, then the formula will represent as p = 2(l + b)
3. The volume of a cube is V and its side is a. Then, the formula is V = a^3.
4. We can also write a formula to express the relation between force, mass, and acceleration. Force of an object F is the product of mass “m” and acceleration “a” of that object. The formula is F = ma.
5. If the sum of two unknown variables is 15, then the formula is a + b = 15, where a and b are unknown variables.

Framing a Formula

Framing a formula is arranging the formula of the given mathematical statements with the help of symbols and literals.

1. Firstly, select variables need to form an equation. Also, decide the symbols that need to use for the equation. Some of the symbols and letters are already in the use to represent certain quantities. For example, p is used to represent the principal.
2. Finally, understand the conditions to write an equation and frame the formula.

Subject of a formula

When a quantity is expressing in terms of other quantities, then that particular quantity expressed is defined as the Subject of a formula. Generally, the Subject of the formula is written on the left-hand side and other constants are written on the right-hand side of the equality sign in a formula.

Example:
If z = x + y, then z is expressed in terms of the sum of the x and y. Here, z is the subject of this formula.
x = z – y. Here x is the subject of this formula.

Substitution in a formula

If the variable of an algebraic expression is assigned with certain values, then the given expression gets a particular value. This process is known as substitution.
1. Note down the unknown quantity as the subject of the formula.
2. Substitute different values of the known quantity in the formula to find the value of the subject.

Examples of Framing of a Formula:

1. The total amount A is equal to the sum of the Interest (I) and Principal (P).

Solution:
Formula: A =  I + P

2. One-third of a number subtracted from 4 gives 2.

Solution:
Formula: 4 – 1/3 x = 2

3. The sum of the three angles (∠a, ∠b, ∠c) of a triangle is equal to two right angles.

Solution:
Formula: ∠a + ∠b + ∠c = 180°

4. The area of the rectangle (A) is equal to the product of the breadth (B) and length (L) of the rectangle.

Solution:
Formula: A = B × L

Solved Examples on Formula and Framing the Formula

1. Express the following as an equation. Arun’s father’s age is 3 years more than 4 times Arun’s age. Father’s age is 39 years.

Solution:
Given that Arun’s father’s age is 3 years more than 4 times Arun’s age. Father’s age is 39 years.
Let Arun’s age is s years.
Four times his age = 4s.
Father’s age 3 + 4s
Given father’s age = 39 years
3 + 4s = 39

The final equation is 3 + 4s = 39

2. Write the formula for the following statement. One-fourth of the weight of an apple (A) is equal to one-fifth of the difference between Orange (O) and 2.

Solution:
Given that One-fourth of the weight of an apple (A) = 1/4
Difference between Orange (O) and 2 = O – 2
One-fifth of the difference between Orange (O) and 2. = 1/5(O – 2)
One-fourth of the weight of an apple (A) is equal to one-fifth of the difference between Orange (O) and 2.
1/4 = 1/5(O – 2)

3. Change the following statement using expression into a statement in ordinary language.
(a) Cost of a Desk is Rs. X and Cost of a Box is Rs. 4X
(b) Sam’s age is p years. His brother’s age is (5p + 2) years.

Solution:
(a) Given that Cost of a Desk is Rs. X and Cost of a Box is Rs. 4X
The Cost of a Box is 4 times the Cost of a Desk.
(b) Sam’s age is p years. His brother’s age is (5p + 2) years.
Sam’s brother’s age is two years more than five times his age.

4. A rectangular box is of height h cm. Its length is 7 times its height and the breadth is 5 cm less than the length. Express the length, breadth, and height.

Solution:
Given that A rectangular box is of height h cm. Its length is 7 times its height and the breadth is 5 cm less than the length. Express the length, breadth, and height.
Height = h, length = l, breadth = b
The length of the rectangle is 7 times the height.
The Length of the rectangle = 7h
The breadth of the rectangle is 5 cm less than the length
The Breadth of rectangle = l – 5
As l = 7h, b = 7h – 5.

Height h
Length l = 7h
Breadth b = 7h – 5

The post Formula and Framing the Formula | Framing Formulas Solved Examples appeared first on Learn CBSE.

In this article, you will learn about How to Calculate Compound Interest with Periodic Deductions or Additions to the Amount. Practice the Problems in Periodic Compound Interest and learn how to solve the related problems. To help you understand the concept better we even listed the Formula along with Step by Step Solutions. Check out the Solved Examples for finding the Compound Interest with Periodic Deductions and learn the concept behind them.

