What is simplify the known numbers before you solve?

After substituting, collect the known numbers into a single coefficient before you start undoing operations. In A = 12bh with A = 96 and h = 12, substituting gives 96 = 12 × b × 12; tidy the right side to 96 = 6b first, and the solve is a single division. In the trapezium formula A = 12(a + b)h with the parallel sides known, work out the bracket and the half before touching the unknown. Simplifying first turns an equation that looks complicated into one of the one-step or two-step forms you already recognise, and it removes most of the places a sign or a factor can go astray.

What is not turning a percentage rate into a decimal?

In I = Prn a rate of 4\% must go in as 0.04, not 4. Forgetting this scales the answer by 100.

NESA 2022 HSC-style-style practice: The surface area of a closed cylindrical tank is approximated by S = 2π r(r + h), where r is the base radius and h is the height, both in metres. A tank has a radius of r = 2 m and a surface area of S = 40π m^2. Find the height h of the tank.

**Substitute the known values.** The unknown is h, which is not the subject, so substitute S = 40π and r = 2: $40π = 2π × 2 × (2 + h)$ **Simplify the known numbers first.** On the right, 2π × 2 = 4π, so the equation becomes $40π = 4π(2 + h)$ **Peel off the coefficient.** Divide both sides by 4π to clear it in one step: $40π4π = 2 + h \;\; 10 = 2 + h$ **Undo the addition.** Subtract 2 from both sides: $h = 10 - 2 = 8$ **State and check.** The height is 8 m. Check: 2π × 2 × (2 + 8) = 4π × 10 = 40π m^2, which matches, so h = 8 m. Markers reward simplifying 2π × 2 to 4π before solving and dividing the whole side by 4π. Dividing by 4π leaves a clean number because the π cancels.

NSWMaths Standard 2Syllabus dot point

How do you find an unknown that is not the subject of a formula by substituting the known values first and then solving the equation that is left?

Substitute known values into a formula and then solve the resulting equation to find an unknown quantity that is not the subject of the formula

A focused answer to the HSC Maths Standard 2 dot point on solving equations after substitution. Substitute the known values into a formula first, then solve the equation that is left to find an unknown that is not the subject, with the order of operations, units and rounding, the substitute-or-rearrange choice, and 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 find an unknown quantity in a formula when that unknown is not the subject. The subject is the letter sitting alone on the left of the equals sign. You do it in two clean stages. First substitute every value you know into the formula, which leaves an equation with just one unknown in it. Then solve that equation with inverse operations, exactly as you would any linear equation. This is the meeting point of two skills you have already built: substituting into a formula, and solving an equation. The whole difficulty is recognising which job you are doing. If the unknown is the subject, you only substitute and evaluate. If the unknown is buried somewhere else in the formula, you substitute the knowns first and then have an equation to solve. The deeper idea is that you do not need to rearrange the formula at all. Once the known numbers are in, what is left is an ordinary equation, and you already know how to peel the operations off the unknown.

The answer

Substitute first, then solve

A formula is a relationship between several variables, like $I = Prn$ (simple interest) or $A = \tfrac{1}{2}bh$ (the area of a triangle). When a question gives you values for some of the variables and asks for one that is missing, follow a fixed routine:

Write the formula down exactly as it is.
Substitute every value you are given, replacing those letters with their numbers.
If the unknown you want is not the subject, you now have an equation in that one unknown, so solve it using inverse operations.
Evaluate and write the answer to the accuracy and units the question asks for.

The schematic above runs through this with $I = Prn$ . Substituting $I = 420$ , $P = 3500$ and $r = 0.04$ leaves $420 = 140n$ , an ordinary equation in $n$ that you solve by dividing both sides by $140$ to get $n = 3$ . The key realisation is that step 3 is not a new skill: once the knowns are in, you are simply solving a linear equation, which you already know how to do.

