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Math Lesson 6.5.2 - Operations with Algebraic Fractions

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Welcome to our Math lesson on Operations with Algebraic Fractions, this is the second lesson of our suite of math lessons covering the topic of Algebraic Fractions, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Operations with Algebraic Fractions

We can do with algebraic fractions everything we did with numerical fractions in the 3rd chapter of this course, including addition, subtraction, multiplication, division, raise in power, etc. Moreover, we can apply with them the new things explained in the current chapter such as simplification, expanding, factorisation, etc. Let's have a closer look at each of these operations.

a. Simplifying algebraic fractions

Simplifying an algebraic fraction means removing the same number, variable, group of variables, number-variable combination or expression from both numerator and denominator in order to obtain a simpler algebraic fraction.

An algebraic fraction can be simplified only when its numerator and denominator represent a single factor or a product of factors. For example, the algebraic fraction

10 - 2x/6x

cannot be simplified in the actual form unless its denominator is written as a product of factors. We achieve this through factorisation, i.e.

10 - 2x/6x = 2(5 - x)/6x

Now, both parts of the given algebraic fraction are expressed as a products of factors. Therefore, we can simplify 2 from both sides to obtain

2(5 - x) ÷ 2/6x ÷ 2 = 5 - x/3x

Obviously, the common factor must not be a disallowed value.

Example 2

Simplify the following algebraic fractions

12x - 4y/9x - 3y
4x3 y2 - 8xy3/12x2 y2

Solution 2

Let's try to identify common factors in the numerator and denominator. First, we spot the number 4 as a common factor for the numerator and 3 for the denominator, that is

12x - 4y/9x - 3y = 4(3x - y)/3(3x - y)

Now, it is easy to see that for 3x - y ≠ 0, we can simplify the expression in brackets to ultimately obtain

4(3x - y)/3(3x - y) = 4/3

It is easy to identify 4xy as a common factor for this algebraic fraction. Thus,

4x3 y2 - 8xy3/12x2 y2 = 4xy(x2 y - 2y2)/4xy ∙ 3xy

Simplifying the common factor yields

4xy(x2 y - 2y2)/4xy ∙ 3xy = x2 y - 2y2/3xy

Obviously, this simplification is valid for x ≠ 0 and y ≠ 0.

It may seem that the new algebraic expression obtained after simplification is as equally long as the original expression. However, if you look carefully at the powers, you will notice that the variables are in a lower power than before, which means that the expression is successfully simplified.

b. Simplifying algebraic fractions using the special algebraic identities

It is good to check whether any of the eight special algebraic identities is present in any part of algebraic fraction or not. If it is, this would help a lot in the process of simplification. For example, in the algebraic fraction

x2 - 4x + 4/x2 - 4

it is easy to see that the numerator contains the expanded form of the square of a difference, while the denominator contains the expanded form of the product of conjugates. Hence,

x2 - 4x + 4/x2 - 4 = x2 - 2 ∙ x ∙ 2 + 22/x2 - 4
= (x - 2)2/(x - 2)(x + 2)
= (x - 2)(x - 2)/(x - 2)(x + 2)

Thus, for x ≠ 2, we can simplify (x - 2) from both sides to obtain

(x - 2)(x - 2)/(x - 2)(x + 2)
= (x - 2)/(x + 2)

Example 3

Factorise the following algebraic fractions

x3 - 1/x2 + x + 1
3x3 y + 12x2 y + 9xy/x2 y - 9y

Solution 3

The numerator shows the expanded form of the expression (x - 1)(x2 + x + 1). Thus, we can write

x3 - 1/x2 + x + 1
= (x - 1)(x2 + x + 1)/x2 + x + 1

Simplifying x2 + x + 1 from both sides yields

(x - 1)(x2 + x + 1)/x2 + x + 1
= x - 1/1
= x - 1

The greatest common factor of the terms in the numerator is 3xy, while in the denominator it is just y. Thus, we have

3x3 y + 12x2 y + 9xy/x2 y - 9y
= 3xy(x2 + 4x + 3)/y(x2 - 9)
= 3x(x2 + 4x + 3)/x2 - 9

