2.10 - Using and rearranging formulae

A formula can calculate a value or describe a relationship in either direction. You will substitute carefully, make a chosen letter the subject and handle formulae in which that letter appears twice or is raised to a power.

Substituting into a formula

If $\frac1f=\frac1u+\frac1v$ , how can you calculate $v$ directly from $f$ and $u$ ? Taking reciprocals of the three terms separately looks tempting, but it changes the relationship. We will build a formula that actually works.

First make sure you can use a formula without changing its structure. For $p=2a^2$ , putting $a=-3$ gives $p=2(-3)^2=18$ , because the square applies to the complete value of $a$ .

Formula and subject

A formula is an equation expressing a relationship between quantities. Its subject is the quantity written alone on one side, expressed in terms of the others. In $p=2a^2$ , $p$ is the subject; $a$ is not.

Substitution replaces a letter everywhere it occurs. Put a negative replacement in brackets, calculate powers before products, and finish the numerator and denominator separately.

Worked example 1: a formula containing a fraction

Question. Work out $p$ when $p=\dfrac{2a^2-b}{a+b}$ , $a=2$ and $b=5$ .

Plan. Substitute into the whole fraction, then evaluate the two parts.

\begin{aligned} p&=\frac{2(2)^2-5}{2+5} &&\text{replace both letters}\\ &=\frac{8-5}{7} &&\text{square, then multiply}\\ &=\frac37 &&\text{simplify} \end{aligned}

Why this step? The denominator is the complete sum $a+b$ ; the fraction bar groups it together.

Answer. $p=\dfrac37$ .

Check. The numerator is positive and smaller than the positive denominator, so the answer should lie between $0$ and $1$ .

The next example changes only the sign of $a$ . Its square stays the same, but the unsquared $a$ in the denominator changes.

Worked example 2: substituting a negative value

Question. Work out $p$ when $p=\dfrac{2a^2-b}{a+b}$ , $a=-2$ and $b=5$ .

Plan. Bracket $-2$ in every replacement and evaluate the square first.

\begin{aligned} p&=\frac{2(-2)^2-5}{-2+5} &&\text{replace every occurrence}\\ &=\frac{8-5}{3} &&\text{the square is positive}\\ &=1 &&\text{divide} \end{aligned}

Why this step? $(-2)^2=4$ , whereas $-2^2=-4$ . Brackets preserve the value being substituted.

Answer. $p=1$ .

Check. Both numerator and denominator equal $3$ , so their quotient is $1$ .

The supplied formula in these examples is not on the formulae sheet; it is given in the question. It is usable only when $a+b\ne0$ . For instance, $a=-5$ , $b=5$ would make the calculation undefined.

Watch out

Wrong: substituting $a=-2$ gives $2a^2=-8$ . Right: $2a^2=2(-2)^2=8$ . The negative sign is part of the value of $a$ .

The check changes the coefficients and a negative input. Keep the same substitution habits.

Undoing operations to isolate a letter

When the required letter appears once, work backwards through the operations surrounding it. Perform each inverse operation on both sides of the equation.

You already know

An equation remains true when you add or subtract the same quantity on both sides, or multiply or divide both sides by the same non-zero quantity. Clearing a fraction means multiplying both sides by its whole denominator.

Think of $3(2x-5)/4$ as a chain: double $x$ , subtract $5$ , multiply by $3$ , divide by $4$ . Reverse the chain from its last operation.

Worked example 3: reversing a chain

Question. Rearrange $T=\dfrac{3(2x-5)}4$ to make $x$ the subject.

Plan. Clear the denominator, divide by $3$ , then undo the subtraction and doubling.

\begin{aligned} 4T&=3(2x-5) &&\text{multiply by }4\\ \frac{4T}{3}&=2x-5 &&\text{divide by }3\\ 2x&=\frac{4T}{3}+5 &&\text{add }5\\ x&=\frac{4T+15}{6} &&\text{divide the whole sum by }2 \end{aligned}

Why this step? Dividing the entire right-hand side by $2$ gives $\frac12(4T/3+5)$ , not $4T/3+5/2$ .

Answer. $x=\dfrac{4T+15}{6}$ .

Check. Choose $x=4$ . The original gives $T=9/4$ , and the rearranged formula gives $(9+15)/6=4$ .

Change only the bracket's position: it now divides rather than multiplies. The unknown is in a denominator, so clear that denominator first.

