NSWMaths Standard 2Syllabus dot point

How do you rearrange a formula to make a different variable the subject by undoing the operations around it?

Change the subject of a formula by rearranging it to isolate any chosen variable, including a subject that appears in a power, a root or a fraction, and check the rearrangement by substitution

A focused answer to the HSC Maths Standard 2 dot point on changing the subject of a formula. Rearrange a formula to isolate any variable using inverse operations and keeping it balanced, with the subject in a power, a root or a fraction, multi-step transposing, and checking by substitution, with worked Australian examples.

Generated by Claude Opus 4.815 min answerUpdated 2026-06-21

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

What this dot point is asking

NESA wants you to take a formula and rearrange it so a different variable stands alone on one side. This is called changing the subject, or transposing the formula (transposing just means rearranging). The subject is the variable written by itself, with no other numbers or letters on its side of the equals sign. In $C = 18n + 250$ the subject is $C$ , and the task might be to rewrite it with $n$ as the subject instead. You do this with the same toolkit you use to solve a linear equation: inverse operations (the opposite of an operation, so subtraction undoes addition), applied to both sides so the formula stays balanced, peeling the operations off the chosen variable in reverse order. The only new idea is that you carry the other pronumerals (the letters) along as symbols instead of numbers, so the answer is a formula, not a single value. This matters because a formula written one way answers one question, and rearranging it lets the same relationship answer a different question. Distance $D = ST$ finds distance; transposed to $T = \dfrac{D}{S}$ it finds time. The marks are usually lost in predictable ways: dividing by only part of the other side, forgetting that a square root undoes a square and an $n$ th power, splitting a root over a sum, or losing a sign.

The answer

Transposing is solving an equation you do not finish numerically

Changing the subject uses the same steps as solving a linear equation. So if you can solve $18n + 250 = 1330$ for $n$ , you can make $n$ the subject of $C = 18n + 250$ . The only difference is that the right side stays as the letter $C$ instead of a number, so you finish with a formula rather than a value. Picture how the chosen variable was built up, then undo each operation in reverse, doing the same thing to both sides each time. The diagram above reads in two directions. Along the top, $n$ is multiplied by $18$ and then $250$ is added to build $C$ . Along the bottom you reverse that: you subtract $250$ first and divide by $18$ last, which leaves $n$ by itself. Whatever was done last to the variable is undone first.

Which variable is the subject

The subject is the variable sitting alone on one side with a coefficient of $1$ and nothing added to it. "Make $r$ the subject" means rearrange until the formula reads $r = \ldots$ with everything else on the other side. A useful first move is to locate where the target variable currently is and list the operations stacked around it from the inside out, because you will undo them from the outside in. In $V = \pi r^2 h$ , the variable $r$ is squared (innermost), then multiplied by $\pi$ and by $h$ (outermost); to free $r$ you remove the multiplications first and the square last.

Undo in reverse order, the same to both sides

The two rules that never change are:

Inverse operations. Addition is undone by subtraction, multiplication by division, a square by a square root, and an $n$ th power by an $n$ th root. Each operation has an opposite, and applying the opposite cancels it.
Balance. Whatever you do to one side, do to the other side as well. Subtract the same term from both sides, divide both sides by the same factor, take the root of both sides. This keeps the two sides equal, so the relationship is unchanged.

Work from the outside in. The term added or subtracted on the subject's side comes off first; the factor multiplying the whole subject comes off next; a power or root on the subject itself is undone last. This is the same reverse order as a two-step equation, just carried out with symbols.

When the subject is inside a fraction

If the subject sits in the numerator (the top of the fraction), like $r$ in $r = \dfrac{I}{Pn}$ being made the subject of $I = Prn$ , treat the bottom as a factor and clear it by multiplying both sides by it. If the subject sits in the denominator (the bottom of the fraction), like $t$ in $v = \dfrac{d}{t}$ , it is trapped underneath. So your first move is to multiply both sides by that bottom to lift the subject up onto the top line. Once $v t = d$ , the subject is an ordinary factor, and you divide both sides by $v$ to finish, giving $t = \dfrac{d}{v}$ . The trap is trying to "divide by $\dfrac{d}{t}$ " or flipping only one side. Lift the variable out of the bottom first, then isolate it.