Example Questions on Compound Interest with Periodic Deductions

1. Jasmine borrows $ 10,000 at a compound interest rate of 4% per annum. If she repays $ 2,000 at the end of each year, find the sum outstanding at the end of the second year?

Solution:

From the Given Data

Principal = $ 10, 000

Interest Rate = 4% Per Annum

Time = 1 Year

Interest = PTR/100

= 10,000*1*4/100

= $400

Amount of the loan after 1 year = Principal + Interest

= $10,000 + $400

= $10, 400

Jasmine Repays $2,000 after the first year

Therefore, New Principal for the 2nd year = $10, 400 – $2,000

= $8,400

Thus for 2nd Year Principal = $8, 400

Interest = 4%

Time = 1 Year

Interest = PTR/100

= 8,400*1*4/100

= $336

Amount after 2nd Year = Principal + Interest

= $8400+$336

= $8736

Therefore, Jasmine needs to pay an outstanding amount of $8736 dollars by the end of the 2nd Year.

2. David invests $ 10,000 at the beginning of every year in a bank and earns 5 % annual interest, compounded at the end of the year. What will be his balance in the bank at the end of two years?

Solution:

From the Given Data

Principal = $10, 000

Rate of Interest = 5%

Time = 1 Year

Interest = PTR/100

= 10,000*1*5/100

= $500

Therefore Amount after 1 Year = Principal + Interest

= $10,000+$500

= $10, 500

David deposits 10,000 at the beginning of the second year

New Principal becomes = $10, 500+ $10, 000

= $20, 500

Thus, for the 2nd Year

Principal = $20, 500

Interest Rate = 5%

Time = 1 Year

Interest = PTR/100

= $20, 500*1*5/100

= $1025

Amount after the 2nd Year = Principal + Interest

= $20, 500 + $1025

= $21, 525

Therefore, David will have $21, 525 in the bank after the end of 2 Years.

3. John lent $ 5,000 at a compound interest rate of 10% per annum. If he repays $ 500 at the end of the first year and $ 1,000 at the end of the second year, find his outstanding loan at the beginning of the third year?

Solution:

From the Given Data

Principal = $ 5,000

Interest rate = 10%

Time = 1 Year

Interest = PTR/100

= 5000*1*10/100

= $500

Amount after 1 year = Principal + Interest

= $5000+$500

= $5500

John repays $500 at the end of first year thus new principal = $5500 – $500

= $5000

Thus, For 2nd Year

New Principal = $5000

Time = 1 Year

Interest Rate = 10%

Interest = PTR/100

= $5000*1*10/100

= $500

Amount after 2nd Year = Principal + Interest

= $5,000+$500

= $5500

John repays $1000 after the end of 2nd year

Thus for third year New Principal = $5500 – $1000

= $4500

Therefore the outstanding loan at the beginning of the third year is $4500.

The post Compound Interest with Periodic Deductions | How to Calculate Periodic Compound Interest? appeared first on Learn CBSE.

Do you feel any difficulty in calculating the Compound Interest? Not anymore after going through this article. Finding Compound Interest using the Formula is quite simple and you don’t have to do hectic calculations, unlike the manual methods. You just need to substitute the inputs and perform basic maths to obtain the Calculate Compound Interest instantly.

For the sake of your convenience, we have listed the Formulas for finding Compound Interest Annually, Half-Yearly, Quarterly along with Solved Examples. Refer to the Step by Step Solutions provided and understand the concept easily.

Compound Interest Formula in Different Cases

In general, Compound Interest is the Interest calculated on the Principal and the Interest accumulated over the previous period. We have listed the Compound Interest Formulas in various cases like When Interest Rate is Compounded Annually, Half-Yearly, Quarterly in the coming modules by taking enough examples.

• Compound Interest Formula when Rate is Compounded Annually
• Compound Interest when Rate is Compounded Half-Yearly
• Compound Interest Quarterly Formula
• When the Interest is Compounded Annually but rates are different for different years
• Interest is compounded annually but time is a fraction

Annually Compound Interest Formula

We know the Formula to Calculate the Amount is A = P(1+r/n)^nt

Where A= Amount

P= Principal

R= Rate of Interest

n= Number of times interest is compounded per year

If the Interest Rate is Compounded Annually we have the Formula as A = P(1+R/100)^t

CI = A – P

Examples

1. Find the amount of $4000 for 2 years, compounded annually at 5% per annum. Also, find the compound interest?

Solution:

We know the formula to calculate Amount is A = P(1+R/100)ⁿ

P = $4000, R = 5%, n = 2 years

Substitute the input data we get the equation as such

A = 4000(1+5/100)²

= 4000(105/100)²

= $4410

CI = A – P

= $4410- $4000

= $410

Therefore, Compound Interest is $410.

