# Basic Algebra/Rational Expressions and Equations/Adding and Subtracting When the Denominators are Different

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The latest reviewed version was checked on

*27 December 2017*. There is 1 pending change awaiting review.## Vocabulary[edit]

## Lesson[edit]

If you have to add two rational fractions with different denominators, as the first step, you have to find the LCM:

3+3x+1 x-1

LCM = (x+1)(x-1)

Now divide the LCM by both denominators and multiply by their respectives numerators:

(x+1)(x-1) / (x+1) = (x-1) . (3) = 3x-3 (x+1)(x-1) / (x-1) = (x+1) . (3) = 3x+3

The sum of the two results would be the new nominator:

3x= (x+1)(x-1)~~-3~~+3x~~+3~~

6x(x+1)(x-1)

This is another example:

6x+9x2x-6 x^{2}-6x+9

We factorize both denominators and find the LCM

2x-6 = 2(x-3) x^{2}-6x+9 = (x-3)^{2}LCM = 2(x-3)^{2}

Now we divide and multiply:

2(x-3)^{2}/ 2(x-3) = 2x^{2}-12x+18 / 2x-6 = x-3 (x-3) . 6x = 6x^{2}-18x

2(x-3)^{2}/ (x-3)^{2}= 2x^{2}-12x+18 / x^{2}-6x+9 = 2 (2) . (9x) = 18x

We add the results to obtain the nominator; the denominator is the LCM:

6x= 2(x-3)^{2}~~-18x~~~~+18x~~^{2}

6x2(x-3)^{2}^{2}

We can factorize the nominator to simplify the result:

=~~2~~(3x^{2})~~2~~(x-3)^{2}

3x(x-3)^{2}^{2}

## Example Problems[edit]

## Practice Games[edit]

## Practice Problems[edit]

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**Medium**

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