When the subject is in a power or a root

A square is undone by a square root, and a square root by squaring; more generally an $n$ th power is undone by an $n$ th root. Two cautions matter here. First, isolate the power before you take the root: in $L^2 = h^2 + b^2$ , subtract $b^2$ to get $h^2 = L^2 - b^2$ before rooting, because the square root does not distribute over a sum or difference ( $\sqrt{L^2 - b^2}$ is not $L - b$ ). Second, when you square-root both sides of a real measurement, keep only the positive root, since lengths, radii, times and masses cannot be negative. If the subject is already under a root, like $r$ in $r = \sqrt{\dfrac{A}{\pi}}$ , you reach it by squaring both sides to remove the root, then isolating as usual.

How exam questions ask about changing the subject

The wording tells you the target variable and how many layers you must peel:

"Make $h$ the subject of the formula" or "write the formula with $h$ as the subject" or "rearrange to make $h$ the subject" all mean the same thing: rewrite it as $h = \ldots$ .
"Express $r$ in terms of the other variables" is the same instruction phrased differently; isolate $r$ .
A two-part question ("(a) make $T$ the subject; (b) find $T$ when ...") wants the rearranged formula in part (a), then a substitution in part (b). The cleanest method is almost always to transpose first and substitute second, so do the algebra once and reuse it.
A formula with a squared variable ("make $r$ the subject of $A = \pi r^2$ ") signals: isolate the square, then take the positive square root.
A formula with the variable under a root ("make $L$ the subject of $T = \sqrt{\dfrac{L}{g}}$ "-style) signals: square both sides first to clear the root.
A formula with the variable in the denominator ("make $t$ the subject of $v = \dfrac{d}{t}$ ") signals: multiply both sides by that denominator first.
"Answer correct to ... decimal places" is a presentation instruction for part (b); transpose exactly, then round only the final substituted value.

Changing the subject in four Australian contexts

Four formulae span the cases NESA tests: a venue-hire cost (a linear subject), an exam average (a subject inside a fraction), a circular garden (a subject under a square root after isolating a square), and a simple-interest total (the subject appearing on both sides, cleared by factoring). Each one peels the operations off the target variable in reverse order, doing the same to both sides, then checks by substitution.

Make the number of guests the subject of a venue-hire cost

A community hall charges a fixed booking fee of $250 plus $18 per guest, so the total cost of a function for $n$ guests is $C = 18n + 250$ dollars. Make $n$ the subject, then find how many guests a budget of $C = 1330$ dollars covers.

Identify what wraps the subject. Reading the right side, $n$ is multiplied by $18$ and then $250$ is added. To free $n$ you undo these in reverse: remove the $+250$ first, then the $\times 18$ .

Subtract 250 from both sides. This peels off the added fee:

C - 250 = 18n

Divide both sides by 18. This removes the factor multiplying $n$ :

\frac{C - 250}{18} = n

Write the subject on the left. Swapping sides gives the rearranged formula:

n = \frac{C - 250}{18}

Use it and check. Substituting the budget $C = 1330$ :

n = \frac{1330 - 250}{18} = \frac{1080}{18} = 60 \text{ guests.}

Check by putting $n = 60$ into the original: $18 \times 60 + 250 = 1080 + 250 = 1330$ , which matches the budget, so $1330 covers $60$ guests.

Make a missing exam mark the subject of an average

A student's average across three exams is $m = \dfrac{a + b + c}{3}$ , where $a$ , $b$ and $c$ are the three marks. Make $c$ the subject, then find the mark needed in the third exam to reach an average of $m = 80$ given $a = 72$ and $b = 85$ .