2. Calculate the compound interest (CI) on Rs. 3000 for 1 year at 10% per annum compounded annually?

Solution:

The Formula to Calculate the Amount is A = P(1+R/100)ⁿ

P = Rs. 3000, n = 1 year R = 10%

Substituting the input data in the formula and we get

A = 3000(1+10/100)¹

= 3000(1.1)

= Rs. 3300

CI = A – P

= 3300 – 3000

= Rs. 300

Half-Yearly Compound Interest Formula

Let us calculate the Compound Interest on a Principal P kept for 1 Year and at an Interest Rate R% compounded half-yearly

As the Interest is compounded half-yearly Interest Amount will vary after the first 6 months. The Interest for the next 6 months will be calculated on the remaining amount after the first 6 months.

Principal = P, Rate = R/2 %, Time = 2n

A = P(1+R/100)ⁿ

Substitute R/2 and 2n in terms of Rate and Time in the above formula

A = P(1+R/2*100)²ⁿ

CI = A – P

Example

Calculate the compound interest to be paid on a loan of Rs. 20,000 for 3/2 years at 10% per annum compounded half-yearly?

Solution:

From the given data

Principal = 20,000

R = 10%

n = 3/2

A= P(1+R/2*100)²ⁿ

Substitute the input values in the formula and we have

A = 20,000(1+10/200)^2*3/2

= 20, 000(1+10/200)³

= 20,000(1.157)

= Rs. 23152

CI = A – P

= 23,152 – 20,000

= Rs. 3152

Compound Interest Quarterly Formula

Let us find the Compound Interest Kept on a Principal for 1 year and a Rate of R% compounded quarterly. Since CI is compounded quarterly principal amount will change after the first 3 months. The next 3 months(second quarter) interest will be calculated on the amount left after the first 3 months. Third-quarter Interest will be calculated on the amount left after the first 6 months. Last quarter will be found on the amount left after the first 9 months.

The formula of Compound Interest Compounded Quarterly is given as

A = P(1+R/4/100)⁴ⁿ

CI = A – P

Example

Find the compound interest on $12,000 if Nick took a loan from a bank for 6 months at 8 % per annum, compounded quarterly?

Solution:

From the given data P = $12, 000

R = 8% per Annum, (8/4)% per quarter = 2% per quarter

T = 6 months = 2 Quarters

A = P(1+R/100)ⁿ

= 12,000(1+2/100)²

= 12,000(1+0.02)²

= 12,000(1.02)²

= Rs. 12,484

CI = A – P

= 12,484 – 12,000

= Rs. 484

When the Interest is Compounded Annually but rates are different for different years

Suppose the Interest Compounded Annually be different in different years. In the first year if the Interest Rate is p % per annum and for the second year if it is q % then

Amount Formula is given by = P*(1+p/100)*(1+q/100)

This formula can be extended for any number of years.

To get the Compound Interest Subtract Principal from Amount i.e. CI = A – P

Example

Find the amount of $10, 000 after 2 years, compounded annually, if the rate of interest being 3 % p.a. during the first year and 4 % p.a. during the second year. Also, find the compound interest?

Solution:

Formula to Calculate the Amount is A = P*(1+p/100)*(1+q/100)

From the given data P = $10, 000, p = 3%, q = 4%

Substitute the inputs in the formula to calculate the Amount and the equation is as under

A = 10,000(1+3/100)(1+4/100)

= 10, 000(1.03)(1.04)

= $10712

CI = A – P

= $10712 – $10, 000

= $712

Interest is Compounded Annually but time is a fraction

For instance, if time is 5 3/4 years then Amount is given as under

A = P * (1 + R/100)⁵ * [1 + (3/4 × R)/100]

Example

Find the compound interest on $ 30,000 at 6 % per annum for 3 years. Solution Amount after  3 3/4 years?