Worked example 4: a letter in the denominator

Question. Rearrange $T=\dfrac3{4(2x-5)}$ to make $x$ the subject.

Plan. Multiply by the complete denominator, then divide by $4T$ .

\begin{aligned} 4T(2x-5)&=3 &&\text{clear the denominator}\\ 2x-5&=\frac3{4T} &&\text{divide by }4T\\ 2x&=\frac3{4T}+5 &&\text{add }5\\ x&=\frac{3+20T}{8T} &&\text{combine and halve} \end{aligned}

Why this step? The factor $2x-5$ belongs to the denominator as a whole. Multiplying by it removes the fraction without splitting that bracket.

Answer. $x=\dfrac{3+20T}{8T}$ , where $T\ne0$ . The original also requires $x\ne\frac52$ .

Check. At $x=4$ , the original gives $T=1/4$ ; the rearranged formula gives $(3+5)/2=4$ .

These rearranging methods are not on the formulae sheet: learn them. A subject is a letter alone, so an answer that stops at $2x=\ldots$ is unfinished.

Watch out

Wrong: $T=\frac3{4(2x-5)}$ becomes $4T(2x)-5=3$ . Right: $4T(2x-5)=3$ , because the complete denominator multiplies the left-hand side.

The new formula again puts the required letter inside a denominator bracket.

Collecting a letter that appears twice

When the required letter appears twice, first bring its terms to the same side. Factorising then turns two occurrences into one factor that you can isolate.

For example, $ab-3b=b(a-3)$ . Dividing by $a-3$ is now useful because the whole other side can be divided by it too. Trying to divide the original equation by $b$ would introduce new fractions without isolating $b$ .

Worked example 5: clearing, collecting and factorising

Question. Rearrange $a=\dfrac{3b+2}{b-4}$ to make $b$ the subject.

Plan. Clear the denominator, collect both $b$ terms and factorise.

\begin{aligned} a(b-4)&=3b+2 &&\text{clear the fraction}\\ ab-4a&=3b+2 &&\text{expand}\\ ab-3b&=4a+2 &&\text{collect }b\text{ terms}\\ b(a-3)&=4a+2 &&\text{factor out }b\\ b&=\frac{4a+2}{a-3} &&\text{divide by }a-3 \end{aligned}

Why this step? Factoring $b$ from its two terms leaves just one occurrence of the required subject.

Answer. $b=\dfrac{4a+2}{a-3}$ , with $a\ne3$ . The original requires $b\ne4$ .

Check. At $b=5$ , the original gives $a=17$ ; the new formula gives $(68+2)/14=5$ . If $a=3$ , the collected equation would say $0=14$ , so no value of $b$ works.

The next example changes only the sign of the constant in the numerator. The method stays the same, but carry that sign carefully.

Worked example 6: a changed sign

Question. Rearrange $a=\dfrac{3b-2}{b-4}$ to make $b$ the subject.

Plan. Clear the denominator, collect the two $b$ terms and divide after factorising.

\begin{aligned} a(b-4)&=3b-2 &&\text{clear the fraction}\\ ab-4a&=3b-2 &&\text{expand}\\ ab-3b&=4a-2 &&\text{collect }b\text{ terms}\\ b(a-3)&=4a-2 &&\text{factorise}\\ b&=\frac{4a-2}{a-3} &&\text{divide} \end{aligned}

Why this step? Moving $-4a$ adds $4a$ to the other side; it does not change the sign of the separate $-2$ .

Answer. $b=\dfrac{4a-2}{a-3}$ , with $a\ne3$ . The original requires $b\ne4$ .

Check. At $b=5$ , $a=13$ , and the answer gives $(52-2)/10=5$ .

The opening reciprocal formula uses exactly this collection method after clearing all denominators:

\begin{aligned} \frac1f=\frac1u+\frac1v &\Longrightarrow uv=fv+fu &&\text{multiply by }fuv\\ &\Longrightarrow v(u-f)=fu &&\text{collect and factorise}\\ &\Longrightarrow v=\frac{fu}{u-f} &&\text{divide} \end{aligned}

Here $f,u,v\ne0$ and $u\ne f$ . For $f=6$ and $u=18$ , $v=108/12=9$ , and $1/6=1/18+1/9$ . This formula is not on the formulae sheet; derive it from the given relationship.