Clear the fraction first. The whole right side is divided by $3$ , so multiply both sides by $3$ to remove the denominator:

3m = a + b + c

Subtract the known terms. The subject $c$ has $a$ and $b$ added to it, so subtract both from each side:

3m - a - b = c \;\Rightarrow\; c = 3m - a - b

Use it and check. Substituting $m = 80$ , $a = 72$ , $b = 85$ :

c = 3 \times 80 - 72 - 85 = 240 - 157 = 83.

Check by averaging all three: $\dfrac{72 + 85 + 83}{3} = \dfrac{240}{3} = 80$ , which matches the target average, so the student needs $83$ in the third exam.

Make the radius the subject of a circle's area

A circular garden bed has area $A = \pi r^2$ square metres, where $r$ is the radius. Make $r$ the subject, then find the radius of a bed with area $A = 50$ m $^2$ , correct to two decimal places.

Isolate the squared subject. The variable $r$ is squared and then multiplied by $\pi$ , so divide both sides by $\pi$ to leave $r^2$ alone:

\frac{A}{\pi} = r^2

Undo the square last. Take the positive square root of both sides; a radius is a positive length, so the negative root is discarded:

r = \sqrt{\frac{A}{\pi}}

Use it and check. Substituting $A = 50$ :

r = \sqrt{\frac{50}{\pi}} = \sqrt{15.915\ldots} \approx 3.99 \text{ m.}

Check by squaring back: $\pi \times 3.99^2 \approx \pi \times 15.92 \approx 50.0$ m $^2$ , which matches the area, so the radius is $3.99$ m. Note that isolating $r^2$ before rooting is essential; you cannot take the square root of $A$ and of $\pi$ separately.

Make the principal the subject when it appears on both sides

The value of a deposit after simple interest is $A = P(1 + rn)$ , where $P$ is the principal, $r$ the annual rate and $n$ the number of years. Make $P$ the subject, then find the principal that grows to $A = 5900$ dollars at $r = 0.06$ over $n = 3$ years.

Spot the subject in two places, then factor. Expanding gives $A = P + Prn$ , so $P$ appears in both terms. Rather than expand, keep $P$ as the common factor it already is: $(1 + rn)$ is a single quantity multiplying $P$ . Factoring the subject out is what lets you isolate it in one division.

Divide both sides by the whole bracket. Treating $(1 + rn)$ as one factor:

\frac{A}{1 + rn} = P \;\Rightarrow\; P = \frac{A}{1 + rn}

Evaluate the bracket, then use it. With $r = 0.06$ and $n = 3$ , the bracket is $1 + 0.06 \times 3 = 1 + 0.18 = 1.18$ . Substituting $A = 5900$ :

P = \frac{5900}{1.18} = 5000.

Check by putting $P = 5000$ into the original: $5000 \times (1 + 0.06 \times 3) = 5000 \times 1.18 = 5900$ , which matches, so the principal is $5000. When a variable appears more than once, gather it as a common factor before dividing; you cannot isolate it while it is in two separate terms.

Final answers: the venue budget of $1330 covers $60$ guests; the third exam mark needed is $83$ ; the garden's radius is $3.99$ m; and the principal required is $5000.

Rearranging $C = 18n + 250$ stage by stage

The "do the same to both sides" rule is easiest to see laid out as a before-and-after sequence. The four panels below make $n$ the subject of the venue-hire formula $C = 18n + 250$ , one inverse operation per stage. In each panel the top row is the formula before the move, the accent badge in the middle is the operation applied to both sides, and the bottom row is the result that carries into the next panel. The formula stays balanced at every stage because the same thing is done to each side.

Stage 1, identify what wraps the subject. Start from $C = 18n + 250$ . On the subject's future side, $n$ has been multiplied by $18$ and then had $250$ added. You will undo these in reverse: the $+250$ first, then the $\times 18$ .