Solution:

Amount after 3 3/4 years is given by A = $ [30,000 × (1 + 6/100)³ × (1 + (3/4 × 8)/100)]

= $[30,000 * (1 + 0.06)³ * (1 + 6/100)]

=$[30,000*(1.06)³*(1.06)]

= $37874

CI = A – P

= $37874 – $30, 000

= $874

The post Compound Interest by Using Formula | How to Calculate Compound Interest using Formula? appeared first on Learn CBSE.

Let us learn How to Calculate Compound Interest when Interest Rate is Compounded Yearly in the coming modules. Check out Formula, Solved Examples on finding the Compound Interest Annually. We tried explaining each and every step on how to find Compound Interest. Computing the Compound Interest using Growing Principal can be difficult if the time period is long.

How to find Compound Interest when Interest is Compounded Annually?

If the Interest is Compounded Annually then Formula to Calculate the Compound Interest is given by

A = P(1+r/100)ⁿ

Where A is the Amount

P = Principal

r = rate of interest per unit time

n = Time Duration

CI can be obtained by subtracting the Principal from the Amount

CI = A – P

= P(1+r/100)ⁿ – P

= P{1+r/100)ⁿ – 1}

Solved Examples on Compound Interest when Interest is Compounded Annually

1. Find the amount and the compound interest on $8, 000 in 2 years and at 5% compounded yearly?

Solution:

Principal = $8, 000

r = 5%

n = 2

A = P(1+r/100)ⁿ

Substitute the Input Values in the formula of Amount

A = 8,000(1+5/100)²

= 8000(1+0.05)²

= 8000(1.05)²

= $8820

CI = A – P

= $8820 – $8000

= $820.

2. Find the amount of $12,000 for 2 years compounded annually, the rate of the interest being 5 % for the 1st year and 6 % for the second year?

Solution:

A = P*(1+p/100)*(1+q/100)

= 12,000(1+5/100)(1+6/100)

= 12,000(1.05)(1.06)

= $13356

Amount after 2 years is $13356.

3. Calculate the compound interest (CI) on Rs. 10, 000 for 3 years at 8% per annum compounded annually?

Solution:

Principal = Rs. 10,000

n = 3

r = 8%

A = P(1+r/100)ⁿ

= 10,000(1+8/100)³

= 10,000(1.08)³

= Rs. 12,597

CI = A – P

= 12, 597 – 10, 000

= Rs. 2, 597

The post Compound Interest when Interest is Compounded Yearly| How to Calculate Compound Interest Annually? appeared first on Learn CBSE.

Learn How to Calculate Compound Interest when Interest is Compounded Half-Yearly. Computation of Compound Interest by the growing principal can be complicated. Check out Solved Examples explaining the step by step process for finding the compound interest when compounded half-yearly. To help you better understand we have given the Compound Interest Formula when Interest Rate is Compounded Half-Yearly.

How to find Compound Interest when Interest is Compounded Half-Yearly?

If the rate of interest is annual and interest is compounded half-yearly then the annual interest rate is halved(r/2) and the number of years is doubled i.e. 2n. The Formula to Calculate the Compound Interest when Interest Rate is Compounded Half Yearly is given by

Let Principal = P, Rate of Interest = r/2 %, time = 2n, Amount = A, Compound Interest = CI then

A = P(1+r/2/100)²ⁿ

In the Case of the Half-Yearly Compounding, Rate Interest is divided by 2 and the number of years is multiplied by 2.

CI = A – P

= P(1+r/2/100)²ⁿ – P

=P{(1+r/2/100)²ⁿ-1}

If any three of the terms are given the fourth one can be found easily.

Problems on Compound Interest when Interest is Compounded Half-Yearly

1. Find the amount and the compound interest on $12,000 at 8 % per annum for 3 1/2 years if the interest is compounded half-yearly?

Solution:

Given Principal = $12, 000

r = 8% per annum

rate of interest half-yearly = 8/2 %

= 4%

n = 3 1/2

= 7/2 years

when compounded half yearly multiply by 2 i.e. 2n

= 7/2*2

= 7

We know Amount A = P(1+r/100)ⁿ

= 12,000(1+4/100)⁷

=12,000(1+0.04)⁷

= 12,000(1.04)⁷

A = Rs. 15791

We know CI = A – P

= 15791 – 12,000

= Rs. 3791

2. Find the compound interest on Rs 5000 for 3/2 years at 5% per annum, interest is payable half-yearly?

Solution:

A = P(1+r/100)ⁿ

P = 5000

n = 3/2

2n = 3/2*2

= 3

r = 5%

A = 5000(1+5/100)³

A = 5000(1.05)³

A = Rs. 5788

CI = A – P

= 5788 – 5000

= Rs. 788

The post Compound Interest when Interest is Compounded Half-Yearly | Half Yearly Compounding Examples appeared first on Learn CBSE.

