A5 - Rearranging formulae

A formula can be rewritten to calculate a different quantity. Learn to reverse operations, manage brackets and fractions, handle powers and roots, and collect a subject that appears more than once.

Changing the subject with inverse operations

A machine multiplies an input by four, then adds seven. If its output is $31$ , how can you recover the input, and write a rule that works for any output?

Undo the last operation first: $(31-7)\div4=6$ . For an arbitrary output $y$ , the same route gives $x=\frac{y-7}{4}$ . Rearranging writes this reverse calculation as a formula.

You already know

An equation stays true when you apply the same operation to both complete sides, provided any divisor is non-zero. Substitution can check a result; valid algebra explains why it works generally.

The letter you isolate has a special name.

Subject of a formula

The subject is the quantity isolated on one side, with no occurrences of it on the other side. In $v=u+at$ , the subject is $v$ ; in $u=v-at$ , it is $u$ .

Changing the subject uses inverse operations in reverse order. This method is not on the formula sheet: learn it.

Worked example 1: undoing addition then multiplication

Question. Make $x$ the subject of $y=4x+7$ .

Plan. Subtract seven from both sides, then divide both sides by four.

\begin{aligned} y-7&=4x &&\text{undo the final addition}\\ x&=\frac{y-7}{4} &&\text{undo multiplication by four} \end{aligned}

Why this step? The original calculation multiplies first and adds last, so the reverse calculation subtracts first and divides last.

Answer. $x=\frac{y-7}{4}$ .

Check. For $x=2$ , the original gives $y=15$ ; $(15-7)/4=2$ .

Change only the sign of the constant. Undo subtraction by adding.

Worked example 2: undoing a subtraction

Question. Make $x$ the subject of $y=4x-7$ .

Plan. Add seven to both sides before dividing by four.

\begin{aligned} y+7&=4x &&\text{undo subtraction}\\ x&=\frac{y+7}{4} &&\text{divide the complete total} \end{aligned}

Why this step? Both $y$ and the added $7$ belong in the numerator because the whole total equals $4x$ .

Answer. $x=\frac{y+7}{4}$ .

Check. For $x=2$ , $y=1$ , and $(1+7)/4=2$ .

Letters follow the same rules as numbers. From the supplied velocity relation $v=u+at$ , subtract $u$ , then divide by $a$ : $t=\frac{v-u}{a}$ , for $a\ne0$ . This scientific relation is provided in the question when needed; it is not on the formula sheet.

For rectangle area $A=lw$ , divide by $l$ to get $w=A/l$ , with $l\ne0$ . The rectangle area formula is not on the formula sheet: learn it. If $A=30\,\mathrm{cm}^2$ and $l=6\,\mathrm{cm}$ , the width is $5\,\mathrm{cm}$ ; area divided by length gives length units.

Watch out

Wrong: $y=4x+7$ gives $x=y/4-7$ . That divides too early. Right: subtract $7$ first, giving $x=(y-7)/4$ .

Change only the constant in Worked example 2.

Rearranging brackets and fractions

Brackets and fraction bars group terms, so inverse operations must act on whole groups.

For $P=2(l+w)$ , divide first to get $P/2=l+w$ , then subtract $w$ : $l=P/2-w$ . Expanding first also works and gives $l=(P-2w)/2$ . Both routes preserve equality. The rectangle perimeter formula is not on the formula sheet: learn it.

Worked example 3: clearing a numerical denominator

Question. Make $x$ the subject of $y=\frac{3x-2}{5}$ .

Plan. Multiply by five, add two, then divide by three.

\begin{aligned} 5y&=3x-2 &&\text{undo division by five}\\ 5y+2&=3x &&\text{undo subtraction}\\ x&=\frac{5y+2}{3} &&\text{undo multiplication by three} \end{aligned}

Why this step? The denominator divides the entire numerator, so multiplying by five removes that division from the complete group.

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

Check. At $x=4$ , the original gives $y=2$ , and $(5(2)+2)/3=4$ .

Change only the sign of the numerator's constant.

Worked example 4: preserving the grouped numerator

Question. Make $x$ the subject of $y=\frac{3x+2}{5}$ .