Why you do not need to rearrange first

There are two valid ways to find a non-subject unknown, and it is worth knowing both. You can change the subject of the formula first (rearrange it algebraically so the unknown is alone) and only then substitute. Or you can substitute first and solve the equation that is left. For Maths Standard, substitute first is almost always the safer choice. Once the numbers are in you are working with an equation full of digits rather than letters, and arithmetic slips are easier to catch than algebra slips. Rearranging first pays off only when you have to find the same unknown for several different sets of numbers, because then one rearrangement serves every case. Both routes give the identical answer. This page teaches the substitute-first route, and changing the subject is its own dot point.

Tell which job you are doing

Before you start, look at where the unknown sits in the formula:

The unknown is the subject (alone on the left, like $A$ in $A = \tfrac{1}{2}bh$ when you are asked for the area). Then you only substitute and evaluate, with no equation to solve. This is the plain substitution dot point.
The unknown is not the subject (anywhere else, like $b$ or $h$ in $A = \tfrac{1}{2}bh$ ). Then you substitute the knowns and solve the equation that remains. This is the dot point on this page.

Both start with substitution; the difference is whether any solving is left to do afterwards. Spotting which case you are in stops you "solving" when there was nothing to solve, and stops you stopping early when there was.

Simplify the known numbers before you solve

After substituting, collect the known numbers into a single coefficient before you start undoing operations. In $A = \tfrac{1}{2}bh$ with $A = 96$ and $h = 12$ , substituting gives $96 = \tfrac{1}{2} \times b \times 12$ ; tidy the right side to $96 = 6b$ first, and the solve is a single division. In the trapezium formula $A = \tfrac{1}{2}(a + b)h$ with the parallel sides known, work out the bracket and the half before touching the unknown. Simplifying first turns an equation that looks complicated into one of the one-step or two-step forms you already recognise, and it removes most of the places a sign or a factor can go astray.

Then it is just inverse operations

Once the knowns are in and tidied, the equation is linear in the unknown and you solve it exactly as in the solving-linear-equations dot point: peel the operations off the unknown in reverse order, doing the same operation to both sides to keep it balanced. If the unknown has been multiplied and then had a number added, undo the addition first and the multiplication second. The hire-cost stage sequence further down shows this happening one line at a time after the substitution.

Rounding and units

Two finishing habits earn the last marks. Round only at the very end, and only to the accuracy the question states, for example "to two decimal places" or "to the nearest dollar". If you round partway through and then use that rounded number, the final digit can come out wrong. Also write the unit on the answer, taken from what the variable measures: a length comes out in metres or centimetres, an interest amount in dollars, a time in hours, a rate as a percentage. A bare number with no unit is an incomplete answer in an applied question.

How exam questions ask about solving after substitution

The wording tells you both which formula to use and that the unknown is not the subject, so solving will be needed:

"Use the formula ... to find ..." is the standard instruction. Write the formula, substitute the given values, and if the quantity asked for is not the subject, solve for it.
A formula is given with values for some letters and one missing. That missing letter is the unknown; substitute the rest, then solve.
"Find the value of $h$ / the length of the base / the rate / the number of years..." names the unknown explicitly. Check whether it is the subject (just evaluate) or not (substitute then solve).
An applied context that supplies a formula ("the cost is given by $C = \ldots$ ", "the area is $A = \ldots$ ") and then gives the result ("a job cost $425", "the area is $96$ m $^2$ ") wants you to substitute the result and the other knowns, then solve for the remaining quantity.
"Answer correct to two decimal places / the nearest whole number" is a rounding instruction: keep full accuracy while solving and round only the final value.
A two-part question often asks you to solve for a non-subject unknown in one part and to evaluate the subject in another. Read each part to see which job it is.

Solving for a non-subject unknown in six Australian contexts

Six problems, each giving a familiar formula with one value missing where the unknown is not the subject. Every one follows the same routine: write the formula, substitute the knowns, simplify the numbers, then solve the equation that is left and check.

Find the travel time from the distance-speed formula

A delivery van must cover $405$ km on a country run at an average speed of $90$ km/h. Using $D = ST$ , find the time $T$ the run will take.

