Rearranging formulae: change the subject

To rearrange a formula is to write it in an equivalent form, usually with a different variable as the subject. The subject is the variable isolated on one side of the equals sign.

For example, in

$v=u+at,$

$v$ is the subject. Rearranging to

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

makes $t$ the subject. Both formulae describe the same relationship, provided $a\ne0$.

This skill appears throughout A-level Mathematics, including coordinate geometry, trigonometry, calculus, statistics and mechanics. The central principle is simple: preserve equality by performing valid operations on both sides.

Prerequisites

You should already be able to:

• use negative numbers and fractions accurately;
• simplify expressions and expand brackets;
• factorise by taking out a common factor;
• solve linear equations;
• use index laws and inverse operations.

Review algebraic manipulation and factorisation, linear equations or indices, roots and surds if needed.

The balance principle

An equation remains true when the same valid operation is applied to both sides. If $A=B$, then, for example,

$A+c=B+c,$

$cA=cB,$

and, when $c\ne0$,

$\frac{A}{c}=\frac{B}{c}.$

Terms do not literally move across the equals sign and change sign. That language is only shorthand for applying an inverse operation to both sides.

Worked example 1: addition and multiplication

Make $x$ the subject of

$y=5x-7.$

Undo the subtraction, then undo the multiplication:

$$\begin{aligned} y&=5x-7\\ y+7&=5x\\ \frac{y+7}{5}&=x. \end{aligned}$$

It is conventional to place the new subject on the left:

$\boxed{x=\frac{y+7}{5}}.$

The order matters. In $5x-7$, multiplication by $5$ happens before subtraction of $7$, so the inverse operations are applied in reverse order.

Worked example 2: the subject has a negative coefficient

Make $p$ the subject of

$q=8-3p.$

$$\begin{aligned} q-8&=-3p\\ p&=\frac{q-8}{-3} =\frac{8-q}{3}. \end{aligned}$$

The final form with a positive denominator is usually easier to read.

Self-check 1

Make the stated variable the subject.

1. $a=4b+9$, for $b$.
2. $m=7-5n$, for $n$.
3. $P=2l+2w$, for $w$.

Answers

1. $a-9=4b$, so $\displaystyle b=\frac{a-9}{4}$.
2. $m-7=-5n$, so $\displaystyle n=\frac{7-m}{5}$.
3. $P-2l=2w$, so $\displaystyle w=\frac{P-2l}{2}=\frac P2-l$.

Keep grouped expressions together

If the required subject lies inside a bracket or is the numerator of a fraction, first isolate the entire grouped expression. This avoids disturbing its internal structure too early.

Worked example 3: subject inside a bracket

Make $r$ the subject of

$A=\pi r(r+s).$

This example is not linear in $r$, because $r$ occurs twice and expanding would produce $r^2$. For now, consider the simpler grouped formula

$A=k(r+s).$

First divide by $k$, then subtract $s$:

$$\begin{aligned} \frac Ak&=r+s\\ r&=\frac Ak-s, \end{aligned} \qquad k\ne0.$$

The condition $k\ne0$ is necessary because the rearrangement divides by $k$.

Worked example 4: a fractional coefficient

Make $h$ the subject of the trapezium area formula

$A=\frac12(a+b)h.$

Treat $\frac12(a+b)$ as the coefficient of $h$:

$$\begin{aligned} 2A&=(a+b)h\\ h&=\frac{2A}{a+b}, \end{aligned} \qquad a+b\ne0.$$

Misconception: dividing only one term

From

$y=a+bx,$

subtracting $a$ gives $y-a=bx$. Dividing by $b$ gives

$x=\frac{y-a}{b},$

not $x=y-\frac ab$. The division applies to the whole numerator $y-a$.

The subject in a denominator

When the required variable is in a denominator, eliminate the fraction by multiplying both sides by that denominator. State any values excluded by division.

