Why must an algebraic fraction be factorised before anything is cancelled?

Cancelling removes a factor , meaning something the whole top and the whole bottom are multiplied by. Until each side is written as a product, there is no way of telling which parts are factors and which are merely terms.

When is an algebraic fraction in its simplest form ?

An algebraic fraction is in simplest form once the top and the bottom share no common factor at all. Look again after cancelling once, because a second factorisation can sometimes reveal a further cancel.

In what order do you handle a product of two algebraic fractions?

Factorise both fractions, cancel any common factors, multiply the tops and the bottoms, then check once more for factorising and cancelling. That last check matters, because a product can factorise in a way that neither of the original fractions did.

What are the two ways of solving an equation containing algebraic fractions?

Either combine the fractions into one and then cross-multiply, or clear the fractions first by multiplying every term by each denominator. Both routes reach the same equation, so use whichever is easier for the fractions in front of you.

Do the two denominators have to be dealt with one at a time?

No, multiplying by both denominators at once reaches the same equation in fewer steps. Taking them one at a time is easier to follow, because each cancelling can be seen as it happens.

National 5MathsSQAFlashcardsAlgebraic SkillsAlgebraic Fractions

Algebraic Fractions (SQA National 5 Maths): Flashcards

Exam code: X847 75

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Define an algebraic fraction.

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Cards in this collection (28)