Write the formula and substitute. The unknown is $T$ , which is not the subject, so substitute $D = 405$ and $S = 90$ :

405 = 90 \times T

Solve for $T$ . Divide both sides by the speed $90$ :

T = \frac{405}{90} = 4.5

State and check. The run takes $4.5$ hours, that is $4$ hours and $30$ minutes. Check: $D = 90 \times 4.5 = 405$ km, which matches. The units are consistent: kilometres divided by km/h leaves hours.

Find the hours worked from a tradie's cost formula

A plumber charges a $70 call-out fee plus $90 an hour, so a job lasting $n$ hours costs $C = 90n + 70$ dollars. A particular job cost $340. How many hours did it take?

Write the formula and substitute. The unknown is $n$ , which is not the subject, so substitute $C = 340$ :

340 = 90n + 70

Undo the addition first. Subtract the call-out fee $70$ from both sides:

340 - 70 = 90n \;\Rightarrow\; 270 = 90n

Undo the multiplication. Divide both sides by the hourly rate $90$ :

n = \frac{270}{90} = 3

State and check. The job took $3$ hours. Check: $90 \times 3 + 70 = 270 + 70 = 340$ , which matches the bill, so $n = 3$ hours.

Find a rectangle's width from the perimeter formula

A rectangular swimming pool has a perimeter of $68$ m and a length of $22$ m. Using $P = 2(L + W)$ , find the width $W$ .

Write the formula and substitute. The unknown is $W$ , which is not the subject, so substitute $P = 68$ and $L = 22$ :

68 = 2(22 + W)

Choose a first move. Because $68$ divides exactly by $2$ , divide both sides by $2$ to remove the bracket in one step:

\frac{68}{2} = 22 + W \;\Rightarrow\; 34 = 22 + W

Undo the addition. Subtract $22$ from both sides:

W = 34 - 22 = 12

Confirm with the expand route. Expanding instead gives $68 = 44 + 2W$ , so $2W = 68 - 44 = 24$ and $W = 24 \div 2 = 12$ , the same answer.

State and check. The width is $12$ m. Check: $P = 2(22 + 12) = 2 \times 34 = 68$ m, which matches, so $W = 12$ m.

Find a triangle's height from the area formula

A triangular shade sail has an area of $84$ cm $^2$ and a base of $24$ cm. Using $A = \tfrac{1}{2}bh$ , find the perpendicular height $h$ .

Write the formula and substitute. The unknown is $h$ , which is not the subject, so substitute $A = 84$ and $b = 24$ :

84 = \frac{1}{2} \times 24 \times h

Simplify the known numbers first. Work out $\tfrac{1}{2} \times 24 = 12$ , so the equation becomes

84 = 12h

Solve for $h$ . Divide both sides by $12$ :

h = \frac{84}{12} = 7

State and check. The height is $7$ cm. Check: $A = \tfrac{1}{2} \times 24 \times 7 = 12 \times 7 = 84$ cm $^2$ , which matches, so $h = 7$ cm.

Find the number of years from the simple interest formula

An investment of $3500 earns $420 of simple interest at a rate of $4\%$ per year. Using $I = Prn$ , find the number of years $n$ the money was invested.

Write the rate as a decimal. A rate of $4\%$ is $4 \div 100 = 0.04$ .

Write the formula and substitute. The unknown is $n$ , which is not the subject, so substitute $I = 420$ , $P = 3500$ and $r = 0.04$ :

420 = 3500 \times 0.04 \times n

Simplify the known numbers. Multiply $3500 \times 0.04 = 140$ , so

420 = 140n

Solve for $n$ . Divide both sides by $140$ :

n = \frac{420}{140} = 3

State and check. The money was invested for $3$ years. Check: $3500 \times 0.04 \times 3 = 140 \times 3 = 420$ , which matches. This is the scenario in the diagram at the top of the page.