Plan. Clear the denominator, then subtract two before dividing by three.

\begin{aligned} 5y&=3x+2 &&\text{multiply both sides by five}\\ 5y-2&=3x &&\text{subtract two}\\ x&=\frac{5y-2}{3} &&\text{divide by three} \end{aligned}

Why this step? The inverse of adding $2$ is subtracting $2$ ; the fraction bar does not change that sign rule.

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

Check. At $x=1$ , $y=1$ , and $(5-2)/3=1$ .

A fractional coefficient can be removed by its reciprocal. From triangle area $A=\frac12bh$ , multiply by $2$ , then divide by $b$ : $h=\frac{2A}{b}$ , with $b\ne0$ . This triangle area formula is not on the formula sheet: learn it.

If the desired subject is in a denominator, clear that complete denominator first. The next pair shows why this differs from dividing by a fixed number.

Worked example 5: a subject in the denominator

Question. Make $x$ the subject of $y=\frac4{x-1}$ .

Plan. Multiply by $x-1$ , then divide by $y$ and add one.

\begin{aligned} y(x-1)&=4 &&\text{clear the original denominator}\\ x-1&=\frac4y &&\text{divide by non-zero }y\\ x&=1+\frac4y &&\text{add one} \end{aligned}

Why this step? Multiplying by $x-1$ removes the fraction without splitting its denominator into separate terms.

Answer. $x=1+\frac4y$ , where $y\ne0$ ; the original requires $x\ne1$ .

Check. At $x=3$ , $y=2$ , and $1+4/2=3$ .

Change only the denominator's constant sign.

Worked example 6: reversing the denominator shift

Question. Make $x$ the subject of $y=\frac4{x+1}$ .

Plan. Clear $x+1$ , divide by $y$ , then subtract one.

\begin{aligned} y(x+1)&=4 &&\text{multiply by the whole denominator}\\ x+1&=\frac4y &&\text{divide by non-zero }y\\ x&=\frac4y-1 &&\text{subtract one} \end{aligned}

Why this step? The $+1$ is part of the original denominator; it is undone only after that denominator has been isolated.

Answer. $x=\frac4y-1$ , for $y\ne0$ ; originally $x\ne-1$ .

Check. At $x=1$ , $y=2$ , and $4/2-1=1$ .

These outputs cannot be zero because a non-zero numerator divided by a permitted denominator is non-zero. Restrictions explain the formula's behaviour; they are not just extra symbols.

Watch out

Wrong: $y=\frac{3x-2}{5}$ becomes $y=3x-10$ . Multiplication must act on both whole sides: the correct first line is $5y=3x-2$ .

Change only the numerator in Worked example 6.

Rearranging powers and roots

Squaring loses the sign of an input: $3^2=(-3)^2=9$ . Undoing a square therefore needs attention to the permitted values.

The symbol $\sqrt y$ means the non-negative square root. For $y=x^2$ , write $x=\pm\sqrt y$ , with $y\ge0$ , unless a condition selects one sign. At $y=0$ , the two written signs give the same value.

Undoing a square

x^2=k\quad\Longrightarrow\quad x=\pm\sqrt k\qquad(k\ge0)

If $x\ge0$ is required, retain only the non-negative root. If $k<0$ , there is no real solution.

Not on the formula sheet: derive it from the definition of square root.

Isolate the square before taking its root. Taking a root of only one term in a sum changes the relationship.

Worked example 7: retaining both signs

Question. Make $x$ the subject of $y=3x^2+2$ , where $x$ can be any real number.

Plan. Remove the addition and coefficient, then take both square roots.

\begin{aligned} y-2&=3x^2 &&\text{subtract two}\\ x^2&=\frac{y-2}{3} &&\text{divide by three}\\ x&=\pm\sqrt{\frac{y-2}{3}} &&\text{take both roots} \end{aligned}

Why this step? Both signs square to the same non-negative quotient.

Answer. $x=\pm\sqrt{\frac{y-2}{3}}$ , for $y\ge2$ .

Check. For $y=14$ , the roots are $2$ and $-2$ ; each gives $3(4)+2=14$ .

Change only the permitted domain: require $x\ge0$ . The algebra up to the square stays the same, but the final sign is selected.