We can use the method explained in the previous tutorial to factorise the numerator. Since it contains a quadratic expression where a = 1, b = 4 and c = 3 (p · q = 1), we have

m ∙ n = c = 3

and

-(m + n) = b = 4

Thus, m = -1 and n = -3. Hence, the expression in the numerator becomes

ax2 + bx + c = (x - m)(x - n)

or

x2 + 4x + 3 = [x - (-1)][x - (-3)]
= (x + 1)(x + 3)

As for the denominator, it is clear that the expression in brackets represents the expanded form of (x - 3)(x + 3). Hence, we obtain

3x(x2 + 4x + 3)/x2 - 9 = 3x ∙ (x - 1)(x + 3)/(x - 3)(x + 3)

Simplifying (x + 3) from both sides (with the condition that x ≠ 3), yields

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

c. Adding and subtracting algebraic fractions

We can add or subtract algebraic fractions in the same way as we do with numerical fractions. Thus, if fractions have the same denominator, we simply add or subtract the numerators while the denominator doesn't change. On the other hand, if the denominators are different we turn all fractions at the same denominator and only then do we continue with the rest. Let's explain these procedures through examples.

Example 4

Do the following operations. Simplify when necessary.

5xy2/2xz + 6xy3/2xyz
6m3 n/5 - 4m5 n3/(m2 n)2
5x2 - 15x/x2 - 9 + x/x + 3

Solution 4

First, we check for any possible simplification; then we continue with the rest of steps. We have to simplify x from the first algebraic fraction and xy from the second. You may think that we can also simplify 2 from the second fraction but we don't include 2 in simplifications to obtain two fractions with the same denominator. Thus,

5xy2/2xz + 6xy3/2xyz
= 5y2/2z + 6y2/2z
= 5y2 + 6y2/2z
= 11y2/2z

We do the necessary simplifications in the second algebraic fraction first. We have to simplify m2n2 from both terms to obtain

6m3 n/5 - 4m5 n3/(m2 n)2
= 6m3 n/5 - 4m3 n/1

Now, we can multiply both terms of the second algebraic fraction by 5 to turn it at the same denominator to the first fraction. Thus,

6m3 n/5 - 5 ∙ 4m3 n/5 ∙ 1
= 6m3 n/5 - 20m3 n/5
= -14m3 n/5

The denominator of the first fraction is the expanded form of (x - 3)(x + 3). Moreover, we can factorise 5x from the numerator of the same fraction to obtain

5x2 - 15x/x2 - 9 + x/x + 3
= 5x(x - 3)/(x - 3)(x + 3) + x/x + 3

Now, we can simplify x - 3 from the first fraction with the condition that x must be different from 3. In this way, we obtain

5x(x - 3)/(x - 3)(x + 3) + x/x + 3
= 5x/x + 3 + x/x + 3
= 6x/x + 3

d. Multiplying and dividing algebraic fractions

It is worth re-emphasizing that the multiplication and division of algebraic fractions is carried out in the same way as we do with numerical fractions. Thus, in multiplication we multiply the numerators with each other; the same procedure is also carried out with denominators. Then, you can check for any possible simplification.

As for the division of algebraic fractions, first the second fraction is inverted down to turn the division into multiplication, followed up by the multiplication procedures.

Example 5

Do the following operations:

3x - 2/618x/9x2 - 4
2x/8x3 + 27 ÷ 4x2/4x2 - 6x + 9

Solution 5

It is easy to see that the denominator of the second algebraic fraction represents the expanded form of the product of two conjugates: 3x - 2 and 3x + 2. Hence, we write

3x - 2/618x/9x2 - 4
= 3x - 2/618x/(3x - 2)(3x + 2)
= 18x ∙ (3x - 2)/6 ∙ (3x - 2)(3x + 2)

At this point, we can simplify 6 and 3x - 2 from both sides of the resulting fraction to obtain

18x ∙ (3x-2)/6 ∙ (3x-2)(3x + 2) = 3x/3x + 2

First, we have to invert down the second fraction to turn the division into multiplication.