Find the Fahrenheit reading from the temperature formula

The formula to convert a Fahrenheit temperature $F$ to Celsius is $C = \tfrac{5}{9}(F - 32)$ . A recipe specifies an oven temperature of $35\degree$ C. What is this in degrees Fahrenheit?

Write the formula and substitute. The unknown is $F$ , which is not the subject, so substitute $C = 35$ :

35 = \frac{5}{9}(F - 32)

Undo the multiplication by $\tfrac{5}{9}$ . Multiply both sides by $9$ then divide by $5$ (the same as multiplying by $\tfrac{9}{5}$ ):

35 \times \frac{9}{5} = F - 32 \;\Rightarrow\; 63 = F - 32

Undo the subtraction. Add $32$ to both sides:

F = 63 + 32 = 95

State and check. The temperature is $95\degree$ F. Check: $\tfrac{5}{9}(95 - 32) = \tfrac{5}{9} \times 63 = 5 \times 7 = 35\degree$ C, which matches, so $F = 95\degree$ F.

Final answers: the delivery run takes $4.5$ hours; the plumbing job took $3$ hours; the pool is $12$ m wide; the shade sail is $7$ cm high; the money was invested for $3$ years; and $35\degree$ C is $95\degree$ F.

Solving after substitution: a stage-by-stage example

When the unknown has more than one operation on it, the solve after substituting is a two-step (or multi-step) equation, undone in reverse order. The four panels below work through a hire-cost formula end to end. A landscaping firm hires out a turf cutter for a $120 fixed fee plus $45 an hour, so a hire of $t$ hours costs $C = 45t + 120$ dollars. A customer was charged $480; the panels find how many hours $t$ they hired it for.

Stage 1, substitute the known cost. The unknown is $t$ , which is not the subject, so put the known total $C = 480$ into the formula. That leaves a two-step equation in $t$ .

Stage 2, undo the addition. The constant $120$ is added last, so it comes off first. Subtract $120$ from both sides, keeping the equation balanced. The right side leaves $45t$ and the left becomes $480 - 120 = 360$ .

Stage 3, undo the multiplication. The unknown $t$ is multiplied by the hourly rate $45$ , so divide both sides by $45$ . This isolates $t$ on the left, with $360 \div 45$ on the right.

Stage 4, state and check. Evaluating $360 \div 45$ gives $t = 8$ , so the turf cutter was hired for $8$ hours. Substitute back into the original to confirm: $45 \times 8 + 120 = 360 + 120 = 480$ , which matches the charge.

Always check by substituting back

Every equation you solve after substitution can be verified the same way. Put your answer back into the original formula (with all the known values in place), evaluate it, and confirm it produces the result the question gave. For the hire-cost example, $t = 8$ gives $45 \times 8 + 120 = 480$ , the stated charge, so the answer is right. The check costs a few seconds and catches almost every slip, whether in the substitution or in the solving, which is why it is worth doing in the exam.

Common traps

Stopping after substituting when there is still solving to do: If the unknown is not the subject, substituting the knowns only gets you to an equation; you must then solve it. Stopping early leaves the question unfinished.
Trying to solve when the unknown was the subject: If the unknown sits alone on the left, just substitute and evaluate. There is nothing to "solve".
Substituting the result in the wrong place: When a context gives a result (a total cost, an area, an interest amount), that value replaces the subject of the formula, not the unknown you are solving for.
Not turning a percentage rate into a decimal: In $I = Prn$ a rate of $4\%$ must go in as $0.04$ , not $4$ . Forgetting this scales the answer by $100$ .
Undoing operations in the wrong order: After substituting, undo addition and subtraction before multiplication and division, just as in any equation. For $480 = 45t + 120$ , subtract $120$ first, then divide by $45$ .
Multiplying only the first term in a bracket: If you expand a bracket like $2(L + W)$ after substituting, the multiplier hits every term: $2(22 + W) = 44 + 2W$ , not $44 + W$ .
Rounding too early: Keep full accuracy while solving and round only the final answer, to the accuracy the question states.
Dropping the units: Read the unit from what the variable measures and write it on the answer: metres, dollars, hours, degrees, or a percentage.

