Algebra

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Algebraic Fractions: Addition and Subtraction

Adding and subtracting algebraic fractions together allows us to express them as a single fraction.

Adding and subtracting algebraic fractions is much like adding and subtracting numerical fractions.

The objective is to be able to write them as a single fraction.

Adding fractions is simple when the denominators are the same, we just add the numerators. What is $\dfrac{5}{x+1}+\dfrac{4}{x+1}$ ?

Subtraction uses the same principle, what is $\dfrac{3}{n}-\dfrac{1+n}{n}$ ?

What happens when the denominators are not the same?

$\dfrac{x}{5} + \dfrac{x}{6}$

Just like with numerical fractions, the first step is to find a common denominator. What does this mean?

Find the lowest common denominator for $\dfrac{x}{5} + \dfrac{x}{6}$ .

We can't just change the denominator as this would change the value of the fraction.

We need to multiply the numerator in the same way as the denominator.

Our first fraction is $\dfrac{x}{5}$ , and we have multiplied top and bottom by 6

$\dfrac{x\times 6}{5\times 6}=\dfrac{6x}{30}$

What does the second fraction $\dfrac{x}{6}$ become to give it a denominator of $30$ ?

We have converted our fractions with different denominators to fractions with common denominators and can add them together as before. What is $\dfrac{x}{5} + \dfrac{x}{6} = \dfrac{6x}{30} + \dfrac{5x}{30}$ ?

Nice! The final answer is $\dfrac{11x}{30}$

This can't be simplified any further.

What is $\dfrac{x}{4} + \dfrac{x}{7}$ ?

Find the sum $\dfrac{2}{3b} + \dfrac{1}{2b}$

We can apply the same principles when the denominator contains a letter.

Let's work through this one $\dfrac{2}{3b} + \dfrac{1}{2b}$

What is the common denominator? $\dfrac{2}{3b} + \dfrac{1}{2b}$

We then multiply both fractions to make $6b$ the denominator. For the first fraction this is $\dfrac{2\times 2}{3b\times 2}=\dfrac{4}{6b}$

Now we can add our fractions. $\dfrac{4}{6b} + \dfrac{3}{6b}$

Well done - you've solved a fraction with an algebraic denominator!

The final answer is $\dfrac{7}{6b}$

What is $\dfrac{4}{3x} + \dfrac{2}{5x}$ ?

$\dfrac{6}{x+1}+\dfrac{x}{x+4}$

Let's try something a little harder. We need to take it one step at a time.

Let's try $\dfrac{6}{x+1}+\dfrac{x}{x+4}$ .

We use the same principle.

We just need to be careful with the denominators as they each have two terms.

First we need to find the common denominator. This is found by multiplying the denominators together, make sure to use brackets!

Our common denominator is $(x+1)(x+4)$ . We must multiply the numerator and denominator of the first fraction by $(x+4)$ .

This makes the first fraction $\dfrac{6(x+4)}{(x+1)(x+4)}$ .

We have converted our first fraction to $\dfrac{6(x+4)}{(x+1)(x+4)}$ . What does the second fraction become?

We can now add the fractions together. $\dfrac{6(x+4)}{(x+1)(x+4)}+\dfrac{x(x+1)}{(x+1)(x+4)}$

We now have a very busy fraction $\dfrac{6(x+4)+x(x+1)}{(x+1)(x+4)}$ !

We can multiply out the brackets in the numerator to simplify it a little.

Focusing on the numerator, multiply out and simplify $6(x+4)+x(x+1)$ .

We can leave this as our final answer. Well done!

$\dfrac{6}{x+1}\dfrac{x}{x+4}=\dfrac{x^2+7x+24}{(x+1)(x+4)}$

Summary! Add algebraic fractions in the same way as numerical fractions.

Find the common denominator first. For $\dfrac{5x}{6}+\dfrac{5}{3x}$ the common denominator is $6x$ .

Cross multiply the fractions to express them in terms of the common denominator.

$\dfrac{5x}{6}+\dfrac{5}{3x}$ $=\dfrac{5x\times x}{6x}+\dfrac{5\times 2}{6x}$ $=\dfrac{5x^2}{6x}+\dfrac{10}{6x}$

With a common denominator we just add the numerator.

$\dfrac{5x^2}{6x}+\dfrac{10}{6x}$ $=\dfrac{5x^2+10}{6x}$

Summary of a more complex addition: $\dfrac{x}{x+2}+\dfrac{8}{x-4}$

First cross multiply to get the common denominator.

$\dfrac{x(x-4)}{(x+2)(x-4)}$ $+$ $\dfrac{8(x+2)}{(x+2)(x-4)}$

Add the numerators

$\dfrac{x(x-4)+8(x+2)}{(x+2)(x-4)}$

Multiply out the numerator

$\dfrac{x^2-4x+8x+16)}{(x+2)(x-4)}$

Gather the like terms together for the final answer

$\dfrac{x^2+4x+16}{(x+2)(x-4)}$