A useful surprise follows when $u$ and $v$ are positive: $f$ is smaller than either of them. Indeed, $f=uv/(u+v)$ , and its denominator is greater than either positive term alone. With $u=v=10$ , the result is $f=5$ , not $20$ .

Exam tip

For a repeated subject, show the line with that subject factorised. Clearing fractions, collecting terms and factorising are separate useful stages: write each one before the final answer.

Now the denominator has a plus sign, so the collected factor will differ.

Isolating a power before taking a root

When the repeated subject is a square or a cube, collect and factorise that whole power first. Take the root only after the power stands alone.

Undoing a square or cube

For real numbers,

s^2=K\Longrightarrow s=\pm\sqrt K\qquad K\geqslant0

and

s^3=K\Longrightarrow s=\sqrt[3]{K}

A positive-value condition selects the positive square root. A cube has one real cube root, including when $K<0$ .
Not on the formulae sheet: learn it.

The root symbol must cover the entire expression isolated on the other side.

Worked example 7: the subject is squared

Question. Rearrange $r=\dfrac{2s^2+3}{s^2-4}$ to make $s$ the subject.

Plan. Treat $s^2$ as one quantity, isolate it, then take both square roots.

\begin{aligned} r(s^2-4)&=2s^2+3 &&\text{clear the fraction}\\ rs^2-2s^2&=4r+3 &&\text{expand and collect}\\ s^2(r-2)&=4r+3 &&\text{factor out the power}\\ s^2&=\frac{4r+3}{r-2} &&\text{divide}\\ s&=\pm\sqrt{\frac{4r+3}{r-2}} &&\text{root the whole fraction} \end{aligned}

Why this step? Both $s$ and $-s$ have the same square. Without a sign condition, both are needed when the fraction is positive; when it is zero, they coincide.

Answer. $s=\pm\sqrt{\dfrac{4r+3}{r-2}}$ , with $r\ne2$ and the fraction non-negative. The original requires $s\ne\pm2$ .

Check. At $s=3$ , the original gives $r=21/5$ ; the fraction under the root becomes $(99/5)/(11/5)=9$ , giving $s=\pm3$ . Both values satisfy the original.

Change only the power from a square to a cube. Collection is identical, but the final root has one real value.

Worked example 8: the subject is cubed

Question. Rearrange $r=\dfrac{2s^3+3}{s^3-4}$ to make $s$ the subject.

Plan. Isolate $s^3$ by collection and factorisation, then take the cube root.

\begin{aligned} r(s^3-4)&=2s^3+3 &&\text{clear the denominator}\\ rs^3-2s^3&=4r+3 &&\text{collect the cubed terms}\\ s^3(r-2)&=4r+3 &&\text{factorise}\\ s^3&=\frac{4r+3}{r-2} &&\text{divide}\\ s&=\sqrt[3]{\frac{4r+3}{r-2}} &&\text{cube root of the whole fraction} \end{aligned}

Why this step? Cubing preserves the sign: $(-2)^3=-8$ . A cube root therefore has no extra $\pm$ .

Answer. $s=\sqrt[3]{\dfrac{4r+3}{r-2}}$ , with $r\ne2$ . The original requires $s^3\ne4$ .

Check. At $s=2$ , $r=19/4$ ; the isolated fraction is $22/(11/4)=8$ , so its cube root is $2$ .

Rearranging has now answered the opening problem and extended the method to powers: clear, collect, factorise, divide, then undo the power if necessary.

Watch out

Wrong: $s=\frac{\sqrt[3]{4r+3}}{r-2}$ . The root covers only the numerator there. Right: $s=\sqrt[3]{\frac{4r+3}{r-2}}$ , or equivalent roots of both numerator and denominator.

Use a cube root in the final check; the changed signs require a fresh collection step.

Quick Recap

Substitute a complete value wherever its letter occurs, using brackets for negatives.
Undo operations on both sides; clear a complete denominator before manipulating its terms.
If the required letter appears twice, collect its terms and factorise it out.
Isolate a square or cube before taking a root; respect sign conditions and non-zero denominators.

The opening relationship gives $v=fu/(u-f)$ , with non-zero original denominators and $u\ne f$ .

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2.10 - Using and rearranging formulae

Substituting into a formula

Undoing operations to isolate a letter

Collecting a letter that appears twice

Isolating a power before taking a root

Quick Recap

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