Stage 2, subtract 250 from both sides. Removing the added fee from each side peels off the outermost operation, leaving $18n$ on the right: $C - 250 = 18n$ .

Stage 3, divide both sides by 18. Dividing each side by the factor multiplying $n$ frees the variable, giving $\dfrac{C - 250}{18} = n$ . The whole left side, not just part of it, is divided by $18$ .

Stage 4, write $n$ as the subject. Swapping the sides puts the subject on the left in the conventional way, giving the finished formula $n = \dfrac{C - 250}{18}$ , highlighted below. This is the rearranged formula you substitute into.

Always check the rearrangement by substitution

A rearranged formula must describe the same relationship as the original, and you can confirm it in seconds. Pick easy numbers, substitute them into the original formula to get the subject's value, then substitute the same numbers into your rearranged formula and check it gives that same value. For $n = \dfrac{C - 250}{18}$ , take $n = 60$ : the original gives $C = 18 \times 60 + 250 = 1330$ , and the rearranged formula gives $\dfrac{1330 - 250}{18} = 60$ , which returns the original $n$ , so the transposition is correct. This catches a dropped term, a sign slip or dividing by only part of a side, the errors that cost transposing marks.

Common traps

Dividing by only part of the other side: When you divide to remove a factor, the whole other side is divided. From $C - 250 = 18n$ , dividing by $18$ gives $\dfrac{C - 250}{18} = n$ , not $\dfrac{C}{18} - 250$ .
Splitting a root over a sum or difference: $\sqrt{L^2 - b^2}$ is not $L - b$ . Isolate the squared term first, then take the root of the single quantity. The same applies to $\sqrt{a^2 + b^2}$ , which is not $a + b$ .
Taking the root before isolating the power: In $A = \pi r^2$ , divide by $\pi$ to reach $r^2 = \dfrac{A}{\pi}$ before square-rooting; rooting the whole right side as $\sqrt{\pi r^2}$ does not free $r$ cleanly.
Leaving the subject in the denominator: If the target variable is underneath, multiply both sides by that denominator first to lift it up. You cannot isolate $t$ in $v = \dfrac{d}{t}$ until $t$ is on the top line.
Failing to gather a repeated subject: If the variable appears more than once, collect it as a common factor before dividing. You cannot isolate $P$ in $A = P + Prn$ until it is written as $A = P(1 + rn)$ .
Keeping the negative root for a measurement: Lengths, radii, times and masses are positive, so take only the positive square root in those contexts.
Undoing in the wrong order: Remove what is added or subtracted before the factor that multiplies the subject, and undo a power or root last; it is the reverse of how the formula was built.

Exam technique

Set the work out vertically, one inverse operation per line, with the equals signs aligned and the operation named beside each line (" $-250$ both sides", " $\div 18$ both sides", "square root both sides"); markers follow the chain of inverse steps and award method marks even if the final arithmetic slips. For a two-part "(a) make the subject, (b) find the value" question, transpose in part (a) and substitute in part (b), because doing the algebra once is cleaner and the rearrangement carries its own marks. Decide your route before writing: if the subject is in a denominator, multiply up first; if it is squared, isolate the square then take the positive root; if it appears twice, factor it out. Keep the negative root out of any length, radius, time or mass answer. Round only at the very end of part (b), to the precision asked, and finish by checking the rearrangement on easy numbers so both the original and the rearranged formula give the same value.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 HSC-style3 marksA plumber charges a call-out fee of $95 plus $75 for each hour worked, so the total bill is

C = 95 + 75h

dollars for

h

hours. (a) Make

h

the subject of the formula. (b) A job costs

C = 470

dollars. Find how many hours the plumber worked.

Show worked answer →

Part (a) - identify what wraps $h$ . On the right side $h$ is multiplied by $75$ , then $95$ is added. Undo these in reverse: remove the $+95$ first, then the $\times 75$ .