Worked example 8: a condition choosing the sign

Question. Make $x$ the subject of $y=3x^2+2$ , where $x\ge0$ .

Plan. Isolate $x^2$ and take only the non-negative root.

\begin{aligned} x^2&=\frac{y-2}{3} &&\text{subtract two and divide by three}\\ x&=\sqrt{\frac{y-2}{3}} &&\text{use }x\ge0 \end{aligned}

Why this step? The negative root satisfies the squared equation but fails the stated condition when $y>2$ .

Answer. $x=\sqrt{\frac{y-2}{3}}$ , for $y\ge2$ .

Check. At $y=14$ , the answer is $2$ , which is permitted; $-2$ is excluded by the condition.

A radius gives a familiar sign condition. Circle area $A=\pi r^2$ gives $r=\sqrt{A/\pi}$ , since a radius is non-negative. For $A=49\pi\,\mathrm{cm}^2$ , $r=7\,\mathrm{cm}$ . The circle area formula is on the formula sheet (2026 version; the 2027 sheet will have the same content).

When the formula already contains a root, square the whole side: $y=\sqrt{x+5}$ gives $y^2=x+5$ , then $x=y^2-5$ . Keep $y\ge0$ and $x\ge-5$ from the original formula. A negative $y$ might fit the squared equation but cannot be the output of the original square root.

Odd powers preserve signs: $V=s^3$ gives $s=\sqrt[3]{V}$ , without a plus-or-minus pair. In a cube-volume context, side $s$ and volume $V$ are non-negative. The cube-volume formula is not on the formula sheet: learn it.

Watch out

Wrong: $y=3x^2+2$ gives $x=\sqrt{y/3}+2$ . Isolate the complete square first. The correct radicand is $(y-2)/3$ , with both signs unless the domain selects one.

Change only the additive constant in Worked example 8.

Collecting a repeated subject

When the desired subject occurs more than once, collect its terms and factor it out before dividing.

This is the common-factor method used in reverse to isolate a letter. It is not on the formula sheet: learn it. Dividing after just one occurrence usually leaves the subject somewhere else in the formula.

Worked example 9: factoring a repeated letter

Question. Make $x$ the subject of $p=3x+qx$ .

Plan. Factor $x$ from both terms, then divide by the complete coefficient.

\begin{aligned} p&=x(3+q) &&\text{take out the shared }x\\ x&=\frac p{3+q} &&\text{divide by }3+q \end{aligned}

Why this step? Factoring turns two occurrences into one occurrence multiplied by a bracket.

Answer. $x=\frac p{3+q}$ , provided $q\ne-3$ .

Check. For $q=2$ , $x=4$ , the original gives $p=20$ ; $20/(3+2)=4$ .

Change only the left side by adding another term containing $x$ . Collect that term before factorising.

Worked example 10: a subject on both sides

Question. Make $x$ the subject of $p+cx=3x+qx$ .

Plan. Subtract $cx$ , factor the resulting right side, then divide.

\begin{aligned} p&=3x+qx-cx &&\text{collect the subject on one side}\\ &=x(3+q-c) &&\text{extract the common factor}\\ x&=\frac p{3+q-c} &&\text{divide by its coefficient} \end{aligned}

Why this step? Moving all the $x$ terms together allows one shared factor, with no $x$ left on the other side.

Answer. $x=\frac p{3+q-c}$ , provided $3+q-c\ne0$ .

Check. For $q=2$ , $c=1$ , $x=4$ , the relation is $p+4=20$ , so $p=16$ ; the answer gives $16/4=4$ .

The same steps give $x=\frac{y-b}{a-c}$ from $y+cx=ax+b$ , where $a-c\ne0$ . Subtract $b$ and $cx$ , then factor $ax-cx=x(a-c)$ .

A zero coefficient needs separate thought. In $p=x(3+q)$ , setting $q=-3$ gives $p=0$ . If $p=0$ , every real $x$ works; if $p\ne0$ , no $x$ works. The divided formula covers neither case. This is why you state the non-zero divisor.

Watch out

Wrong: $p=3x+qx$ gives $x=p/3-qx$ . Dividing by $3$ has not isolated $x$ , and every term would need dividing. Factor first to obtain $x=p/(3+q)$ .

