Here, in this Simplifying Algebraic Fractions Worksheet, you will learn about how to solve the questions on reducing and simplifying algebraic fractions. This helps you to have a better practice for exams. Worksheet on Simplifying Algebraic Fractions includes different types of questions with answers. You can also get some solved example questions on algebraic fractions in the below sections. We have mentioned the step by step solutions for all problems so that it would be easy for you to understand the concept.

1. Simplify by adding and subtracting algebraic fractions:

(a) [(5a – 1) / 8] – [(3a – 2) / 7] + [(a – 5) / 4]

(b) [5a / (a – 5)] – [a² / (a² – 25)]

(c) [2(x – 3) / (x² -5x +6)] + [3(x – 1) / (x² – 4x + 3)] + [5(x – 2) / (x² – 3x + 2)]

## Solution:

(a) [(5a – 1) / 8] – [(3a – 2) / 7] + [(a – 5) / 4]

L.C.M of denominators 8, 7, 4 is 56

= [7(5a – 1) / 56] – [8(3a – 2) / 56] + [14(a – 5) / 56]

= [(35a – 7) / 56] – [(24a – 16) / 56] + [(14a – 70) / 56]

= (35a – 7 – (24a – 16) + 14a – 70) / 56

= (35a – 7 – 24a + 16 + 14a – 70) / 56

= (25a – 61) / 56

(b) [5a / (a – 5)] – [a² / (a² – 25)]

The lowest common multiple of denominators is (a – 5), (a² – 25) = (a – 5), (a + 5)(a – 5) = (a + 5)(a – 5)

To make the fractions having common denominator both numerator and denominator of these are to be multiplied by (a + 5) / (a + 5) in case of [5a / (a – 5)], 1/ 1 in case of [a² / (a² – 25)].

Therefore, [5a / (a – 5)] – [a² / (a² – 25)]

= [5a(a + 5) / (a + 5)(a – 5)] – [a² / (a + 5)(a – 5)]

= [(5a² + 25a – a²] / (a + 5)(a – 5)

= (4a² + 25a) / (a + 5)(a – 5)

(c) [2(x – 3) / (x² -5x +6)] + [3(x – 1) / (x² – 4x + 3)] + [5(x – 2) / (x² – 3x + 2)]

The least common multiple of (x² -5x +6), (x² – 4x + 3), (x² – 3x + 2) is calculated by finding the factors.

Factors of (x² -5x +6), (x² – 4x + 3), (x² – 3x + 2) are (x² – 3x – 2x + 6), (x² – 3x – x + 3), (x² – 2x – x + 2)

= (x(x – 3) – 2(x – 3)), (x(x – 3) – 1(x – 3)), (x(x – 2) -1(x – 2))

= (x – 3)(x – 2), (x – 3)(x – 1), (x – 2)(x – 1)

The L.C.M of (x² -5x +6), (x² – 4x + 3), (x² – 3x + 2) is (x – 3)(x – 2)(x – 1).

To make the fractions having common denominator both numerator and denominator of these are to be multiplied by (x – 1) / (x – 1) in case of [2(x – 3) / (x² -5x +6)], (x – 2) / (x – 2) in case of [3(x – 1) / (x² – 4x + 3)], (x – 3) / (x – 3) in case of [5(x – 2) / (x² – 3x + 2)].

Therefore, [2(x – 3) / (x² -5x +6)] + [3(x – 1) / (x² – 4x + 3)] + [5(x – 2) / (x² – 3x + 2)]

= [2(x – 3)(x – 1) / (x – 3)(x – 2)(x – 1)] + [3(x – 1)(x – 2) / (x – 3)(x – 2)(x – 1)] + [5(x – 2)(x – 3) / (x – 3)(x – 2)(x – 1)]

= [2(x – 3)(x – 1) + 3(x – 1)(x – 2) + 5(x – 2)(x – 3)] / (x – 3)(x – 2)(x – 1)

= [2(x² – x – 3x + 3) + 3(x² -x – 2x +2) +5(x² – 2x – 3x + 6)] / (x – 3)(x – 2)(x – 1)

= [2(x² – 4x + 3) + 3(x² – 3x + 2) + 5(x² – 5x + 6)] / (x – 3)(x – 2)(x – 1)

= [2x² – 8x + 6 + 3x² – 9x + 6 + 5x² – 25x + 30] / (x – 3)(x – 2)(x – 1)

= [10x² – 42x + 42] / (x – 3)(x – 2)(x – 1)

= 2[5x² – 21x + 21] / (x – 3)(x – 2)(x – 1)

2. Simplify by multiplying and dividing algebraic fractions:

(a) [(x + y) / (2x – 3)] x [(x – y) / (2x + 3)]

(b) [(14a² – 7a) / (12a³ + 24a²)] ÷ [(2a – 1) / (a² + 2a)]

(c) [(m² – m – 12) / 5m] x [(m³ – m) / (m² – 9)]

## Solution:

(a) [(x + y) / (2x – 3)] x [(x – y) / (2x + 3)]

Multiply the numerator and denominator separately.

