A5 - Using and rearranging formulae | AQA GCSE Mathematics

A formula describes a relationship between quantities. You will read formulae in words and symbols, use them with values, and rearrange them to make a different letter the subject. Doing the same thing to both sides lets one relationship answer several different questions.

Reading and using formulae

A journey costs a fixed £4 plus £1.80 per kilometre. The formula is $C=4+1.8d$ , where $C$ is the cost in pounds and $d$ is the distance in kilometres. Can it tell you how far a budget of £22 will take you? Keep that question: changing the subject will make it easy.

You already know

Substitution replaces each letter by its value. Brackets, powers, multiplication and division come before addition and subtraction. In $3t$ , the number $3$ multiplies $t$ .

A word formula and a symbol formula can say exactly the same thing. “Distance equals average speed multiplied by time” becomes $d=st$ . The letters stand for quantities, so their units matter. With distance in kilometres and time in hours, average speed is in kilometres per hour.

Subject of a formula

The letter isolated on one side of a formula. In $d=st$ , the subject is $d$ . In $s=d/t$ , it is $s$ . These describe the same relationship, written to give different quantities directly.

Use the stated meanings of the letters. In another formula, $s$ might stand for something different.

Average speed

\text{average speed}=\frac{\text{distance travelled}}{\text{time taken}}

The time must be positive. Use compatible units for distance, time and speed.

Not on the formulae sheet: learn it.

Here are two non-calculator examples. Write the substitution before working out its value.

Worked example 1: using a word formula

Question. A cyclist travels $42\,\mathrm{km}$ in $3$ hours. Work out the average speed.

Plan. Divide the distance by the time, keeping kilometres and hours.

\begin{aligned} \text{average speed}&=\frac{42}{3}&&\text{substitute distance and time}\\ &=14\,\mathrm{km/h}&&\text{divide} \end{aligned}

Why this step? Average speed measures the distance per one hour, so share the total distance between the hours.

Answer. $14\,\mathrm{km/h}$ .

Check. $14\times3=42$ , recovering the distance.

Change only the time to a decimal number of hours. The formula still uses the whole journey time.

Worked example 2: a decimal time

Question. A cyclist travels $42\,\mathrm{km}$ in $1.5$ hours. Work out the average speed.

Plan. Divide by $1.5$ , multiplying both numbers by $10$ to make a whole-number division.

\begin{aligned} \text{average speed}&=\frac{42}{1.5}&&\text{substitute}\\ &=\frac{420}{15}&&\text{multiply top and bottom by }10\\ &=28\,\mathrm{km/h}&&\text{divide} \end{aligned}

Why this step? Multiplying numerator and denominator by the same number preserves the quotient.

Answer. $28\,\mathrm{km/h}$ .

Check. The time has halved while the distance stays fixed, so the average speed has doubled from $14$ to $28$ .

Formulae also come from other subjects. For example, if a question supplies $E=Pt$ , with energy $E$ in joules, power $P$ in watts and time $t$ in seconds, the word version is “energy equals power times time”. That supplied formula is given in the question; you can use it without learning a new law of algebra.

Watch out

An hour and a half is $1.5$ hours, not $1.30$ hours. Convert units before substituting: $30$ minutes is $30/60=0.5$ hours.

Try the same division with a different distance.

Changing the subject with inverse operations

Rearranging keeps the relationship true while isolating the letter you want. Think about which operation was done last, then undo it first.

For $y=3x+7$ , the route from $x$ to $y$ is “multiply by $3$ , then add $7$ ”. The route back is “subtract $7$ , then divide by $3$ ”. Subtracting $7$ from both sides keeps them equal; dividing both sides by $3$ does the same.

Keeping an equality balanced

If $L=R$ , then

L+k=R+k,\qquad L-k=R-k,\qquad \frac Lc=\frac Rc\quad(c\ne0)

Multiplying both sides by the same non-zero number also preserves an equivalent relationship.