Subtract $95$ from both sides. This peels off the added call-out fee:

C - 95 = 75h

Divide both sides by $75$ . This removes the factor multiplying $h$ , and swapping sides gives the rearranged formula:

h = \dfrac{C - 95}{75}

Part (b) - substitute. With $C = 470$ :

h = \dfrac{470 - 95}{75} = \dfrac{375}{75} = 5 \text{ hours}

Check. Put $h = 5$ into the original: $95 + 75 \times 5 = 95 + 375 = 470$ , which matches the bill, so the plumber worked $5$ hours.

Markers reward subtracting $95$ from the whole side before dividing, dividing the entire left side by $75$ (not just $C$ ), and the substituted answer with units. A common slip is writing $\dfrac{C}{75} - 95$ , which divides only part of the side.

2022 HSC-style4 marksThe distance an object falls from rest is

d = 5t^2

metres after

t

seconds. (a) Make

t

the subject of the formula. (b) A stone is dropped from a bridge and falls

d = 80

metres before hitting the water. Find the time taken, in seconds.

Show worked answer →

Part (a) - isolate the squared subject. The subject $t$ is squared, then multiplied by $5$ . Divide both sides by $5$ to leave $t^2$ alone:

\dfrac{d}{5} = t^2

Undo the square last. Take the positive square root of both sides; a time is positive, so the negative root is discarded:

t = \sqrt{\dfrac{d}{5}}

Isolate $t^2$ before rooting; you cannot take the square root of $d$ and of $5$ separately.

Part (b) - substitute. With $d = 80$ :

t = \sqrt{\dfrac{80}{5}} = \sqrt{16} = 4 \text{ seconds}

Check. Put $t = 4$ into the original: $5 \times 4^2 = 5 \times 16 = 80$ metres, which matches, so the stone falls for $4$ seconds.

Markers reward dividing by $5$ before taking the root, keeping only the positive root for a time, and the final value with units. Half marks for $\sqrt{\dfrac{d}{5}}$ written as $\dfrac{\sqrt{d}}{5}$ , which mishandles the root.

2023 HSC-style4 marksA car's fuel consumption is

R = \dfrac{100V}{D}

litres per

100

km, where

V

is the litres of petrol used over a distance of

D

km. (a) Make

D

the subject of the formula. (b) On a country trip a car used

V = 52

litres at a consumption of

R = 8

litres per

100

km. Find the distance travelled, in kilometres.

Show worked answer →

Part (a) - lift the subject out of the denominator. The subject $D$ is underneath, so multiply both sides by $D$ first to bring it onto the top line:

R D = 100V

Divide both sides by $R$ . Now $D$ is an ordinary factor, so divide by $R$ to isolate it:

D = \dfrac{100V}{R}

Part (b) - substitute. With $V = 52$ and $R = 8$ :

D = \dfrac{100 \times 52}{8} = \dfrac{5200}{8} = 650 \text{ km}

Check. Put $D = 650$ into the original: $\dfrac{100 \times 52}{650} = \dfrac{5200}{650} = 8$ litres per $100$ km, which matches $R$ , so the car travelled $650$ km.

Markers reward multiplying both sides by $D$ to clear the denominator first, then dividing by $R$ , and the substituted distance with units. A common error is trying to divide by $\dfrac{100V}{D}$ or flipping only one side without lifting $D$ up first.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksMake

x

the subject of

y = x + 9

, then use your rearranged formula to find

x

when

y = 23

Show worked solution →

Isolate the chosen variable. The variable $x$ has $9$ added to it, so subtract $9$ from both sides:

y - 9 = x \;\Rightarrow\; x = y - 9

Use the rearranged formula. Substitute $y = 23$ :

x = 23 - 9 = 14

Check. Put $x = 14$ back into the original: $14 + 9 = 23$ , which matches $y$ , so $x = 14$ is correct.

foundation3 marksThe distance travelled is

D = ST

, where

S

is the speed and

T

is the time. (a) Make

T

the subject of the formula. (b) Find the time, in hours, for a trip of

D = 315

km at a speed of

S = 90

km/h.