2x/8x3 + 27 ÷ 4x2/4x2 - 6x + 9
= 2x/8x3 + 274x2 - 6x + 9/4x2

Now, if we look carefully at the denominator of the first fraction, we can detect a familiar expression. Indeed, it is the expanded form of the special algebraic identity

(a + b)(a2 - ab + b2 ) = a3 + b3

where a = 2x and b = 3. Hence, we write

2x/8x3 + 274x2 - 6x + 9/4x2
= 2x/(2x + 3)(4x2 - 6x + 9)4x2 - 6x + 9/4x2
= 2x ∙ (4x2 - 6x + 9)/4x2 ∙ (2x + 3)(4x2 - 6x + 9)

Now, it is easy to understand that we can simplify 2x and 4x2 - 6x + 9 from both sides of the resulting fraction to obtain

2x ∙ (4x2 - 6x + 9)/4x2 ∙ (2x + 3)(4x2 - 6x + 9)
= 1/(x ∙ (2x + 3)

e. Splitting an algebraic fraction in two component algebraic fractions

Sometimes, it is necessary to do the inverse action of the addition or subtraction of two fractions with different denominators. In other words, we may need to split an algebraic fraction into two other algebraic fractions which, when combined together (through addition or subtraction), give the original fraction. We will explain how to do this by taking the result of the last example as an illustration. Thus, for example, let's suppose that we are required to split the algebraic fraction

1/2x ∙ (2x + 3)

in two algebraic fractions that are related through addition. In other words, we want to express the given algebraic fraction in the form

A/2x + B/2x + 3

where A and B are (for now) unknown algebraic expressions.

In general, if the original fraction is written in the form

E/M ∙ N

where E, M and N are three different algebraic expressions, we can express it in the form

E/M ∙ N = A/M + B/N

where

E = A ∙ N + B ∙ M

From the properties of fractions addition, it is clear that

A ∙ (2x + 3) + B ∙ 2x = 1

Expanding the new expression obtained yields

2Ax + 3A + 2Bx = 1

or

(2A + 2B) ∙ x + 3A = 1

Since the numerator of the original fraction is 1, it is clear that there is no variable x involved there. So, we can write

2A + 2B = 0

or

2A = -2B
A = -B

In addition, we have

3A = 1
A = 1/3

Hence,

B = -1/3

Therefore, we can write the two component fractions of the original as

A/2x + B/2x + 3 = 1/3/2x + -1/3/2x + 3
= 1/3 ∙ 2x - 1/3(2x + 3)
= 1/6x - 1/6x + 9

Proof: Let's start from the last expression obtained. Thus,

1/6x - 1/6x + 9 = 1 ∙ (6x + 9) - 1 ∙ 6x/6x ∙ (6x + 9)
= 6x + 9 - 6x/6x ∙ (6x + 9)
= 9/6x(6x + 9)
= 9/6x ∙ 3(2x + 3)
= 9/18x ∙ (2x + 3)
= 1/2x ∙ (2x + 3)

This is identical to the original algebraic fraction we took as example.

Example 6

Write the following algebraic fractions as sum or difference of other algebraic fractions.

2x - 3/3x(x - 1)
5x/x2 - 16

Solution 6

We have

2x - 3/3x(x - 1) = A/3x + B/x - 1

In addition,

A ∙ (x - 1) + B ∙ 3x = 2x - 3
Ax - A + 3Bx = 2x - 3
(A + 3B) ∙ x - A = 2x - 3

Combining the like terms yields

A + 3B = 2

and

-A = -3

From the last equation, we obtain

A = 3

Substituting this value to the previous one, yields

3 + 3B = 2
3B = 2 - 3 = -1
B = -1/3

Hence, the original algebraic expression can be written as

2x - 3/3x(x - 1) = A/3x + B/x- 1
= 3/3x + -1/3/x - 1
= 1/x - 1/3(x - 1)
= 1/x - 1/3x - 3

Proof:

1/x - 1/3x - 3 = 1 ∙ (3x - 3) - x/x(3x - 3)
= 3x - 3 - x/3x(x - 1)
= 2x - 3/3x(x - 1)

which is identical to the original algebraic fraction.

