Why is it useful to leave an algebraic fraction in factorised form ?

Factorised form means leaving the top and bottom of the fraction each expressed as the product of its factors . This is useful because it makes it easier to see if anything cancels at the end.

True or False? If asked to simplify an algebraic fraction in an exam question, one factor will likely be the same on the top and bottom.

True. If asked to simplify an algebraic fraction in an exam question, one factor will likely be the same on the top and bottom. Factorise the easier expression, then use this fact to help you to factorise the more difficult quadratic.

IGCSEMathsEdexcelMaths A (Modular)Higher Unit 1FlashcardsAlgebraAlgebraic Fractions

Algebraic Fractions (Edexcel IGCSE Maths A (Modular))

Flashcards

1/15

0Still learning

Know0

FrontSimplifying Algebraic Fractions
What is an algebraic fraction?

Enjoying Flashcards?
Tell us what you think

Cards in this collection (15)

What is an algebraic fraction?
An algebraic fraction is a fraction with an algebraic expression on the top (numerator) and/or the bottom (denominator).
E.g. $\frac{3 x}{2}$ , $\frac{7}{x - 4}$ and $\frac{2 x}{x + 5}$ are all examples of algebraic fractions.
What is a common factor?
A common factor is a factor that is shared between two or more expressions.
E.g. $x$ is a common factor for the expressions $3 x + 8$ and $x y$ .
What is cancelling in algebraic fractions?
Cancelling in algebraic fractions means removing common factors that appear in both the numerator and denominator.
E.g. The algebraic fraction $\frac{2 (x + 3)}{6}$ can have the common factor of $2$ removed from both the numerator and denominator, leaving $\frac{x + 3}{3}$ .
Why is it useful to leave an algebraic fraction in factorised form?
Factorised form means leaving the top and bottom of the fraction each expressed as the product of its factors.
This is useful because it makes it easier to see if anything cancels at the end.
What are the common steps for simplifying an algebraic fraction such as $\frac{6 x - 24}{5 x - 20}$ ?
To simplify an algebraic fraction such as $\frac{6 x - 24}{5 x - 20}$ , you should:
1. Factorise fully top and bottom, $\frac{6 (x - 4)}{5 (x - 4)}$ .
2. Then cancel common factors (including common brackets), $\frac{6}{5}$ .
True or False?
If asked to simplify an algebraic fraction in an exam question, one factor will likely be the same on the top and bottom.
True.
If asked to simplify an algebraic fraction in an exam question, one factor will likely be the same on the top and bottom.
Factorise the easier expression, then use this fact to help you to factorise the more difficult quadratic.
How can you find the lowest common denominator (LCD) for a pair of algebraic fractions?
With algebraic fractions, the lowest common denominator (LCD) is found by multiplying the denominators together if they do not share any factors.
E.g. The LCD of $\frac{3}{x + 2}$ and $\frac{5 x}{x - 3}$ is $(x + 2) (x - 3)$ .
If they do share factors, find the lowest common denominator by taking the denominator that already includes the other(s).
E.g. The LCD of $\frac{4}{x}$ and $\frac{9}{x (x + 1)}$ is $x (x + 1)$ .
True or False?
If $x$ and $2 x$ are the denominators of two algebraic fractions, then the lowest common denominator is found by multiplying $x$ and $2 x$ together.
False.
$2 x$ already includes the factor $x$ .
If $x$ and $2 x$ are the denominators of two algebraic fractions, then the lowest common denominator is $2 x$ .