Show worked solution →

Part (a) - isolate $T$ . The subject $T$ is multiplied by $S$ , so divide both sides by $S$ :

\frac{D}{S} = T \;\Rightarrow\; T = \frac{D}{S}

Part (b) - substitute. With $D = 315$ and $S = 90$ :

T = \frac{315}{90} = 3.5 \text{ hours}

Check. $S \times T = 90 \times 3.5 = 315$ km, which matches $D$ , so the trip takes $3.5$ hours.

core3 marksSimple interest is

I = Prn

, where

P

is the principal,

r

the annual rate (as a decimal) and

n

the number of years. (a) Make

r

the subject. (b) Find the rate, as a percentage, when

I = 540

P = 4500

and

n = 2

Show worked solution →

Part (a) - isolate $r$ . The subject $r$ is multiplied by both $P$ and $n$ , so divide both sides by $Pn$ :

\frac{I}{Pn} = r \;\Rightarrow\; r = \frac{I}{Pn}

Part (b) - substitute. With $I = 540$ , $P = 4500$ and $n = 2$ :

r = \frac{540}{4500 \times 2} = \frac{540}{9000} = 0.06

Convert to a percentage. $0.06 \times 100 = 6\%$ .

Check. $Prn = 4500 \times 0.06 \times 2 = 540$ , which matches $I$ , so the rate is $6\%$ per year.

core3 marksThe area of a square is

A = s^2

, where

s

is the side length in metres. (a) Make

s

the subject. (b) Find the side length of a square of area

A = 2.56

^2

Show worked solution →

Part (a) - undo the square. The subject $s$ is squared, and the opposite of squaring is the square root, so take the positive square root of both sides:

s = \sqrt{A}

A side length is a positive measurement, so only the positive root is kept.

Part (b) - substitute. With $A = 2.56$ :

s = \sqrt{2.56} = 1.6 \text{ m}

Check. $s^2 = 1.6^2 = 2.56$ m $^2$ , which matches $A$ , so each side is $1.6$ m.

core3 marksAverage speed is

v = \dfrac{d}{t}

, where

d

is the distance and

t

is the time. (a) Make

t

the subject. (b) Find the time, in hours, when

v = 80

km/h and

d = 200

km.

Show worked solution →

Part (a) - get the subject out of the denominator. The subject $t$ is underneath, so multiply both sides by $t$ first to lift it up:

vt = d

Now $t$ is multiplied by $v$ , so divide both sides by $v$ :

t = \frac{d}{v}

Part (b) - substitute. With $d = 200$ and $v = 80$ :

t = \frac{200}{80} = 2.5 \text{ hours}

Check. $\dfrac{d}{t} = \dfrac{200}{2.5} = 80$ km/h, which matches $v$ , so the trip takes $2.5$ hours.

exam4 marksThe volume of a cylinder is

V = \pi r^2 h

, where

r

is the radius and

h

the height. (a) Make

r

the subject. (b) A cylindrical tank has volume

V = 282.74

^3

and height

h = 10

m. Find its radius, correct to two decimal places.

Show worked solution →

Part (a) - peel the layers off $r$ in reverse. The subject $r$ is squared, then multiplied by $\pi$ and by $h$ . Undo the multiplications first by dividing both sides by $\pi h$ :

\frac{V}{\pi h} = r^2

Undo the square last. Take the positive square root of both sides (a radius is positive):

r = \sqrt{\frac{V}{\pi h}}

Part (b) - substitute. With $V = 282.74$ and $h = 10$ :

r = \sqrt{\frac{282.74}{\pi \times 10}} = \sqrt{\frac{282.74}{31.4159\ldots}} = \sqrt{8.999\ldots} \approx 3.00 \text{ m}