Worked example 5: a simple denominator

Make $t$ the subject of

$v=\frac dt.$

Since the original formula requires $t\ne0$,

$$\begin{aligned} vt&=d\\ t&=\frac dv, \end{aligned}$$

where this final form also requires $v\ne0$. If $d\ne0$, the original relationship already guarantees $v\ne0$.

Worked example 6: the subject is part of a denominator

Make $x$ the subject of

$y=\frac{3}{2x-5}.$

Multiply by $2x-5$:

$$\begin{aligned} y(2x-5)&=3\\ 2xy-5y&=3\\ 2xy&=3+5y\\ x&=\frac{3+5y}{2y}, \end{aligned}$$

with $y\ne0$. A shorter route keeps the bracket intact:

x=\frac12\left(5+\frac3y\right).$$ These final expressions are equivalent. ### Self-check 2 Make the stated variable the subject. 1. $I=\dfrac V R$, for $R$. 2. $q=\dfrac{k}{x+c}$, for $x$. 3. $f=\dfrac{uv}{u+v}$, for $v$. <details> <summary>Answers</summary> 1. $IR=V$, so $\displaystyle R=\frac VI$, where $I\ne0$. 2. $q(x+c)=k$, so $\displaystyle x=\frac kq-c$, where $q\ne0$. 3. $f(u+v)=uv$, so $fu+fv=uv$. Hence $fu=v(u-f)$ and $\displaystyle v=\frac{fu}{u-f}$, where $u\ne f$. </details> ## Roots and powers Squaring and taking a square root are inverse operations only when signs and domains are handled carefully. If $$x^2=A,$$ then algebraically $$x=\pm\sqrt A$$ for real $x$, with $A\ge0$. In a geometric context, a length may be restricted to the positive root. ### Worked example 7: square root formula Make $x$ the subject of $$y=\sqrt{3x+4}.$$ The principal square root is non-negative, so the original formula requires $y\ge0$. Square both sides:

\begin{aligned} y^2&=3x+4\ x&=\frac{y^2-4}{3}. \end{aligned}

The condition $y\ge0$ must remain. Without it, the rearranged equation would permit values that the original formula cannot produce. ### Worked example 8: squared subject Make $r$ the subject of $$S=4\pi r^2.$$

\begin{aligned} r^2&=\frac{S}{4\pi}\ r&=\pm\sqrt{\frac{S}{4\pi}}. \end{aligned}

If $r$ is a radius, then $r\ge0$, so context selects $$\boxed{r=\sqrt{\frac{S}{4\pi}}}.$$ ### Worked example 9: a general power Make $x$ the subject of $$y=a(x-b)^3+c.$$ Assuming $a\ne0$,

\begin{aligned} y-c&=a(x-b)^3\ \frac{y-c}{a}&=(x-b)^3\ \sqrt[3]{\frac{y-c}{a}}&=x-b\ x&=b+\sqrt[3]{\frac{y-c}{a}}. \end{aligned}

Unlike an even root, a real cube root does not introduce a $\pm$ choice. ## When the subject appears more than once If the required variable occurs in several terms, collect those terms on one side and **factorise**. The factorisation turns several occurrences into one multiplicative factor that can be isolated. ### Worked example 10: collect and factorise Make $x$ the subject of $$y=ax+bx+c.$$

\begin{aligned} y-c&=ax+bx\ y-c&=x(a+b)\ x&=\frac{y-c}{a+b}, \end{aligned}

provided $a+b\ne0$. ### Worked example 11: occurrences on both sides Make $x$ the subject of $$p(x+q)=rx+s.$$ Expand, then collect all $x$ terms on one side:

\begin{aligned} px+pq&=rx+s\ px-rx&=s-pq\ x(p-r)&=s-pq\ x&=\frac{s-pq}{p-r}, \end{aligned}

where $p\ne r$. ### Worked example 12: reciprocal sum Make $R_2$ the subject of $$\frac1R=\frac1{R_1}+\frac1{R_2}.$$ Isolate the term containing $R_2$:

\begin{aligned} \frac1{R_2}&=\frac1R-\frac1{R_1}\ &=\frac{R_1-R}{RR_1}. \end{aligned}

