2.3 - Changing the subject of a formula

Rewrite a formula so the required letter stands alone. Start with a letter that appears once, including a squared quantity such as a radius. The Higher extensions develop cubic powers, roots inside formulae and letters appearing in more than one term.

Undoing operations

A circular design has area $49\pi\,\mathrm{cm}^2$ . How can you find its radius directly from its area? We will turn $A=\pi r^2$ into a radius formula. First, practise undoing simpler operations.

You already know

An equality stays true when you apply the same valid operation to both sides. Multiplication is undone by division by a nonzero number; addition is undone by subtraction.

The subject is the letter isolated on one side, expressed using the other quantities. In $v=u+at$ , the subject is $v$ . To make $t$ the subject, undo its operations in reverse order: remove the added $u$ , then the multiplier $a$ . A formula normally rearranges to another formula, rather than to a numerical answer.

Worked example 1: undo addition then multiplication

Question. Make $t$ the subject of $v=u+at$ , where $a\ne0$ .

Plan. Subtract $u$ from both sides, then divide by $a$ .

\begin{aligned} v-u&=at &&\text{subtract }u\\ \frac{v-u}{a}&=t &&\text{divide both sides by }a \end{aligned}

Why divide the whole difference? The entire expression $v-u$ equals $at$ .

Answer. $t=\frac{v-u}{a}$ , for $a\ne0$ .

Check. Substitution gives $u+a\frac{v-u}{a}=u+v-u=v$ .

Change only the sign of the $at$ term. Division must now undo a negative multiplier.

Worked example 2: a negative multiplier

Question. Make $t$ the subject of $v=u-at$ , where $a\ne0$ .

Plan. Subtract $u$ , then divide by $-a$ .

\begin{aligned} v-u&=-at &&\text{subtract }u\\ t&=\frac{v-u}{-a}=\frac{u-v}{a} &&\text{divide by }-a \end{aligned}

Why reverse the numerator's signs? Multiplying its numerator and denominator by $-1$ gives an equivalent fraction.

Answer. $t=\frac{u-v}{a}$ .

Check. $u-a\frac{u-v}{a}=u-(u-v)=v$ .

Fractions work the same way. From $C=\frac59(F-32)$ , multiply by $\frac95$ , then add $32$ , giving $F=\frac95C+32$ . This reverses the multiplication before undoing the subtraction inside the bracket. Neither this temperature conversion nor a rearrangement rule is supplied on the Foundation or Higher formulae sheet.

Watch out

Wrong: $t=v-\frac ua$ from $v=u+at$ . It divides only $u$ . Correct: $t=\frac{v-u}{a}$ , with the whole difference over $a$ .

Now replace the multiplier by a known number. Keep the same reverse-order reasoning.

Subjects in fractions

Clear a fraction by multiplying both sides by its denominator. A denominator must be nonzero. This turns a subject in a denominator into part of a product, which can then be isolated. There is no special fraction-rearrangement rule on either tier's formulae sheet.

Worked example 3: the subject is in a numerator

Question. Make $x$ the subject of $y=\frac{3x-4}{5}$ .

Plan. Clear the denominator before undoing the subtraction and multiplication.

\begin{aligned} 5y&=3x-4 &&\text{multiply both sides by }5\\ 5y+4&=3x &&\text{add }4\\ x&=\frac{5y+4}{3} &&\text{divide by }3 \end{aligned}

Why multiply by $5$ first? It removes the denominator from the whole numerator.

Answer. $x=\frac{5y+4}{3}$ .

Check. $\frac{3(5y+4)/3-4}{5}=\frac{5y}{5}=y$ .

Move the same subject-containing expression from numerator to denominator, taking the reciprocal of the original right side. Keep track of the new nonzero conditions.

Worked example 4: the subject is in a denominator

Question. Make $x$ the subject of $y=\frac{5}{3x-4}$ , where $y\ne0$ .

Plan. Multiply by $3x-4$ , then divide by $y$ .

\begin{aligned} y(3x-4)&=5 &&\text{clear the denominator}\\ 3x-4&=\frac5y &&\text{divide by nonzero }y\\ x&=\frac{5/y+4}{3} &&\text{add }4\text{, divide by }3 \end{aligned}

Why may we divide by $y$ ? A nonzero numerator $5$ means the original fraction cannot equal zero.