We have

5x/x2 - 16 = 5x/(x - 4)(x + 4)
= A/x - 4 + B/x + 4

Moreover,

A(x + 4) + B(x-4) = 5x
Ax + 4A + Bx - 4B = 5x
(A + B) ∙ x + 4A - 4B = 5x
A + B = 5
4A - 4B = 0

So, A = B. Thus A + A = 5 which yields A = B = 5/2.

In this way, the expression becomes

5x/x2 - 16 = 5/2/x - 4 + 5/2/x + 4
= 5/2(x - 4) + 5/2(x + 4)
= 5/2x - 8 + 5/2x + 8

Proof:

5/2x - 8 + 5/2x + 8 = 5 ∙ (2x + 8) + 5 ∙ (2x - 8)/(2x - 8)(2x + 8)
= 10x + 40 + 10x - 40/4x2 - 64
= 20x/4(x2 - 16)
= 5x/x2 - 16

which is identical to the original expression.

f. Raising in power algebraic fractions

We can raise an algebraic fraction in power by raising the the numerator and denominator separately. In this way, we may apply the following property of powers

(x/y)n = xn/yn

Any negative exponent makes the algebraic fraction invert down and the new exponent has the same value but positive, i.e.

(x/y)-n = (y/x)n = yn/xn

We will discuss the properties of exponents more extensively in the next chapter, so for the moment we are stopping here with them.

Example 7

Complete the following operations.

[(2x)3 + 2x3/3x5]2
(3x - 5y/15xy)-3

Solution 7

We have

[(2x)3 + 2x3/3x5]2 = (23 ∙ x3 + 2x3/3x5)2
= (8x3 + 2x3/3x5)2
= (8x3 + 2x3/3x5)2
= (10x3/3x5)2
= (10/3x2)2
= 102/3x22
= 102/32 (x2 )2
= 100/9x4

We have

(3x - 5y/15xy)-3 = (15xy/3x - 5y)3
= (15xy)3/(3x - 5y)3
= 153 x3 y3/(3x)3 - 3 ∙ (3x)2 ∙ (5y) + 3 ∙ (3x) ∙ (5y)2 - (5y)3
= 153 x3 y3/27x3 - 3 ∙ 9x2 ∙ 5y + 3 ∙ 3x ∙ 25y2 - 125y3
= 3,375x3 y3/27x3 - 135x2 y + 175xy2 - 125y3

Sometimes, raising an algebraic expression by a given power may be beneficial because, as we will see in the next chapter, raising a negative number at an even power makes it positive. This property (as well as many others) may help in simplifying part of the algebraic fraction (that could have not otherwise been possible to simplify) after having raised it in an even power. Let's consider an example to clarify this point.

Example 8

Complete the operations and simplify where possible.

(1 - x/x2 - 1)2

Solution 8

At first glance it looks that there is nothing to simplify after raising the expression in the given power. However, if we express the numerator as

1 - x = - [-(1-x)]
= -[-1 + x]
= -(x - 1)

we can write

(1 - x/x2 - 1)2 = [-(x - 1)/(x - 1)(x + 1)]2

We can simplify (x - 1) from both sides first for easier operations (obviously, with the condition that x ≠ 1). Thus,

[-(x - 1)/(x - 1)(x + 1) ]2 = [-1/(x + 1)]2
= (-1)2/(x + 1)2
= 1/(x + 1)2
= 1/x2 + 2x + 1

The last expression (or the penultimate one if you want) is/are simplified version(s) of the original expression. Such a simplification could not have been possible without the 'trick' of the negative sign. Sometimes we must check for other options to simplify an expression in addition to the traditional options.

More Algebraic Fractions Lessons and Learning Resources

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Tutorial IDMath Tutorial TitleTutorialVideo
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Notes
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6.5Algebraic Fractions
Lesson IDMath Lesson TitleLessonVideo
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6.5.1What Are Algebraic Fractions?
6.5.2Operations with Algebraic Fractions

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