GCSE Maths · Algebra

Algebraic Fractions flashcards

11 cards on Algebraic Fractions: the key points, worked examples and the slips to watch. Swipe through them, or print them.

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Algebraic Fractions

Simplifying algebraic fractions

How do you simplify an algebraic fraction such as $\frac{x^{2} - 16}{x^{2} + 2 x - 8}$ ?

Answer it in your head, then turn the card over.

Simplifying algebraic fractions

Key points

Factorise the top and the bottom fully first.
Cancel any bracket that is a factor of both the top and the bottom.
What is left is the simplest form: leave it in brackets.

$\frac{x^{2} - 16}{x^{2} + 2 x - 8}$

Top: x² − 16 = (x + 4)(x − 4), a difference of two squares

Bottom: 4 and −2 multiply to −8 and add to 2, so (x + 4)(x − 2)

Cancel (x + 4): the answer is $\frac{x - 4}{x - 2}$

Check with x = 6: $\frac{20}{40}$ = $\frac{1}{2}$ , and $\frac{2}{4}$ = $\frac{1}{2}$

Watch out

Crossing out the x² terms is a slip: they are added to other terms, so factorise first.

Algebraic Fractions

Only factors cancel

Why can't you cancel the x's in $\frac{x + 3}{x + 5}$ to get $\frac{3}{5}$ ?

Answer it in your head, then turn the card over.

Only factors cancel

Key points

A factor multiplies the whole top or the whole bottom; a term is only part of a sum.
Only whole factors cancel, so $\frac{x + 3}{x + 5}$ is already as simple as it gets.
$\frac{2 x + 6}{x + 3}$ does simplify: 2x + 6 = 2(x + 3), so it cancels to 2.

Test it with a number

x = 1: $\frac{1 + 3}{1 + 5}$ = $\frac{4}{6}$ = $\frac{2}{3}$

That is not $\frac{3}{5}$ , so cancelling the x's changed the fraction.

Watch out

$\frac{x}{x + 5}$ does not cancel to $\frac{1}{5}$ either: the x on the bottom is only one term of x + 5.

Algebraic Fractions

Spotting what to factorise

Which factorisations should you try on the top and bottom of an algebraic fraction?

Answer it in your head, then turn the card over.

Spotting what to factorise

Key points

A common factor first: 5x − 10 = 5(x − 2).
A quadratic x² + bx + c: two numbers that multiply to c and add to b.
A difference of two squares: x² − 25 = (x + 5)(x − 5).

$\frac{5 x - 10}{x^{2} - 4}$

= $\frac{5 (x - 2)}{(x + 2) (x - 2)}$

Cancel (x − 2): the answer is $\frac{5}{x + 2}$

Check with x = 4: $\frac{10}{12}$ = $\frac{5}{6}$ , and $\frac{5}{4 + 2}$ = $\frac{5}{6}$

Watch out

x² + 16 is not a difference of two squares and does not factorise: look for the minus sign.

Algebraic Fractions

Multiplying algebraic fractions

How do you multiply two algebraic fractions?

Answer it in your head, then turn the card over.

Algebraic Fractions

Dividing algebraic fractions

How do you divide one algebraic fraction by another?

Answer it in your head, then turn the card over.

Dividing algebraic fractions

Key points

Keep the first fraction, change ÷ to ×, and flip the second one.
Then factorise, cancel and multiply, exactly as for multiplying.
It is the same rule as for number fractions: $\frac{3}{4}$ ÷ $\frac{1}{2}$ = $\frac{3}{4}$ × $\frac{2}{1}$ .

$\frac{x^{2} - 1}{x + 3}$ ÷ $\frac{x - 1}{2 x + 6}$

= $\frac{x^{2} - 1}{x + 3}$ × $\frac{2 x + 6}{x - 1}$

= $\frac{(x + 1) (x - 1)}{x + 3}$ × $\frac{2 (x + 3)}{x - 1}$

Cancel (x − 1) and (x + 3): the answer is 2(x + 1)

Check with x = 2: $\frac{3}{5}$ ÷ $\frac{1}{10}$ = 6, and 2 × 3 = 6

Watch out

Flip the second fraction, the one after the ÷ sign, never the first.

Algebraic Fractions

Adding over a number denominator

How do you add algebraic fractions such as $\frac{x + 1}{3}$ + $\frac{x - 2}{4}$ ?

Answer it in your head, then turn the card over.

Adding over a number denominator

Key points

Use the lowest common multiple of the bottoms as the common denominator.
Multiply each top by whatever its bottom was multiplied by.
Put each new top in a bracket, expand, and collect like terms.

$\frac{x + 1}{3}$ + $\frac{x - 2}{4}$

Over 12: $\frac{4 (x + 1)}{12}$ + $\frac{3 (x - 2)}{12}$

Tops: 4x + 4 + 3x − 6 = 7x − 2

So the answer is $\frac{7 x - 2}{12}$

Check with x = 2: $\frac{3}{3}$ + $\frac{0}{4}$ = 1, and $\frac{12}{12}$ = 1

Watch out

Multiply the whole top, not just its first term: 4(x + 1) is 4x + 4, not 4x + 1.