= (x + y)(x – y) / (2x – 3)(2x + 3)

= (x² – y²) / (4x² – 9)

(b) [(14a² – 7a) / (12a³ + 24a²)] ÷ [(2a – 1) / (a² + 2a)]

= [(14a² – 7a) / (12a³ + 24a²)] x [(a² + 2a) / (2a – 1)]

= (14a² – 7a) (a² + 2a) / (12a³ + 24a²) (2a – 1)

= 7a²(2a – 1) (a + 2) / 12a²(a + 2) (2a – 1)

Cancel the common terms a², (2a – 1), and (a + 2) in both numerator and denominator.

= 7 / 12

(c) [(m² – m – 12) / 5m] x [(m³ – m) / (m² – 9)]

= (m² – m – 12) (m³ – m) / 5m(m² – 9)

= m(m² – 4m + 3m – 12) (m² – 1) / 5m (m² – 3²)

Cancel the common term m in both numerator and denominator.

= (m(m – 4) + 3(m – 4) (m² – 1) / 5(m + 3) (m – 3)

= (m + 4) (m – 3) (m² – 1) / 5(m + 3) (m – 3)

Cancel the common term (m – 3) in both numerator and denominator.

= (m + 4) (m² – 1) / 5(m + 3).

3. Simplify algebraic fractions to its lowest terms:

(a) (abc + bc²) / (acd + dc²)

(b) (4a² – 9b²) / (4a² + 6ab)

(c) [(a + 1)³ – (a – 1)³] / (3a³ + a)

## Solution:

(a) (abc + bc²) / (acd + dc²)

= bc(a + c) / dc(a + c)

Cancel the common term c(a + c) in both numerator and denominator.

= b / d.

(b) (4a² – 9b²) / (4a² + 6ab)

= [(2a)² – (3b)²] / (4a² + 6ab)

= (2a + 3b) (2a – 3b) / 2a(2a + 3b)

Cancel the common term (2a + 3b) in both numerator and denominator of the algebraic fraction.

= (2a – 3b) / 2a.

(c) [(a + 1)³ – (a – 1)³] / (3a³ + a)

= [(a³ + 3a² + 3a + 1) – (a³ – 3a² + 3a – 1)] / a(3a² + 1)

= [a³ + 3a² + 3a + 1 – a³ + 3a² – 3a + 1] / a(3a² + 1)

= (6a² + 2) / a(3a² + 1)

= 2(3a² + 1) / a(3a² + 1)

Cancel the common term (3a² + 1) in both numerator and denominator.

= 2 / a.

4. Simplify and Reduce the following Algebraic Fractions:

(a) [(a – b) / b)] + [(a + b) / b] – [(a² – b²) / 2ab]

(b) (ab – 3b²)² / (a²b² – 27b⁵)

(c) [(a³ – 2a)² – (a² – 2)²] / [(a – 1)(a + 1) (a² – 2)]

## Solution:

(a) [(a – b) / b)] + [(a + b) / b] – [(a² – b²) / 2ab]

The least common multiple of denominators b, b, 2ab is 2ab.

To make the fractions having the common denominator both numerator and denominator of these are to be multiplied by 2a / 2a in case of [(a – b) / b)] and [(a + b) / b], 1 / 1 in case of [(a² – b²) / 2ab].

Therefore, [(a – b) / b)] + [(a + b) / b] – [(a² – b²) / 2ab]

= [2a(a – b) / 2ab] + [2a(a + b) / 2ab] – [(a² – b²) / 2ab]

= [2a(a – b) + 2a(a + b) – (a² – b²)] / 2ab

= [2a² – 2ab + 2a² + 2ab – a² + b²] / 2ab

= [3a² + b²] / 2ab.

(b) (ab – 3b²)² / (a²b² – 27b⁵)

= (a²b² – 6ab³ + 9b⁴) / (a²b² – 27b⁵)

= b²(a² – 6ab + 9b²) / b²(a² – 27b³)

Cancel the common term b² in both numerator and denominator.

= (a – 3b)² / (a² – (3b)³)

(c) [(a³ – 2a)² – (a² – 2)²] / [(a – 1)(a + 1) (a² – 2)]

= [(a⁶ – 4a⁴ + 4a²) – (a⁴ – 4a² + 4)] / [(a – 1)(a + 1) (a² – 2)]

= [a⁶ – 4a⁴ + 4a² – a⁴ + 4a² – 4] / [(a – 1)(a + 1) (a² – 2)]

= [a⁶ – 5a⁴ + 8a² – 4] / [(a – 1)(a + 1) (a² – 2)].