Category: Mathematics

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

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)

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)].

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.

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.