Define an algebraic fraction.
An algebraic fraction is a fraction with an algebraic expression on the top, on the bottom, or on both.
$\frac{2 x}{x^{2} + 3 x}$ and $\frac{5}{x + 3}$ are both algebraic fractions.
What is the first move when simplifying an algebraic fraction?
Fully factorise the top and the bottom, before looking for anything to cancel.
Writing $\frac{2 x}{x^{2} + 3 x}$ as $\frac{2 x}{x (x + 3)}$ turns the bottom into a product and makes the shared $x$ visible.
What kinds of thing can be cancelled from the top and bottom?
A common factor may be a single term or a whole bracket.
Cancelling gives $\frac{x (5 x - 1)}{4 x} = \frac{5 x - 1}{4}$ and $\frac{x (x + 2)}{(x + 2) (x - 1)} = \frac{x}{x - 1}$ .
True or False?
$\frac{6 x}{x + 1}$ can be simplified by cancelling the $x$ .
False.
A factor can only be cancelled when it divides every term, and the $x$ in $x + 1$ is not a factor of the $1$ .
The bottom will not factorise either, so $\frac{6 x}{x + 1}$ is already in its simplest form.
Complete the simplification by filling in the missing fraction.
$\frac{4 (y - 3)}{(y - 3) (y + 5)} =______$
The completed simplification is:
$\frac{4 (y - 3)}{(y - 3) (y + 5)} = \frac{4}{y + 5}$
The bracket $(y - 3)$ is a factor of both the top and the bottom, so it cancels completely.
How is $\frac{x^{2} - 5 x}{x^{2} - 2 x - 15}$ reduced to its simplest form?
Factorise the top as $x (x - 5)$ and the bottom as $(x + 3) (x - 5)$ .
Cancelling the shared bracket $(x - 5)$ leaves $\frac{x}{x + 3}$ .
Why must an algebraic fraction be factorised before anything is cancelled?
Cancelling removes a factor, meaning something the whole top and the whole bottom are multiplied by.
Until each side is written as a product, there is no way of telling which parts are factors and which are merely terms.
When is an algebraic fraction in its simplest form?
An algebraic fraction is in simplest form once the top and the bottom share no common factor at all.
Look again after cancelling once, because a second factorisation can sometimes reveal a further cancel.
For $\frac{1}{x + 2}$ and $\frac{1}{x + 5}$ what is the lowest common denominator?
Multiplying the two denominators gives $(x + 2) (x + 5)$ , and here that is the lowest one.
Each bracket has to appear in the denominator, so both are needed and neither can be left out.
True or False?
Multiplying the two denominators together always gives the lowest common denominator.
False.
Multiplying them always gives a common denominator, but not the lowest one when the two denominators already share something.
For $\frac{1}{x}$ and $\frac{1}{2 x}$ the lowest is $2 x$ rather than $2 x^{2}$ , just as $4$ rather than $8$ serves $\frac{1}{2}$ and $\frac{1}{4}$ .
What happens to the numerators when both fractions are put over the lowest common denominator?
Each numerator is multiplied by whatever its own denominator was multiplied by.
So $\frac{x}{x - 4} + \frac{1}{x + 2}$ becomes $\frac{x (x + 2) + (x - 4)}{(x - 4) (x + 2)}$ .
Why do minus signs need extra care when subtracting algebraic fractions?
The minus sign applies to the whole numerator that follows it, so that numerator has to be bracketed before it is expanded.
In $\frac{4}{x - 7} - \frac{3}{x}$ the top becomes $4 x - 3 (x - 7)$ , which expands to $4 x - 3 x + 21$ and not $4 x - 3 x - 21$ .
Complete the addition by filling in the missing numerator in its simplest form.
$\frac{2}{x + 5} + \frac{3}{x - 1} = \frac{______}{(x + 5) (x - 1)}$
The completed addition is:
$\frac{2}{x + 5} + \frac{3}{x - 1} = \frac{5 x + 13}{(x + 5) (x - 1)}$
The numerator comes from $2 (x - 1) + 3 (x + 5) = 2 x - 2 + 3 x + 15$ .
Why is a condition such as $x \neq - 5$ attached to an algebraic fraction?
The condition rules out the value that would make a denominator equal to zero, which is never allowed.
For $\frac{2}{x + 5}$ that value is $x = - 5$ , since the bottom would then be $0$ .
Why is it worth leaving the denominator in factorised form at the end?
A factorised top and bottom make any final common factor easy to spot, so a possible cancel is not missed.
Expanding $(x + 5) (x - 1)$ into $x^{2} + 4 x - 5$ hides whichever brackets were there.
In what order do you handle a product of two algebraic fractions?
Factorise both fractions, cancel any common factors, multiply the tops and the bottoms, then check once more for factorising and cancelling.
That last check matters, because a product can factorise in a way that neither of the original fractions did.
True or False?
A bracket on the top of one algebraic fraction can cancel with the same bracket on the bottom of the other.
True.
Once the two fractions are being multiplied, every top is on the top of the whole product and every bottom on the bottom.
So $\frac{y + 1}{5 (y - 3)} \times \frac{y - 3}{2}$ cancels $(y - 3)$ across the pair, leaving $\frac{y + 1}{10}$ .
Complete the factorised form so that a common bracket can be cancelled.
$\frac{x}{3 x + 6} \times \frac{2 x + 4}{x + 7} = \frac{x}{3 (______)} \times \frac{2 (______)}{x + 7}$
The completed factorisation is:
$\frac{x}{3 (x + 2)} \times \frac{2 (x + 2)}{x + 7}$
Cancelling $(x + 2)$ then leaves $\frac{x}{3} \times \frac{2}{x + 7} = \frac{2 x}{3 (x + 7)}$ .
Why does ${(x + 3)}^{2}$ on the top allow one $(x + 3)$ to be cancelled?
A squared bracket is two copies multiplied together, so ${(x + 3)}^{2} = (x + 3) (x + 3)$ .
One copy cancels with an $(x + 3)$ on the bottom and the other stays, so $\frac{5}{x + 3} \times \frac{{(x + 3)}^{2}}{6}$ becomes $\frac{5 (x + 3)}{6}$ .
How is a division of two algebraic fractions turned into something you can cancel?
Flip the second fraction and change the division into a multiplication, then factorise and cancel as usual.
So $\frac{3 x - 12}{x} \div \frac{2 x + 8}{x + 3}$ becomes $\frac{3 x - 12}{x} \times \frac{x + 3}{2 x + 8}$ .
Can the answer $\frac{5 (x + 3)}{6}$ be written in another way?
Yes, expanding the bracket gives $\frac{5 x + 15}{6}$ , and both forms are equally correct.
Neither is simpler than the other, since $5$ and $6$ share no factor and $x + 3$ does not divide $6$ .
What are the two ways of solving an equation containing algebraic fractions?
Either combine the fractions into one and then cross-multiply, or clear the fractions first by multiplying every term by each denominator.
Both routes reach the same equation, so use whichever is easier for the fractions in front of you.
When clearing fractions from an equation what must every denominator multiply?
Every single term on both sides has to be multiplied, not just the fractions themselves.
In $\frac{4}{x - 3} + \frac{5}{x + 1} = 5$ the $5$ on the right is multiplied too, becoming $5 (x - 3)$ at the first step.
In $\frac{5}{x - 4} + \frac{7}{x + 2} = 6$ every term has been multiplied by $(x - 4)$ so complete the result.
$______+ \frac{7 (x - 4)}{x + 2} = 6 (x - 4)$
The completed step is:
$5 + \frac{7 (x - 4)}{x + 2} = 6 (x - 4)$
The $(x - 4)$ cancels in the first term, while the other two terms simply gain a factor of $(x - 4)$ .
True or False?
Clearing the fractions from an equation with two algebraic fractions usually leaves a quadratic to solve.
True.
Multiplying by both denominators produces a product of two brackets on one side, which expands to give an $x^{2}$ term.
Clearing $\frac{5}{x - 4} + \frac{7}{x + 2} = 6$ leads to $x^{2} - 4 x - 5 = 0$ , giving $x = - 1$ and $x = 5$ .
Why must an expression you multiply through by be kept in brackets?
The whole expression multiplies each term, so without brackets only its first part would.
Multiplying $5$ by $x - 4$ must be written $5 (x - 4)$ , which is $5 x - 20$ rather than $5 x - 4$ .
How is $\frac{8}{x + 1} - \frac{5}{x + 2} = 1$ solved by combining the fractions first?
Subtracting gives $\frac{3 x + 11}{(x + 1) (x + 2)} = 1$ , and cross-multiplying then gives $3 x + 11 = (x + 1) (x + 2)$ .
Expanding and rearranging leaves $x^{2} - 9 = 0$ , so $x = 3$ or $x = - 3$ .
Do the two denominators have to be dealt with one at a time?
No, multiplying by both denominators at once reaches the same equation in fewer steps.
Taking them one at a time is easier to follow, because each cancelling can be seen as it happens.

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