Subtracting Functions with Fractions

In modern mathematics, you will most likely encounter functions. In mathematics, a function is a binary relationship between two or more variables. You can think of the variables as sets, and the relationship is such that an element of one set maps exactly to another element of the other set. A function is a law that defines how one set relates to another set. Another way to also visualize a function is to think of it as a process or machine that outputs a specific result when there is a particular input. An example is

F(x) = x2 + 1

In the above example, the name of the function is F(x) pronounced as ‘f of x’. Other examples of names of functions include g(n), h(x), etc. The function is defined as x2 + 1. The x here serves the same function as in your usual algebra, which is a placeholder for any set of values. This means that x can assume any value, such as 3, 1.5, -5, etc.

So if f(x) = x2 + 1 then f(2) = 22+ 1 = 5.

Sometimes, the name of the function is replaced with a second variable, such as in the example

y = x – 2

This second way of expressing a function clearly illustrates the concept of a relationship between two variables. In the above example, x is the independent variable because it can assume any value, and y is the dependent variable because its values are determined by the values of x.

Now that we understand what functions are, let us look at functions that involve fractions. These functions have a numerator and denominator, the same as normal fractions.

Let us look at an example. Consider the function

f(x) = (x2 – 5) / 3

The function f(x) has a numerator which is x2 – 5 and a denominator, which is 3.

Let us further discuss how you would do the subtraction of functions that involve fractions.

The procedure for subtracting functions that involve fractions is pretty the same as in normal fractions. The same broad categorization applies: fractions with the same denominator and those with different denominators.

Let us start with functions that have the same denominator.

Subtracting Functions With The Same Denominator

Fractions that have the same denominators are very easy to subtract. You simply subtract the numerators without touching the denominator. The denominator remains unchanged in your final result. Let us look at an example.

Example 1

Simplify the expression below.

(( x2 – 3 ) / 3 ) – ( ( x – 3 ) / 3))

Both of the functions have a denominator of 3. The steps below can be used to simplify the expression.

Step 1

Subtract the numerators.

( x2 – 3 ) – ( x – 3 )

Remove the brackets

x2 – 3 – x + 3

= x2 – x …. since -3 + 3 = 0

Since all the terms involve x, we can factorize x

= x ( x – 1 )

Note that we factorized because there is a common factor x. If there is no common factor, you will not be able to factorize. You just move on to the next step.

Step 2

Rewrite the function with your new numerator and the same old denominator

(x ( x – 1 )) / 3

You have successfully subtracted two fractions with the same denominator.

Example 2

Simplify the expression

(x3 / ( x – 2 )) – (2x2 / ( x – 2 ))

Step 1

Subtract the numerators.

x3 – 2x2

The numerators both have single terms. Also, x2 is a common factor so we can go ahead and factorize.

Both terms have x2 in common so we factorize it out.

x2 ( x – 2 )

There is no way to simplify the numerator further so we move on to the denominator.

Step 2

Rewrite the function with your new numerator and the old denominator

x2 ( x – 2 ) / ( x – 2 )

In this example, you can notice that there is something common between the numerator and the denominator which is ( x – 2 ). This means that we can further simplify the function by dividing and hence canceling out the common factor.

x2 ( x – 2 ) / ( x – 2 ) = x2

So the function after the subtraction has simplified to a single term x2.

Subtracting Functions with Different Denominators

When the denominators of fractions are not the same, the addition or subtraction of the fractions requires extra steps. Consider the examples below.

Example 3

Simplify the expression below:

(( 2x + 1 ) / 2x ) – ( 3x / 5)

In the above example, the two functions have different denominators, 2x and 5 respectively. Since the denominators are not the same, we cannot follow the same steps we used in the previous examples.

Before we proceed with the subtraction, let us first revise one of the rules of subtraction.

( a / b ) – ( c / e ) = ( ae – bc ) / be

This rule is general for all fractions. Now that we understand this rule, let us proceed with our simplification. If the rule is new to you, do not worry, you will understand as we apply it to the example.

Step 1

Multiply the numerator of the first fraction with the denominator of the second fraction.

( 2x + 1 ) * 5 = 5 ( 2x + 1 )

This will become the first term of your new numerator.

Step 2

Multiply the numerator of the second fraction with the denominator of the first fraction.

3x * 2x = 6x2

This will become the second term of your new numerator

Step 3

Multiply the denominators of the two fractions.

2x * 5 = 10x

Step 4

Write out your resulting fraction now: the complete numerator and denominator

( 5 ( 2x + 1 ) – 6x2 ) / 10x

You now have your final fraction which is the difference between the two initial functions. Although the result we have above may not be in its simplest form, because one can expand the numerator and solve the resulting quadratic equation which may have common factors with the denominator. However, that is a different topic on its own. So we will stop here in our simplification.

For solving fractions, always remember the rule:

( a / b ) – ( c / e ) = ( ae – bc ) / be

For the addition of fractions, there is just a slight modification in the formula. The idea is the same. Just replace the minus signs with plus signs.

( a / b ) + ( c / e ) = ( ae + bc ) / be

The above general rules can be used to subtract or add functions that involve fractions. The rules of fractions are used the same way as in arithmetic.

I believe that you now understand what functions are and can comfortably subtract functions with the aid of the above formulae.

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