Change only the numerical coefficient in Worked example 10.

A repeated subject in algebraic fractions (Higher)

A subject in both a numerator and denominator needs the denominator cleared before its terms can be collected.

Multiply by the complete denominator, expand, gather every subject term on one side and factorise. This combines the rearrangement methods already taught; it is not a separate formula on the formula sheet.

Worked example 11: a repeated subject across a fraction

Question. Make $x$ the subject of $r=\frac{x+a}{x+b}$ , where $x\ne-b$ and $r\ne1$ .

Plan. Clear $x+b$ , then collect and factor the $x$ terms.

\begin{aligned} r(x+b)&=x+a &&\text{multiply by the whole denominator}\\ rx+rb&=x+a &&\text{expand}\\ rx-x&=a-rb &&\text{collect the subject terms}\\ x(r-1)&=a-rb &&\text{factor out }x\\ x&=\frac{a-rb}{r-1} &&\text{divide by non-zero }r-1 \end{aligned}

Why this step? Factoring is possible only after both occurrences of $x$ are on the same side.

Answer. $x=\frac{a-rb}{r-1}$ , for $r\ne1$ , with the original exclusion $x\ne-b$ .

Check. For $a=1$ , $b=4$ , $x=2$ , the original gives $r=1/2$ ; the new formula gives $(1-2)/(-1/2)=2$ .

Change only the coefficient of $x$ in the numerator from $1$ to $2$ . The final divisor changes accordingly.

Worked example 12: a changed numerator coefficient

Question. Make $x$ the subject of $r=\frac{2x+a}{x+b}$ , where $x\ne-b$ and $r\ne2$ .

Plan. Clear the denominator and factor the two resulting subject terms.

\begin{aligned} r(x+b)&=2x+a &&\text{clear the denominator}\\ rx+rb&=2x+a &&\text{expand}\\ rx-2x&=a-rb &&\text{collect }x\text{ terms}\\ x(r-2)&=a-rb &&\text{factorise}\\ x&=\frac{a-rb}{r-2} &&\text{divide by the complete coefficient} \end{aligned}

Why this step? The numerator's $2x$ makes the collected coefficient $r-2$ , rather than $r-1$ .

Answer. $x=\frac{a-rb}{r-2}$ , for $r\ne2$ , retaining $x\ne-b$ .

Check. For $a=1$ , $b=4$ , $x=2$ , $r=5/6$ ; $(1-4(5/6))/(5/6-2)=2$ .

For numerical coefficients, the same chain gives

z=\frac{2t-1}{t+4} \quad\Longrightarrow\quad zt+4z=2t-1 \quad\Longrightarrow\quad t=\frac{-1-4z}{z-2}=\frac{1+4z}{2-z}.

Here $t\ne-4$ and $z\ne2$ . For $t=1$ , $z=1/5$ , and the rearranged expression returns $1$ .

Exceptional parameter values must be checked in the original formula. In Worked example 11, $r=1$ reduces it to $a=b$ . If $a=b$ , every original permitted $x$ gives $r=1$ ; if $a\ne b$ , no $x$ does. Even when $r\ne1$ , a candidate result equal to $-b$ is forbidden. Keeping the original restriction catches such cases.

Watch out

Wrong: $r(x+b)=rx+b$ . Distribute $r$ to both terms: $r(x+b)=rx+rb$ . A missing parameter product changes the whole rearrangement.

Change only the numerator's $x$ coefficient once more.

Quick Recap

Apply inverse operations to both complete sides, in reverse order.
Brackets and fraction bars group terms; clear whole denominators.
Isolate powers before taking roots. Keep signs and domain restrictions from the original formula.
Gather repeated subject terms, factor out the subject and divide by a non-zero coefficient.

For the opening machine, $x=(y-7)/4$ . Output $31$ therefore came from input $6$ .

Last Accessed

Papers 1, 2 and 3

A5 - Rearranging formulae

Changing the subject with inverse operations

Rearranging brackets and fractions

Rearranging powers and roots

Collecting a repeated subject

A repeated subject in algebraic fractions (Higher)

Quick Recap

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