How do you simplify an algebraic fraction?

Factorise fully on the top and on the bottom, then cancel any factors common to both. Whole brackets can cancel, not just numbers and single letters.

How do you solve an equation containing algebraic fractions?

Multiply every term by each expression on a denominator, which clears the fractions completely. What is left is a linear, quadratic or cubic equation, which is then solved in the usual way.

How can factorising the top of an algebraic fraction help you factorise the bottom?

If the fraction is going to simplify at all, one of the bottom's factors has to match one from the top. So a factor already found on the top tells you what to look for below, which is a real help on an awkward quadratic.

GCSEFurther MathsAQAFlashcards2. Algebra & FunctionsAlgebraic Fractions

Algebraic Fractions (AQA GCSE Further Maths): Flashcards

Exam code: 8365

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FrontAlgebraic Fractions
Define algebraic fraction.

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Cards in this collection (12)

Define algebraic fraction.
An algebraic fraction is a fraction with an algebraic expression on the top, the bottom, or both.
So $\frac{x + 2}{5}$ , $\frac{7}{x^{2}}$ and $\frac{x + 2}{x - 3}$ are all algebraic fractions.
How do you simplify an algebraic fraction?
Factorise fully on the top and on the bottom, then cancel any factors common to both.
Whole brackets can cancel, not just numbers and single letters.
Simplify $\frac{4 x + 6}{2 x^{2} - 7 x - 15}$ .
Factorising the top and the bottom gives $\frac{2 (2 x + 3)}{(2 x + 3) (x - 5)}$ .
The $(2 x + 3)$ cancels, leaving $\frac{2}{x - 5}$ .
True or False?
The lowest common denominator of $x$ and $2 x$ is $2 x^{2}$ .
False.
The lowest common denominator is just $2 x$ , because $2 x$ already contains an $x$ .
This mirrors ordinary numbers, where the lowest common denominator of 2 and 4 is 4 rather than 8.
When is the lowest common denominator found by multiplying the two denominators together?
When they share no factor, as with $x - 2$ and $x + 5$ , whose lowest common denominator is $(x - 2) (x + 5)$ .
Where they share a factor it is not repeated, so $(x + 3) (x - 1)$ and $(x + 4) (x - 1)$ need only the three brackets $(x + 3) (x - 1) (x + 4)$ .
Express $\frac{x}{x + 4} - \frac{3}{x - 1}$ as a single fraction.
Over the common denominator $(x + 4) (x - 1)$ the numerator becomes $x (x - 1) - 3 (x + 4) = x^{2} - 4 x - 12$ .
That factorises, giving $\frac{(x + 2) (x - 6)}{(x + 4) (x - 1)}$ , with nothing left to cancel.
Complete the rule for dividing algebraic fractions:
Dividing by $\frac{a}{b}$ is the same as multiplying by $______$ .
The completed rule is:
Dividing by $\frac{a}{b}$ is the same as multiplying by $\frac{b}{a}$ .
Factorise and cancel before multiplying the tops together and the bottoms together.
Divide $\frac{x + 3}{x - 4}$ by $\frac{2 x + 6}{x^{2} - 16}$ , giving a simplified fraction.
Flipping the second fraction and factorising gives $\frac{x + 3}{x - 4} \times \frac{(x - 4) (x + 4)}{2 (x + 3)}$ .
Both $(x + 3)$ and $(x - 4)$ cancel, leaving $\frac{x + 4}{2}$ .
How do you solve an equation containing algebraic fractions?
Multiply every term by each expression on a denominator, which clears the fractions completely.
What is left is a linear, quadratic or cubic equation, which is then solved in the usual way.
Solve $\frac{4}{x - 3} + \frac{5}{x + 1} = 5$ .
Multiplying every term by $(x - 3)$ and $(x + 1)$ gives $4 (x + 1) + 5 (x - 3) = 5 (x - 3) (x + 1)$ .
That rearranges to $5 x^{2} - 19 x - 4 = 0$ , which factorises to give $x = - \frac{1}{5}$ or $x = 4$ .
Show that $\frac{2}{p + 3} - \frac{5}{p} = 6 p$ can be written as $2 p^{3} + 6 p^{2} + p + 5 = 0$ .
Multiplying through by $(p + 3)$ and then by $p$ gives $2 p - 5 (p + 3) = 6 p^{2} (p + 3)$ .
Expanding and collecting gives $6 p^{3} + 18 p^{2} + 3 p + 15 = 0$ , and dividing by 3 gives the required form.
How can factorising the top of an algebraic fraction help you factorise the bottom?
If the fraction is going to simplify at all, one of the bottom's factors has to match one from the top.
So a factor already found on the top tells you what to look for below, which is a real help on an awkward quadratic.

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