Exam technique

Always begin by writing the formula, then a substitution line with the known values in place; markers award method marks for the correct formula and substitution even before any solving. Decide whether the unknown is the subject (substitute and evaluate, done) or not (substitute, then solve), and do not stop until you have actually found the unknown. After substituting, simplify the known numbers into a single coefficient first, which turns the equation into a one-step or two-step form, then solve it with one operation per line, equals signs aligned, stating the operation beside each line (" $-120$ both sides", " $\div 45$ both sides"). Convert any percentage rate to a decimal before substituting. Round only at the very end to the stated accuracy, write the correct units, and finish by substituting your answer back into the original formula to confirm it reproduces the given result.

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 mobile phone plan charges a fixed monthly access fee plus a set amount for each gigabyte of data used, giving a monthly cost

C = 19 + 6d

dollars for

d

gigabytes. In March the bill came to $67. Using the formula, find the number of gigabytes of data used that month.

Show worked answer →

Substitute the known cost. The unknown is $d$ , which is not the subject, so put the known bill $C = 67$ into the formula:

67 = 19 + 6d

Undo the addition first. The access fee $19$ is added last, so subtract it from both sides:

67 - 19 = 6d \;\Rightarrow\; 48 = 6d

Undo the multiplication. Divide both sides by the per-gigabyte rate $6$ :

d = \dfrac{48}{6} = 8

State and check. The plan used $8$ GB of data in March. Check by substituting back into the original: $19 + 6 \times 8 = 19 + 48 = 67$ , which matches the bill, so $d = 8$ GB.

Markers award method marks for the substitution line $67 = 19 + 6d$ and reward undoing the $+19$ before dividing by $6$ . A common slip is dividing by $6$ first without removing the fixed fee, which gives a wrong answer.

2022 HSC-style3 marksThe surface area of a closed cylindrical tank is approximated by

S = 2\pi r(r + h)

, where

r

is the base radius and

h

is the height, both in metres. A tank has a radius of

r = 2

m and a surface area of

S = 40\pi

^2

. Find the height

h

of the tank.

Show worked answer →

Substitute the known values. The unknown is $h$ , which is not the subject, so substitute $S = 40\pi$ and $r = 2$ :

40\pi = 2\pi \times 2 \times (2 + h)

Simplify the known numbers first. On the right, $2\pi \times 2 = 4\pi$ , so the equation becomes

40\pi = 4\pi(2 + h)

Peel off the coefficient. Divide both sides by $4\pi$ to clear it in one step:

\dfrac{40\pi}{4\pi} = 2 + h \;\Rightarrow\; 10 = 2 + h

Undo the addition. Subtract $2$ from both sides:

h = 10 - 2 = 8

State and check. The height is $8$ m. Check: $2\pi \times 2 \times (2 + 8) = 4\pi \times 10 = 40\pi$ m $^2$ , which matches, so $h = 8$ m.

Markers reward simplifying $2\pi \times 2$ to $4\pi$ before solving and dividing the whole side by $4\pi$ . Dividing by $4\pi$ leaves a clean number because the $\pi$ cancels.

2023 HSC-style4 marksA spring stretches according to

F = kx

, where

F

is the applied force in newtons,

x

is the extension in metres and

k

is the spring constant. (a) A force of

F = 90

N stretches a spring by

x = 0.15

m. Find the spring constant

k

. (b) Using that value of

k

, find the extension produced by a force of

F = 150

Show worked answer →

Part (a) - substitute the known values. The unknown is $k$ , which is not the subject, so substitute $F = 90$ and $x = 0.15$ :

90 = k \times 0.15

Solve for $k$ . Divide both sides by $0.15$ :

k = \dfrac{90}{0.15} = 600

so the spring constant is $k = 600$ N/m. (Check: $600 \times 0.15 = 90$ N.)