Answer. $x=\frac{5+4y}{3y}$ , for $y\ne0$ .

Check. This gives $3x-4=\frac5y\ne0$ , so the original denominator is valid and $\frac5{5/y}=y$ .

Another useful form is $P=\frac{k}{t+2}$ . For $k\ne0$ and $P\ne0$ , the steps $P(t+2)=k$ , then $t+2=\frac{k}{P}$ , give $t=\frac{k}{P}-2$ . It never gives the forbidden $t=-2$ , because $\frac{k}{P}\ne0$ . Do not impose positivity unless a physical meaning requires it.

Watch out

Wrong: from $y=\frac5{3x-4}$ , write $3xy-4=5$ . Multiplication by $y$ affects both terms. Correct: $y(3x-4)=5$ , then $3xy-4y=5$ .

Change the numerical coefficients in the denominator example. State the condition that makes your rearrangement valid.

A subject that is squared

If the required letter appears once but is squared, isolate its square, then take a square root. This includes the circle-area rearrangement at both tiers. The permitted sign comes from the variable's domain.

Undoing a square

z^2=q\quad\Longrightarrow\quad z=\pm\sqrt q\qquad(q>0)

If $z\ge0$ , choose $z=\sqrt q$ ; if $q=0$ , the only value is zero. There is no real solution when $q<0$ . Foundation and Higher: Not in the formulae sheet: learn it.

For example, $z^2=9$ has solutions $3$ and $-3$ , but a radius cannot be negative. The circle-area formula $A=\pi r^2$ is also absent from both tier sheets: learn it.

Worked example 5: a radius from area

Question. Make $r$ the subject of $A=\pi r^2$ , where $A\ge0$ and $r\ge0$ .

Plan. Divide by $\pi$ , then use the nonnegative root.

\begin{aligned} r^2&=\frac A\pi &&\text{divide by }\pi\\ r&=\sqrt{\frac A\pi} &&\text{radius is nonnegative} \end{aligned}

Why is there no negative root? The formula represents a physical radius.

Answer. $r=\sqrt{\frac A\pi}$ .

Check. At $A=9\pi$ , $r=\sqrt9=3$ , and $\pi(3)^2=9\pi$ , as required.

Replace only the coefficient $\pi$ by $5$ . The root must still cover the entire quotient.

Worked example 6: a different squared coefficient

Question. Make $s$ the subject of $A=5s^2$ , where $A>0$ and $s>0$ .

Plan. Isolate $s^2$ , then select the positive root.

\begin{aligned} s^2&=\frac A5 &&\text{divide by }5\\ s&=\sqrt{\frac A5} &&\text{take the positive root} \end{aligned}

Why not divide by $5$ after taking the root? The coefficient belongs inside the isolated quotient.

Answer. $s=\sqrt{\frac A5}$ .

Check. At $A=20$ , $s=2$ , and $5(2)^2=20$ .

For the opening design, $A=49\pi$ gives $r=\sqrt{49\pi/\pi}=7\,\mathrm{cm}$ . The square-root formula answers it directly.

Watch out

Wrong: $\sqrt{\frac A\pi}=\frac{\sqrt A}{\pi}$ . Squaring the proposed right side gives $A/\pi^2$ . Correct: $\sqrt{\frac A\pi}=\frac{\sqrt A}{\sqrt\pi}$ .

Change the multiplier again, retaining the positive-length condition.

Higher powers of the subject (Higher)

A sphere uses a cubic radius, so a square root will not undo it. Isolate the cube, then take a cube root. Unlike a square, a cube preserves its input's sign: $2^3=8$ and $(-2)^3=-8$ . Every real number has one real cube root.

Sphere volume

V=\frac43\pi r^3

For a sphere of radius $r\ge0$ , volume $V\ge0$ . Higher: In the formulae sheet. The rearranged radius formula is not supplied.

The power determines the inverse operation. Dividing by $3$ does not undo cubing: it only undoes multiplication by $3$ .

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Number

Algebra

Shape, space and measure

Handling data

2.3 - Changing the subject of a formula

Undoing operations

Subjects in fractions

A subject that is squared

Higher powers of the subject (Higher)

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