Now take reciprocals: $$\boxed{R_2=\frac{RR_1}{R_1-R}}.$$ The denominators in the original formula must be non-zero, and the final formula requires $R_1\ne R$. ### Misconception: dividing before factorising From $y=ax+bx$, it is not useful to divide one term by $a$ or $b$. First expose the common factor: $$y=x(a+b),$$ then divide once: $$x=\frac{y}{a+b}.$$ ## When a simple rearrangement is impossible Changing the subject does not always reduce to inverse operations and factorisation. If expanding produces a quadratic in the required subject, use quadratic methods. For example, $$A=\pi r(r+s)$$ becomes $$\pi r^2+\pi sr-A=0.$$ Regarding this as a quadratic in $r$, the quadratic formula gives $$r=\frac{-\pi s\pm\sqrt{\pi^2s^2+4\pi A}}{2\pi} =\frac{-s\pm\sqrt{s^2+4A/\pi}}{2}.$$ If $r$ is a length and the other quantities are positive, context determines the admissible root. See [quadratic equations](/learn/foundations/quadratic-equations/) for the required method. Some equations, such as $y=x+\sin x$, cannot be rearranged into a formula for $x$ using elementary algebraic operations. They may require numerical methods, which is a different task. ## Checking a rearrangement There are three useful checks. 1. **Reverse the algebra.** Starting from your answer, recover the original formula. 2. **Substitute numbers.** Choose permissible values for the letters and test both forms. 3. **Check dimensions or units.** Terms added together must have compatible units. For example, from $s=ut+\frac12at^2$, rearranging for $u$ gives $$u=\frac{s-\frac12at^2}{t},qquad t\ne0.$$ The numerator has units of length, and division by time gives velocity, as required. Also, $$ut=s-\frac12at^2 \quad\Longrightarrow\quad s=ut+\frac12at^2,$$ which recovers the original formula. ## Mixed self-check Make the stated variable the subject. 1. $C=\dfrac59(F-32)$, for $F$. 2. $E=\dfrac12mv^2$, for $v$, assuming $v\ge0$. 3. $y=\dfrac{ax+b}{cx+d}$, for $x$. 4. $T=2\pi\sqrt{\dfrac Lg}$, for $L$. 5. $k=\dfrac{x-a}{x+a}$, for $x$. <details> <summary>Answers</summary> 1. $\dfrac95C=F-32$, so $\displaystyle F=\frac95C+32$. 2. $v^2=\dfrac{2E}{m}$, so $\displaystyle v=\sqrt{\frac{2E}{m}}$, with $m\ne0$ and a non-negative radicand. 3. $y(cx+d)=ax+b$, so $ycx+yd=ax+b$. Hence $x(yc-a)=b-yd$ and $$\displaystyle x=\frac{b-yd}{yc-a},$$ provided $yc\ne a$. 4. $\dfrac{T}{2\pi}=\sqrt{\dfrac Lg}$. Squaring gives $\dfrac{T^2}{4\pi^2}=\dfrac Lg$, so $$\displaystyle L=\frac{gT^2}{4\pi^2}.$$ 5. $k(x+a)=x-a$, so $kx+ka=x-a$. Therefore $x(k-1)=-a(1+k)$ and $$\displaystyle x=\frac{a(1+k)}{1-k},$$ provided $k\ne1$. </details> ## A reliable strategy When changing the subject: 1. identify every occurrence of the required variable; 2. clear outer fractions or brackets where helpful; 3. collect all terms containing the variable on one side; 4. factorise if the variable appears more than once; 5. undo the remaining operations in reverse order; 6. state restrictions introduced by denominators, roots or context; 7. check by reversing the steps or substituting values. Next, apply these ideas when manipulating [algebraic fractions](/learn/foundations/algebraic-fractions/), working with [function notation](/learn/foundations/function-notation/), and using formulae in [units and compound measures](/learn/foundations/units-and-compound-measures/).