Part (b) - substitute the new force with the found $k$ . The unknown is now $x$ , still not the subject, so substitute $F = 150$ and $k = 600$ :

150 = 600 \times x

Solve for $x$ . Divide both sides by $600$ :

x = \dfrac{150}{600} = 0.25

State and check. The extension is $0.25$ m. Check: $600 \times 0.25 = 150$ N, which matches the applied force.

Markers award separate method marks for each substitution line and reward carrying the exact value $k = 600$ into part (b) rather than a rounded figure. Both parts substitute first and then divide, because the unknown is never the subject.

Practice questions

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

foundation2 marksA market stall sells punnets of strawberries. The takings are given by

T = 4 + 1.5n

dollars, where

4

is a fixed setup amount and

n

is the number of punnets sold. One morning the takings were $25. How many punnets were sold?

Show worked solution →

Substitute the known value. The unknown is $n$ , which is not the subject, so put the known takings $T = 25$ into the formula:

25 = 4 + 1.5n

Undo the addition first. Subtract $4$ from both sides:

25 - 4 = 1.5n \;\Rightarrow\; 21 = 1.5n

Undo the multiplication. Divide both sides by $1.5$ :

n = \frac{21}{1.5} = 14

State and check. $14$ punnets were sold. Check: $4 + 1.5 \times 14 = 4 + 21 = 25$ , which matches the takings, so $n = 14$ is correct.

foundation2 marksSimple interest is given by

I = Prn

, where

P

is the principal,

r

is the annual interest rate as a decimal and

n

is the number of years. Find the interest rate

r

when $1200 earns $180 of interest over

3

years.

Show worked solution →

Substitute every known value. The unknown is $r$ , which is not the subject, so substitute $I = 180$ , $P = 1200$ and $n = 3$ :

180 = 1200 \times r \times 3

Simplify the known numbers. Multiply $1200 \times 3 = 3600$ , so

180 = 3600\,r

Solve for $r$ . Divide both sides by $3600$ :

r = \frac{180}{3600} = 0.05

State and check. As a percentage $r = 0.05 = 5\%$ per year. Check: $1200 \times 0.05 \times 3 = 180$ , which matches, so the rate is $5\%$ .

core3 marksThe perimeter of a rectangle is

P = 2(L + W)

. A rectangular courtyard has a perimeter of

54

m and a width of

11

m. Find its length

L

Show worked solution →

Substitute the known values. The unknown is $L$ , which is not the subject, so substitute $P = 54$ and $W = 11$ :

54 = 2(L + 11)

Choose a first move. Because $54$ divides exactly by $2$ , divide both sides by $2$ to peel off the bracket in one step:

\frac{54}{2} = L + 11 \;\Rightarrow\; 27 = L + 11

Undo the addition. Subtract $11$ from both sides:

L = 27 - 11 = 16

State and check. The length is $16$ m. Check: $P = 2(16 + 11) = 2 \times 27 = 54$ m, which matches, so $L = 16$ m.

core3 marksThe area of a triangle is

A = \tfrac{1}{2}bh

, where

b

is the base and

h

is the perpendicular height. A triangular sail has an area of

96

^2

and a perpendicular height of

12

m. Find the length of its base

b

Show worked solution →

Substitute the known values. The unknown is $b$ , which is not the subject, so substitute $A = 96$ and $h = 12$ :

96 = \frac{1}{2} \times b \times 12

Simplify the known numbers. Half of $12$ is $6$ , so the equation becomes

96 = 6b

Solve for $b$ . Divide both sides by $6$ :

b = \frac{96}{6} = 16

State and check. The base is $16$ m. Check: $A = \tfrac{1}{2} \times 16 \times 12 = 8 \times 12 = 96$ m $^2$ , which matches, so $b = 16$ m.

core2 marksDistance, speed and time are linked by

D = ST

, where

D

is distance,

S

is speed and

T

is time. A train covers

150

km in

2.5

hours. Find its average speed

S

in km/h.