In this article, you will learn how to find Compound Interest when Interest is Compounded Quarterly. You may feel the process of calculating the Compound Interest using the growing principal a bit difficult if the time duration is long. Refer to Solved Examples on finding Quarterly Compounded Interest and learn how to solve related problems. We even provided the solutions for the sample problems on calculating the Compound Interest when Interest is Compounded Quarterly in the coming modules.

How to find Compound Interest when Interest is Compounded Quarterly?

If the Rate of Interest is Annual and Interest is Compounded Quarterly then the number of years is multiplied by 4 i.e. 4n and the annual interest rate is cut down by one-fourth. In such cases, Formula for Quarterly Compound Interest is given as under

Let us assume the Principal = P, Rate of Interest = r/4 %, and time = 4n, Amount = A, Compound Interest = CI then

A = P(1+(r/4)/100)⁴ⁿ

In the above formula rate of interest is divided by 4 whereas the time is multiplied by 4.

We know CI = A – P

= P(1+(r/4)/100)⁴ⁿ – P

= P{1+(r/4)/100)⁴ⁿ – 1}

If you are aware of any of the three then you can automatically find the other one.

Solved Problems on finding Compound Interest when Compounded Quarterly

1. Find the compound interest when $1,00, 000 is invested for 6 months at 5 % per annum, compounded quarterly?

Solution:

Principal Amount = $1,00, 000

Rate of Interest = 5% per annum

n = 6 months = 1/2 year

Since Interest Rate is Compounded Quarterly divide the interest rate by 4 i.e. r/4 and multiply the time by 4 i.e. 4n

Amount A = P(1+(r/4)/100)⁴ⁿ

Substitute the Inputs in the above formula to find the amount

A = 1,00,000(1+(5/4)/100)^4*1/2

= 1,00,000(1+5/400)²

= $ 1,02,515

CI = A – P

= $ 1,02,515 – $ 1,00,000

=$2515

2. Find the amount and the compound interest on Rs. 12,000 compounded quarterly for 9 months at the rate of 10% per annum?

Solution:

Principal Amount = Rs.12, 000

Rate of Interest = 10% per annum

n = 9 months = 9/12 = 3/4 year

Since Interest Rate is Compounded Quarterly divide the interest rate by 4 i.e. r/4 and multiply the time by 4 i.e. 4n

Amount A = P(1+(r/4)/100)⁴ⁿ

Substitute the Input Values in the above formula to find the amount

A= 12,000(1+(10/4)/100)^4*3/4

= 12,000(1+10/400)³

= 12,000(1+0.025)³

= 12,000(1.025)³

= Rs. 12922

CI = A – P

= 12922 – 12000

= Rs. 922

3. Calculate the compound interest (CI) on Rs. 4000 for 1 year at 10% per annum compounded quarterly?

Solution:

Principal Amount = Rs. 4,000

Rate of Interest = 10% per annum = 10/4 %

n = 1 year

Since Interest Rate is Compounded Quarterly divide the interest rate by 4 i.e. r/4 and multiply the time by 4 i.e. 4n

Amount A = P(1+(r/4)/100)⁴ⁿ

Substitute the Input Values in the above formula to find the amount

A = 4000(1+(10/4)/100)^4*1

= 4000(1+10/400)⁴

= 4000(1.1038)

= Rs. 4415.25

CI = A – P

= 4415.25 – 4000

= Rs. 415.25

The post Compound Interest when Interest is Compounded Quarterly | Quarterly Compounded Interest Solved Examples appeared first on Learn CBSE.

Find the solved example questions on algebraic fractions. This article includes addition, subtraction, multiplication, division, simplification, and reducing a fraction to its lowest term. By reading this page, you can solve any type of algebraic fraction questions easily & quickly. So, have a look at all the questions and solutions provided below and learn the concepts involved easily.

Questions on Solving Algebraic Fractions

Example 1.

Simplify the algebraic fraction [1 + 1 / (x + 1)] / [x – 4/x]?

