Linear equations and formulas | Year 11

01 / LEARN

The idea, step by step

Keep the equation balanced

An equation states that two expressions have equal values. Undo the operations around the unknown in reverse order, doing the same operation on both sides. Keep each step visible so that a sign or arithmetic error can be found.

A neat sequence of steps is not itself a check. Substitute the proposed value into the original equation and compare both sides. If they differ, return to the first step that changed the equality.

3 x + 7 = 25

Choose the subject of a formula

A formula relates quantities. Before substituting, identify the quantity to find and write the units of every input. In a cost rule, a fixed charge has a different role from a per-unit rate.

To make a different quantity the subject, isolate its letter while treating the others as known. Keep a whole sum or difference together when dividing it. A restriction such as a nonzero rate matters.

C = F + r d

, so

d = (C - F) / r

for

r \neq 0

A common trap

Dividing only one term of a sum while rearranging.

Instead: Subtract the whole fixed part first, then divide the whole remaining expression.

02 / WORKED EXAMPLES

See the thinking

Example 1

Solve and check

Solve $3 x + 7 = 25$ .

Subtract 7 from both sides: $3 x = 18$ .
Divide by 3: $x = 6$ .
Check: $3 (6) + 7 = 25$ .

x = 6

Example 2

Reverse a delivery rule

A delivery costs $C = 4 + 2.5 d$ dollars for a distance of $d$ km. A bill is 24 dollars. Find the distance.

Substitute $C = 24$ : $24 = 4 + 2.5 d$ .
Subtract 4, then divide by 2.5: $d = 20 / 2.5 = 8$ .
The unknown is distance, so attach km. Check $4 + 2.5 (8) = 24$ .

8 km

Check your understanding

Solve $5 x - 4 = 31$ .

Show answer

$x = 7$ .

Add 4 to obtain $5 x = 35$ , then divide by 5.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $4 x + 9 = 37$ .
2 marks
Solve $7 - 2 x = - 5$ .
2 marks
Solve $3 (x + 2) = 21$ and check.
2 marks
Make $h$ the subject of $A = b h / 2$ , where $b \neq 0$ .
2 marks
Use $C = 18 + 6 n$ to find the cost for five items.
2 marks
A tank has $V = 150 - 12 t$ litres after $t$ minutes. When does it hold 90 L?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Solve $4 x + 9 = 37$ .
2 marks
Subtract 9: $4 x = 28$ .
Divide by 4.
$x = 7$ .
Solve $7 - 2 x = - 5$ .
2 marks
Subtract 7: $- 2 x = - 12$ .
Divide by $- 2$ .
$x = 6$ .
Solve $3 (x + 2) = 21$ and check.
2 marks
Divide by 3: $x + 2 = 7$ .
Subtract 2; $3 (5 + 2) = 21$ .
$x = 5$ .
Make $h$ the subject of $A = b h / 2$ , where $b \neq 0$ .
2 marks
Multiply both sides by 2.
Divide the entire expression $2 A$ by $b$ .
$h = 2 A / b$ .
Use $C = 18 + 6 n$ to find the cost for five items.
2 marks
$C = 18 + 6 (5) = 48$ .
48 dollars.
A tank has $V = 150 - 12 t$ litres after $t$ minutes. When does it hold 90 L?
2 marks
$90 = 150 - 12 t$ gives $12 t = 60$ .
Thus $t = 5$ .
5 min.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: algebraic and graphical modelling of linear/exponential change, with interpretation and evaluation.

Official curriculum source ↗

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