Solving Algebraic Fractions | Simplify & Reduce Algebraic Fractions

# Solving Algebraic Fractions | Simplify & Reduce Algebraic Fractions

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.

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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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