Check. $\pi \times 3.00^2 \times 10 = \pi \times 90 \approx 282.74$ m $^3$ , which matches $V$ , so the radius is $3.00$ m.

exam4 marksThe total value of an investment under simple interest is

A = P(1 + rn)

, where

P

is the principal,

r

the rate and

n

the years. (a) Make

P

the subject. (b) Find the principal needed to grow to

A = 7930

when

r = 0.055

and

n = 4

Show worked solution →

Part (a) - notice the subject is in two places, then factor. Expanding the bracket gives $A = P + Prn$ , so $P$ sits in both terms. Because the original formula is already written with $P$ as a common factor outside the bracket, you do not need to expand: treat $(1 + rn)$ as a single number multiplying $P$ . Divide both sides by that whole bracket:

\frac{A}{1 + rn} = P \;\Rightarrow\; P = \frac{A}{1 + rn}

Part (b) - evaluate the bracket first. With $r = 0.055$ and $n = 4$ :

1 + rn = 1 + 0.055 \times 4 = 1 + 0.22 = 1.22

Substitute. With $A = 7930$ :

P = \frac{7930}{1.22} = 6500

Check. $P(1 + rn) = 6500 \times 1.22 = 7930$ , which matches $A$ , so the principal is $6500.

exam4 marksA ladder of length

L

leans against a wall. Its foot is

b

metres from the wall and it reaches a height

h

up the wall, related by

L^2 = h^2 + b^2

. (a) Make

h

the subject. (b) Find the height reached, in metres, by a

6.5

m ladder whose foot is

2.5

m from the wall.

Show worked solution →

Part (a) - isolate the squared subject first. The subject $h$ appears as $h^2$ added to $b^2$ . Subtract $b^2$ from both sides to leave $h^2$ alone:

L^2 - b^2 = h^2

Undo the square. Take the positive square root of both sides (a height is positive):

h = \sqrt{L^2 - b^2}

Subtract inside the root before taking the root; you cannot split the square root over a subtraction.

Part (b) - substitute. With $L = 6.5$ and $b = 2.5$ :

h = \sqrt{6.5^2 - 2.5^2} = \sqrt{42.25 - 6.25} = \sqrt{36} = 6 \text{ m}

Check. $h^2 + b^2 = 6^2 + 2.5^2 = 36 + 6.25 = 42.25 = 6.5^2 = L^2$ , which matches, so the ladder reaches $6$ m.

exam4 marksThe body mass index is

B = \dfrac{m}{h^2}

, where

m

is the mass in kilograms and

h

the height in metres. (a) Make

h

the subject. (b) Find the height, correct to two decimal places, of a person with

B = 24

and

m = 60

kg.

Show worked solution →

Part (a) - lift the subject out of the denominator. The subject $h$ is squared and underneath, so multiply both sides by $h^2$ to bring it up:

B h^2 = m

Get the squared subject alone. Divide both sides by $B$ :

h^2 = \frac{m}{B}

Undo the square. Take the positive square root of both sides (a height is positive):

h = \sqrt{\frac{m}{B}}

Part (b) - substitute. With $m = 60$ and $B = 24$ :

h = \sqrt{\frac{60}{24}} = \sqrt{2.5} \approx 1.58 \text{ m}

Check. $\dfrac{m}{h^2} = \dfrac{60}{1.58^2} = \dfrac{60}{2.4964} \approx 24$ , which matches $B$ , so the height is $1.58$ m.

What this dot point is asking

The answer

Transposing is solving an equation you do not finish numerically

Which variable is the subject

Undo in reverse order, the same to both sides

When the subject is inside a fraction

When the subject is in a power or a root

How exam questions ask about changing the subject

Rearranging C=18n+250C = 18n + 250C=18n+250 stage by stage

Always check the rearrangement by substitution

Exam-style practice questions

Practice questions

Related dot points

Rearranging $C = 18n + 250$ stage by stage