What is the process for adding two algebraic fractions?
E.g. $\frac{3 x}{x + 5} + \frac{4 x + 2}{x}$ .
The process for adding two algebraic fractions is:
1. Find the lowest common denominator, $x (x + 5)$
2. Write each fraction as an equivalent fraction over the lowest common denominator, $\frac{3 x}{(x + 5)} \times \frac{x}{x} + \frac{4 x + 2}{x} \times \frac{(x + 5)}{(x + 5)} = \frac{3 x^{2}}{x (x + 5)} + \frac{(x + 5) (4 x + 2)}{x (x + 5)}$
3. Add the numerators, $\frac{3 x^{2} + (x + 5) (4 x + 2)}{x (x + 5)}$ .
4. Simplify, $\frac{7 x^{2} + 22 x + 10}{x (x + 5)}$ .
What is the process for subtracting two algebraic fractions?
E.g. $\frac{x + 6}{x} - \frac{3}{x - 1}$ .
The process for subtracting two algebraic fractions is:
1. Find the lowest common denominator, $x (x - 1)$
2. Write each fraction as an equivalent fraction over the lowest common denominator, $\frac{x + 6}{x} \times \frac{(x - 1)}{(x - 1)} - \frac{3}{x - 1} \times \frac{x}{x} = \frac{(x + 6) (x - 1)}{x (x - 1)} - \frac{3 x}{x (x - 1)}$
3. Subtract the numerators, $\frac{(x + 6) (x - 1) - 3 x}{x (x - 1)}$ .
4. Simplify, $\frac{x^{2} + 2 x - 6}{x (x - 1)}$ .
What is the process for multiplying algebraic fractions?
E.g. $\frac{6 x}{x + 1} \times \frac{2 x - 4}{3 x^{2} - 6}$ .
To multiply algebraic fractions:
1. Simplify both fractions first by factorising and cancelling common factors $\frac{6 x}{x + 1} \times \frac{2 (x - 2)}{3 x (x - 2)} = \frac{6 x}{x + 1} \times \frac{2}{3 x}$ .
2. Multiply the numerators together, $6 x \times 2 = 12 x$ .
3. Multiply the denominators together, $(x + 1) \times 3 x = 3 x^{2} + 3 x$ .
4. Check for further factorising and cancelling, $\frac{12 x}{3 x^{2} + 3 x} = \frac{12 x}{3 x (x + 1)} = \frac{4}{x + 1}$ .
What is the reciprocal of an algebraic fraction.
The reciprocal of an algebraic fraction is the fraction 'flipped', i.e. with the original denominator divided by the numerator.
For example the reciprocal of $\frac{x - 4}{x^{2} + 3}$ is $\frac{x^{2} + 3}{x - 4}$ .
What is the process for dividing algebraic fractions?
E.g. $\frac{2 x}{7 x - 1} \div \frac{x}{x + 3}$ .
To divide algebraic fractions:
1. Find the reciprocal of the second fraction and replace $\div$ with $\times$ , $\frac{2 x}{7 x - 1} \times \frac{x + 3}{x}$ .
2. Then follow the rules for multiplying two algebraic fractions.
Note that this is the same thing you do when dividing normal fractions.
What are the two main methods for solving equations with algebraic fractions?
E.g. $\frac{2}{x + 1} + \frac{x}{x - 3} = 2$ .
The two main methods for solving equations with algebraic fractions are:
1. Adding/subtract the fractions first, $\frac{2 (x - 3) + x (x + 1)}{(x + 1) (x - 3)} = 2$ .
  Then solve the resulting equation.
2. Multiply everything by the common denominator to eliminate fractions first, $2 (x - 3) + x (x + 1) = 2 (x + 1) (x - 3)$ .
  Then solve the resulting equation.
What are the steps to solve an equation with algebraic fractions by multiplying through by the common denominator?
E.g. $\frac{1}{2 x} + \frac{7 x + 5}{2 x} = 4$ .
The steps to solve an equation with algebraic fractions by multiplying through by the common denominator are:
1. Multiply every term by the common denominator, $1 + 7 x + 5 = 8 x$ .
2. Expand any brackets and collect like terms, $7 x + 6 = 8 x$ .
3. Rearrange and solve the equation, $6 = x$ .

Previous:Completing the SquareNext:Solving Linear Equations

More Revision Notes you might like