Define the subject of a formula.

The subject is the variable you are trying to get on its own on one side of the formula. Rearranging a formula is therefore also called changing the subject .

What is the first thing to do when rearranging a formula containing a fraction?

Multiply both sides by whatever is on the denominator, so that the fraction disappears. After that, use inverse operations to isolate the subject, exactly as when solving a linear equation.

What must you always do when the subject appears twice in a formula?

Factorise , putting the subject outside a bracket so that it appears only once. Before that you may need to expand brackets to release it, and to bring all of its terms to the same side.

GCSEFurther MathsAQAFlashcards2. Algebra & FunctionsRearranging Formulae

Rearranging Formulae (AQA GCSE Further Maths): Flashcards

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FrontRearranging Formulae
Define the subject of a formula.

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Cards in this collection (8)

Define the subject of a formula.
The subject is the variable you are trying to get on its own on one side of the formula.
Rearranging a formula is therefore also called changing the subject.
What is the first thing to do when rearranging a formula containing a fraction?
Multiply both sides by whatever is on the denominator, so that the fraction disappears.
After that, use inverse operations to isolate the subject, exactly as when solving a linear equation.
When making $h$ the subject of $A = \frac{(a + b) h}{2}$ , why should you not expand the bracket?
Because expanding would give $a h + b h$ , putting $h$ in two places and making the rearrangement harder than it needs to be.
Divide by the whole bracket instead, which gives $h = \frac{2 A}{a + b}$ in a single step.
Why must you write $\pm$ when making $x$ the subject of $y = x^{2}$ ?
Because both $\sqrt{y}$ and $- \sqrt{y}$ square to give $y$ , so $x = \pm \sqrt{y}$ .
Squaring loses the sign of $x$ , and the plus-or-minus is what puts it back.
True or False?
Making $x$ the subject of $y = a x^{5}$ gives only one answer.
True.
The fifth root of a number is unique, so $x = \sqrt[5]{\frac{y}{a}}$ and no plus-or-minus is needed.
Plus-or-minus is only required when undoing an even power.
Make $a$ the subject of $m = \sqrt[3]{2 a b}$ .
Cube both sides to undo the cube root, which gives $m^{3} = 2 a b$ .
Then divide by $2 b$ , so $a = \frac{m^{3}}{2 b}$ .
What must you always do when the subject appears twice in a formula?
Factorise, putting the subject outside a bracket so that it appears only once.
Before that you may need to expand brackets to release it, and to bring all of its terms to the same side.
Rearrange $p = \frac{2 - a x}{x - b}$ to make $x$ the subject.
Multiplying by $(x - b)$ and expanding gives $p x - p b = 2 - a x$ .
Collecting the $x$ terms gives $x (p + a) = 2 + p b$ , so $x = \frac{2 + p b}{p + a}$ .

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