Show worked solution →

Substitute the known values. The unknown is $S$ , which is not the subject, so substitute $D = 150$ and $T = 2.5$ :

150 = S \times 2.5

Solve for $S$ . Divide both sides by $2.5$ :

S = \frac{150}{2.5} = 60

State and check. The average speed is $60$ km/h. Check: $D = 60 \times 2.5 = 150$ km, which matches. The units are consistent because kilometres divided by hours gives km/h.

exam4 marksAn electrician charges a fixed call-out fee plus an hourly rate, giving a total cost

C = 110n + 95

dollars for a job lasting

n

hours. (a) A job cost $425. How many hours did it take? (b) Use the formula to find the cost of a job that takes

6

hours.

Show worked solution →

Part (a) - substitute the known cost. The unknown is $n$ , which is not the subject, so substitute $C = 425$ :

425 = 110n + 95

Undo the addition first. Subtract the call-out fee $95$ from both sides:

425 - 95 = 110n \;\Rightarrow\; 330 = 110n

Undo the multiplication. Divide both sides by the hourly rate $110$ :

n = \frac{330}{110} = 3

so the job took $3$ hours. (Check: $110 \times 3 + 95 = 330 + 95 = 425$ .)

Part (b) - here $n$ is known, so just substitute. This time the unknown $C$ is the subject, so no equation needs solving; substitute $n = 6$ and evaluate:

C = 110 \times 6 + 95 = 660 + 95 = 755

so a $6$ hour job costs $755. The contrast is the point: in (a) the unknown was not the subject so you substitute then solve, while in (b) the unknown is the subject so you substitute and evaluate.

exam4 marksThe area of a trapezium is

A = \tfrac{1}{2}(a + b)h

, where

a

and

b

are the two parallel sides and

h

is the perpendicular distance between them. A trapezium-shaped garden bed has an area of

120

^2

and parallel sides of

14

m and

26

m. Find the perpendicular distance

h

between the parallel sides.

Show worked solution →

Substitute every known value. The unknown is $h$ , which is not the subject, so substitute $A = 120$ , $a = 14$ and $b = 26$ :

120 = \frac{1}{2}(14 + 26)h

Simplify inside the bracket first. Add the parallel sides: $14 + 26 = 40$ . Then halve it: $\tfrac{1}{2} \times 40 = 20$ . The equation collapses to

120 = 20h

Solve for $h$ . Divide both sides by $20$ :

h = \frac{120}{20} = 6

State and check. The perpendicular distance is $6$ m. Check: $A = \tfrac{1}{2}(14 + 26) \times 6 = \tfrac{1}{2} \times 40 \times 6 = 20 \times 6 = 120$ m $^2$ , which matches, so $h = 6$ m. Simplifying the known numbers before solving turned a messy-looking equation into a one-step divide.

exam3 marksSimple interest is given by

I = Prn

. An investment earns $525 of interest at a rate of

3.5\%

per year over

5

years. Find the principal

P

that was invested.

Show worked solution →

Write the rate as a decimal. A rate of $3.5\%$ is $3.5 \div 100 = 0.035$ .

Substitute every known value. The unknown is $P$ , which is not the subject, so substitute $I = 525$ , $r = 0.035$ and $n = 5$ :

525 = P \times 0.035 \times 5

Simplify the known numbers. Multiply $0.035 \times 5 = 0.175$ , so

525 = 0.175\,P

Solve for $P$ . Divide both sides by $0.175$ :

P = \frac{525}{0.175} = 3000

State and check. The principal was $3000. Check: $3000 \times 0.035 \times 5 = 3000 \times 0.175 = 525$ , which matches, so $P =$ $3000.

What this dot point is asking

The answer

Substitute first, then solve

Why you do not need to rearrange first

Tell which job you are doing

Simplify the known numbers before you solve

Then it is just inverse operations

Rounding and units

How exam questions ask about solving after substitution

Solving after substitution: a stage-by-stage example

Always check by substituting back

Exam-style practice questions

Practice questions

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