Solution:

Given fraction is [1 + 1 / (x + 1)] / [x – 4/x]

Find the L.C.M of denominators.

[(1. (x + 1)+ 1) / (x + 1)] / [(x² – 4) / x]

= [(x + 2) / (x + 1)] / [(x² – 2²) / x]

= [(x + 2) / (x + 1)] * [x / (x + 2) (x – 2)]

= [x(x + 2)] / [(x + 1) ( x – 2) (x + 2)]

= x / [(x + 1) (x – 2)]

∴[1 + 1 / (x + 1)] / [x – 4/x] = x / [(x + 1) (x – 2)]

Example 2.

Simplify the algebraic fraction [((k² + 1 / k – 1) – k) / ((k² – 1 / k + 1) + 1)] [1 – 2/(1 + 1/k)]?

Solution:

Given algebraic fraction is [((k² + 1 / k – 1) – k) / ((k² – 1 / k + 1) + 1)] [1 – 2/(1 + 1/k)]

Find the L.C.M of denominators of the first fraction and simplify.

= [(k² + 1 – k (k – 1)) / (k – 1)] / [(k² – 1 + 1 (k + 1)) / (k + 1)]

= [(k² + 1 – k² + k)) / (k – 1)] / [(k² – 1 + k + 1)) / (k + 1)]

= [(k + 1) / (k – 1)] / [(k² + k) / (k + 1)]

= [(k + 1) / (k – 1)] / [(k(k + 1) / (k + 1)]

= [(k + 1) / (k – 1)] / k / 1

= (k + 1) / k(k – 1)

= (k + 1) / (k² – 1²)

= (k + 1) / (k + 1) ( k – 1)

= 1 / (k – 1)

Simplification of the second fraction is

[1 – 2/(1 + 1/k)]= [1- 2 / k(k +1)]

= [k(k +1) – 2] / [k(k +1)]

= (k² + k – 2) / [k(k +1)]

= (k² + 2k – k – 2) / (k(k +1))

= (k (k + 2) – 1(k + 2)) / (k(k +1))

= [(k – 1) ( k + 2)] / (k(k +1))

Product of first and second fraction is

= 1 / (k – 1) * [(k – 1) ( k + 2)] / (k(k +1))

= (k + 2) / k(k +1)

∴ [((k² + 1 / k – 1) – k) / ((k² – 1 / k + 1) + 1)] [1 – 2/(1 + 1/k)] = (k + 2) / k(k +1)

Example 3.

Reduce the algebraic fractions [3 / √(1+x) + √(1-x)] : [3 / √(1 – x²) + 1]

Solution:

Given algebraic fraction is [3 / √(1+x) + √(1-x)] : [3 / √(1 – x²) + 1]

Simplification of first fraction is

3 / √(1 + x) + √(1 – x)

= (3 + √(1 + x) * √(1 – x)) / √(1 + x)

= (3 + √(1 + x)(1 – x) / √(1 + x)

Simplification of the second fraction is

3 / √(1 – x²) + 1

= 3 + √(1 – x²) / √(1 – x²)

The division of algebraic fraction is

= [(3 + √(1 + x)(1 – x) / √(1 + x)] : [3 + √(1 – x²) / √(1 – x²)]

= [(3 + √(1 + x)(1 – x) (√(1 – x²))] / [√(1 + x) (3 + √(1 – x²))

= √(1 – x²) / √(1 + x)

= √(1 + x)(1 – x) / √(1 + x)

= √(1 – x)

∴ [3 / √(1+x) + √(1-x)] : [3 / √(1 – x²) + 1] = √(1 – x)

Example 4.

Reduce to lowest terms — if possible 3x / 4a²b – 7 / 6ab⁵ – 5x / 2ab².

Solution:

Given fraction is 3x / 4a²b – 7 / 6ab⁵ – 5x / 2ab²

Find the L.C.M of all terms denominators.

L.C.M of 4a²b, 6ab⁵, 2ab² is 12a²b⁶.

= [3x . 3b⁵]/12a²b⁶ – [7 . 2a]/12a²b⁶ – [5x . 6ab⁴]/12a²b⁶

= [9xb⁵ – 14a – 30xab⁴]/ 12a²b⁶

∴ 3x / 4a²b – 7 / 6ab⁵ – 5x / 2ab² = [9xb⁵ – 14a – 30xab⁴]/ 12a²b⁶

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