Algebraic Fractions

Adding over algebraic denominators

How do you add two fractions whose bottoms are different brackets, like $\frac{4}{x + 3}$ + $\frac{2}{x - 1}$ ?

Answer it in your head, then turn the card over.

Adding over algebraic denominators

Key points

Multiply the two bottoms to get the common denominator.
Multiply each top by the other fraction's bottom.
Add the tops, expand and collect; leave the bottom factorised.

$\frac{4}{x + 3}$ + $\frac{2}{x - 1}$

Over (x + 3)(x − 1): tops 4(x − 1) + 2(x + 3)

= 4x − 4 + 2x + 6 = 6x + 2

So the answer is $\frac{6 x + 2}{(x + 3) (x - 1)}$

Check with x = 2: $\frac{4}{5}$ + $\frac{2}{1}$ = $\frac{14}{5}$ , and $\frac{14}{5 \times 1}$ = $\frac{14}{5}$

Watch out

Never add tops and bottoms separately: $\frac{2}{x}$ + $\frac{3}{y}$ is $\frac{2 y + 3 x}{x y}$ , not $\frac{5}{x + y}$ .

Algebraic Fractions

Subtracting algebraic fractions

When you subtract algebraic fractions, why does the second top go in a bracket?

Answer it in your head, then turn the card over.

Subtracting algebraic fractions

Key points

The minus sign takes away every term of the second top, not just the first.
Keep the bracket until you expand: − (x − 2) = − x + 2, because taking away −2 adds 2.
Then collect like terms, as when adding.

$\frac{3}{x - 1}$ − $\frac{2}{x + 4}$

Over (x − 1)(x + 4): tops 3(x + 4) − 2(x − 1)

= 3x + 12 − 2x + 2 = x + 14

So the answer is $\frac{x + 14}{(x - 1) (x + 4)}$

Check with x = 2: $\frac{3}{1}$ − $\frac{2}{6}$ = $\frac{8}{3}$ , and $\frac{16}{6}$ = $\frac{8}{3}$

Watch out

− 2(x − 1) is − 2x + 2, not − 2x − 2: the minus reaches the last term too.

Algebraic Fractions

Solving with algebraic fractions

How do you solve an equation with a fraction in it, such as $\frac{5}{x}$ + 2 = x − 2?

Answer it in your head, then turn the card over.

Solving with algebraic fractions

Key points

Multiply every term on both sides by the denominator to clear the fraction.
Expand and rearrange: it is often a quadratic, so get everything on one side, = 0.
Solve, then check each answer in the original equation.

$\frac{5}{x}$ + 2 = x − 2

× x: 5 + 2x = x² − 2x

Rearrange: x² − 4x − 5 = 0

(x − 5)(x + 1) = 0, so x = 5 or x = −1

Check x = 5: $\frac{5}{5}$ + 2 = 3, and 5 − 2 = 3

Check x = −1: −5 + 2 = −3, and −1 − 2 = −3

Watch out

Multiply every term, not just the fraction: $\frac{5}{x}$ + 2 = x − 2 does not become 5 + 2 = x − 2.

Algebraic Fractions

Clearing two denominators

What do you multiply by to clear an equation like $\frac{2}{x}$ + $\frac{1}{x - 1}$ = 2?

Answer it in your head, then turn the card over.

Clearing two denominators

Key points

Multiply every term by both denominators together: here x(x − 1).
Each fraction loses its own bottom; a term with no fraction is multiplied by both.
Then expand, rearrange to = 0, and solve.

$\frac{2}{x}$ + $\frac{1}{x - 1}$ = 2

× x(x − 1): 2(x − 1) + x = 2x(x − 1)

3x − 2 = 2x² − 2x, so 2x² − 5x + 2 = 0

(2x − 1)(x − 2) = 0, so x = $\frac{1}{2}$ or x = 2

Check x = 2: $\frac{2}{2}$ + $\frac{1}{1}$ = 2

Check x = $\frac{1}{2}$ : 4 + (−2) = 2

Watch out

The 2 on the right is a term too: it has to be multiplied by x(x − 1) as well.

Algebraic Fractions

Roots that make a denominator zero

Why must you check each root against the denominators in the original equation?

Answer it in your head, then turn the card over.

Roots that make a denominator zero

Key points

You cannot divide by zero, so a root that makes an original denominator 0 is not a solution.
Multiplying through by a bracket can bring in a root like that.
Put each root into every original bottom, and discard any that gives 0.

$\frac{x^{2}}{x - 2}$ = $\frac{4}{x - 2}$

× (x − 2): x² = 4, so x = 2 or x = −2

x = 2 makes x − 2 = 0, and $\frac{4}{0}$ has no value: discard it

x = −2: both sides are $\frac{4}{- 4}$ = −1, so x = −2 is the only solution

Watch out

Giving both roots without checking: x = 2 looks fine until you put it into x − 2.

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