Not on the formulae sheet: learn it.

These non-calculator examples keep the letters throughout. You are finding a formula, not one numerical solution.

Worked example 3: undo addition then multiplication

Question. Make $x$ the subject of $y=3x+7$ .

Plan. Subtract $7$ from both sides, then divide both sides by $3$ .

\begin{aligned} y-7&=3x&&\text{subtract }7\\ \frac{y-7}{3}&=x&&\text{divide by }3 \end{aligned}

Why this step? The addition is outside the multiplication. Undoing it first leaves one complete multiple of $x$ .

Answer. $x=\dfrac{y-7}{3}$ .

Check. Choose $x=4$ . The original gives $y=19$ ; the rearranged formula gives $(19-7)/3=4$ .

Now change only the sign of the constant. Undo subtraction by adding.

Worked example 4: undo subtraction

Question. Make $x$ the subject of $y=3x-7$ .

Plan. Add $7$ to both sides before dividing by $3$ .

\begin{aligned} y+7&=3x&&\text{add }7\\ x&=\frac{y+7}{3}&&\text{divide by }3 \end{aligned}

Why this step? Adding $7$ removes the subtraction. You must add it to the left side as well.

Answer. $x=\dfrac{y+7}{3}$ .

Check. At $x=4$ , $y=5$ , and $(5+7)/3=4$ .

The fraction line groups a whole expression. Keep that grouping visible.

Watch out

Wrong: $x=y+\frac73$ . Only the $7$ has been divided. Right: $x=\frac{y+7}{3}$ , so both terms in $y+7$ are divided by $3$ .

Return briefly to the journey formula: subtracting the fixed cost and dividing by the cost per kilometre gives $d=(C-4)/1.8$ . The formula is supplied by the journey question. Changing its subject has not changed its pricing rule.

Rearranging brackets and powers

Brackets tell you which operations act on a whole expression. Work from the outside in, undoing those operations in reverse order.

For $y=2(x+3)$ , the bracket is multiplied by $2$ . Divide by $2$ to expose it. If the bracket is squared as well, undo the outside multiplication, then the square, then the addition inside the bracket.

Undoing a square

u^2=v\quad\Longrightarrow\quad u=\pm\sqrt v\qquad(v\geqslant0)

If the question tells you $u\geqslant0$ , use $u=\sqrt v$ . A length is normally positive, so its negative value is rejected.

Not on the formulae sheet: learn it.

Both examples are non-calculator. The second adds one extra operation to undo.

Worked example 5: isolating a bracket

Question. Make $x$ the subject of $y=2(x+3)$ .

Plan. Divide by $2$ , then subtract $3$ .

\begin{aligned} \frac y2&=x+3&&\text{divide both sides by }2\\ x&=\frac y2-3&&\text{subtract }3 \end{aligned}

Why this step? Division removes the outer multiplication without requiring an expansion of the bracket.

Answer. $x=\dfrac y2-3$ .

Check. At $x=5$ , $y=16$ , and $16/2-3=5$ .

Now square the bracket. The condition tells us which square root to use.

Worked example 6: undoing a squared bracket

Question. Make $x$ the subject of $y=2(x+3)^2$ , where $x\geqslant-3$ .

Plan. Divide by $2$ , take the non-negative square root, then subtract $3$ .

\begin{aligned} \frac y2&=(x+3)^2&&\text{divide by }2\\ \sqrt{\frac y2}&=x+3&&x+3\geqslant0\\ x&=\sqrt{\frac y2}-3&&\text{subtract }3 \end{aligned}

Why this step? The condition $x\geqslant-3$ makes $x+3$ non-negative. Without that condition, both signs of the root could be needed.

Answer. $x=\sqrt{y/2}-3$ , for $y\geqslant0$ .

Check. At $x=1$ , $y=32$ , and $\sqrt{32/2}-3=4-3=1$ .

