How do you change the subject of a formula?

Remove any fractions first, then use inverse operations to get the new subject on its own. It works exactly like solving an equation, except that the answer is an expression in the other letters rather than a number.

National 5MathsSQAFlashcardsAlgebraic SkillsChanging the Subject

Changing the Subject (SQA National 5 Maths): Flashcards

Exam code: X847 75

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Define the subject of a formula.

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Cards in this collection (14)

Define the subject of a formula.
The subject is the single letter standing on its own on one side of the equals sign.
In $C = 2 π r$ the subject is $C$ , and rearranging so that $r$ stands alone would change the subject to $r$ .
How do you change the subject of a formula?
Remove any fractions first, then use inverse operations to get the new subject on its own.
It works exactly like solving an equation, except that the answer is an expression in the other letters rather than a number.
When rearranging a formula should you always expand the brackets?
Only expand if the new subject is inside the bracket and needs releasing.
To make $x$ the subject of $3 (1 + x) = y$ you expand, but for $(1 + k) x = y$ you divide by the whole bracket instead, giving $x = \frac{y}{1 + k}$ .
Complete the rearrangement that makes $x$ the subject.
$\frac{5 x + 6}{2} = y \Rightarrow 5 x =______\Rightarrow x =______$
The completed rearrangement is:
$\frac{5 x + 6}{2} = y \Rightarrow 5 x = 2 y - 6 \Rightarrow x = \frac{2 y - 6}{5}$
Multiplying both sides by $2$ clears the fraction, and subtracting $6$ then dividing by $5$ releases the $x$ .
True or False?
$\frac{a}{- b}$ , $\frac{- a}{b}$ and $- \frac{a}{b}$ all mean the same thing.
True.
A single minus sign gives the same negative value wherever it is placed in a fraction.
Take care with $\frac{- a}{- b}$ though, since two minus signs cancel to give $\frac{a}{b}$ .
How do you deal with a fraction that has another fraction inside it?
Rewrite it as a division and use the rule for dividing fractions, or multiply the top and bottom by the lowest common denominator and cancel.
So $\frac{\frac{3}{t}}{2}$ is $\frac{3}{t} \div 2 = \frac{3}{t} \times \frac{1}{2} = \frac{3}{2 t}$ .
What changes when the denominator of a formula is a letter rather than a number?
Nothing changes, since you multiply both sides by that letter exactly as you would by a number.
Making $a$ the subject of $c = \frac{3 a + 7}{b}$ starts with $b c = 3 a + 7$ .
Which operation undoes squaring, and which undoes a square root?
Squaring and taking a square root are inverse operations, so each one undoes the other.
To release $v$ from $v^{2}$ you take a square root, and to release $l$ from $\sqrt{\frac{l}{g}}$ you square.
True or False?
Taking the square root of both sides of a formula gives a single answer.
False.
Both a positive and a negative number square to give the same positive value, so the square root gives $\pm$ .
Rearranging $E = \frac{1}{2} m v^{2}$ gives $v = \pm \sqrt{\frac{2 E}{m}}$ .
How do you decide which of the two square roots to keep?
The situation the formula describes decides it, so a quantity that cannot be negative takes the positive root.
Rearranging $A = π r^{2}$ gives $r = \pm \sqrt{\frac{A}{π}}$ , but a radius cannot be negative, so $r = \sqrt{\frac{A}{π}}$ .
Does every square in a formula have to be removed when changing the subject?
Only remove a square or a square root when it is holding the new subject.
Making $m$ the subject of $E = \frac{1}{2} m v^{2}$ leaves the $v^{2}$ alone and simply divides by it, giving $m = \frac{2 E}{v^{2}}$ .
Complete the rearrangement that makes $x$ the subject.
$y = a \sqrt{x} + b \Rightarrow \sqrt{x} =______\Rightarrow x =______$
The completed rearrangement is:
$y = a \sqrt{x} + b \Rightarrow \sqrt{x} = \frac{y - b}{a} \Rightarrow x = {(\frac{y - b}{a})}^{2}$
Subtracting $b$ and dividing by $a$ leaves the root alone, and squaring both sides then releases the $x$ .
Why must the whole of one side be bracketed before it is squared?
Squaring applies to everything on that side, so leaving off the brackets would square only part of it.
Squaring $\frac{p + q}{3}$ must be written ${(\frac{p + q}{3})}^{2}$ , which is not the same as $\frac{p^{2} + q^{2}}{3}$ .
How is $l$ made the subject when $T = 2 π \sqrt{\frac{l}{g}}$ is given?
Divide by $2 π$ to leave the root alone, square both sides, then multiply by $g$ .
That gives $l = g {(\frac{T}{2 π})}^{2}$ , which can also be written $l = \frac{g T^{2}}{4 π^{2}}$ .

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