One of the formulae you can rearrange this way is the area of a trapezium, $A=\frac12(a+b)h$ , where $a$ and $b$ are the parallel side lengths and $h$ is their perpendicular separation. This formula is on the 2027 formulae sheet. Dividing by $(a+b)/2$ gives $h=2A/(a+b)$ , provided $a+b\ne0$ . You will use its geometry later.

Watch out

Wrong: $\sqrt{u+v}=\sqrt u+\sqrt v$ . For example, $\sqrt{9+16}=5$ , whereas $\sqrt9+\sqrt{16}=7$ . Take a square root of the whole quantity on the other side.

Keep the same order of steps with a new multiplier and bracket.

When the subject appears twice

If the letter you want appears twice, bring those terms together and factorise. You then have one copy of the letter multiplying a bracket.

For example, $3x-ax=x(3-a)$ . The bracket is the total coefficient of $x$ . This is the same common-factor idea as $6x-2x=x(6-2)$ , with a letter replacing one number.

Collecting a repeated subject

px-qx=x(p-q)

To isolate $x$ by dividing by $p-q$ , you need $p\ne q$ .

Not on the formulae sheet: learn it.

Watch the role of each letter. In these non-calculator examples, $x$ is the subject being isolated; $a$ and $y$ remain in the answer.

Worked example 7: factorising the wanted letter

Question. Make $x$ the subject of $y+ax=3x$ , where $a\ne3$ .

Plan. Subtract $ax$ , factorise the right side, then divide by its coefficient.

\begin{aligned} y&=3x-ax&&\text{subtract }ax\\ y&=x(3-a)&&\text{take out }x\\ x&=\frac y{3-a}&&\text{divide by }3-a \end{aligned}

Why this step? Dividing before collecting would leave $x$ on both sides. Factorising gathers both copies into one.

Answer. $x=\dfrac y{3-a}$ .

Check. With $a=1$ and $y=8$ , this gives $x=4$ . The original becomes $8+4=12$ .

The next example adds a constant term on the right. Move that as well before factorising.

Worked example 8: moving a constant as well

Question. Make $x$ the subject of $y+ax=3x+4$ , where $a\ne3$ .

Plan. Collect the $x$ -terms on the right and the remaining terms on the left.

\begin{aligned} y-4&=3x-ax&&\text{subtract }4\text{ and }ax\\ y-4&=x(3-a)&&\text{factorise}\\ x&=\frac{y-4}{3-a}&&\text{divide by }3-a \end{aligned}

Why this step? The whole expression $y-4$ must be divided by the whole coefficient $3-a$ .

Answer. $x=\dfrac{y-4}{3-a}$ .

Check. With $a=1$ and $y=10$ , $x=3$ . The original gives $10+3=9+4$ .

The condition $a\ne3$ has a reason. If $a=3$ , the original second formula reduces to $y=4$ . When $y=4$ , every $x$ works; otherwise no $x$ works. There is no single rearranged value obtained by dividing by zero.

Exam tip

Write the line where you factorise the wanted letter. In a rearranging question, the final answer must have that letter alone and contain no further copies of it on the other side.

There is a pleasing example of a rearranged formula that undoes itself: $y=10-x$ becomes $x=10-y$ . Doing “subtract from $10$ ” twice brings every starting number back. At $x=3$ , it gives $7$ , then $3$ again. You will meet this idea again in functions.

Quick Recap

Read what every letter means and use compatible units when substituting.
Undo operations in reverse order, doing the same thing to both sides.
Respect brackets and square roots; use any stated sign conditions.
If the wanted letter appears twice, collect its terms, factorise and divide.

For the opening journey, $d=(C-4)/1.8$ . A £22 budget gives $d=(22-4)/1.8=10\,\mathrm{km}$ .

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A5 - Using and rearranging formulae

Reading and using formulae

Changing the subject with inverse operations

Rearranging brackets and powers

When the subject appears twice

Quick Recap

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