Topics

Linear equations and rearranging formulae

Covers AS91261

Solving a linear equation

A linear equation has the variable to the power 1 only — no $x^2$ , no $\frac{1}{x}$ .
The principle: whatever you do to one side, do to the other. The equation stays balanced.
The order of operations for solving:
- Expand any brackets.
- Clear any fractions by multiplying every term by the common denominator.
- Collect the variable terms on one side and the numbers on the other.
- Divide by the coefficient of the variable.
Worked through — solve $3(2x - 5) = 4x + 7$ :
- Expand: $6x - 15 = 4x + 7$
- Collect: $6x - 4x = 7 + 15$ , so $2x = 22$
- Divide: $x = 11$
Check by substituting back. $3(22 - 5) = 51$ and $4(11) + 7 = 51$ ✓

Equations with fractions

Multiply every term by the lowest common denominator — including terms that are not fractions.
Worked through — solve $\dfrac{x}{3} + \dfrac{x-1}{4} = 5$ :
- The LCD is 12. Multiply every term: $4x + 3(x-1) = 60$
- Expand: $4x + 3x - 3 = 60$
- Collect: $7x = 63$ , so $x = 9$
The whole numerator gets multiplied, which is why $\frac{x-1}{4} \times 12 = 3(x-1)$ and not $3x - 1$ .

Rearranging a formula

Making a letter the subject uses exactly the same moves as solving an equation — the only difference is that the answer contains letters.
The method:
- Identify the letter you want.
- Clear fractions and roots first.
- Collect every term containing that letter on one side.
- Factorise it out if it appears more than once.
- Divide by whatever multiplies it.
Worked through — make $r$ the subject of $A = \pi r^2 h$ :
- Divide: $\dfrac{A}{\pi h} = r^2$
- Square root: $r = \sqrt{\dfrac{A}{\pi h}}$
- Note that $r$ is a radius, so only the positive root is meaningful — say so.

When the letter appears twice

Factorise it out. This is the step that separates Achieved from Merit on rearranging questions.
Worked through — make $x$ the subject of $ax + b = cx + d$ :
- Collect: $ax - cx = d - b$
- Factorise: $x(a - c) = d - b$
- Divide: $x = \dfrac{d-b}{a-c}$
- Note the restriction $a \neq c$ — otherwise you would divide by zero.

Inequalities

Inequalities are solved the same way, with one exception:

$\textbf{Multiplying or dividing by a negative number reverses the inequality sign.}$

$-2x > 6$ gives $x < -3$ , not $x > -3$ .
Adding and subtracting never reverse it, whatever the sign of the number.

Substituting carefully

The 2025 report lists "substituted incorrectly into algebraic expressions" among the Not Achieved behaviours.
Use brackets around every substituted value, especially negatives:
- To evaluate $-x^2$ at $x = -4$ : $-(-4)^2 = -(16) = -16$
- To evaluate $(-x)^2$ at $x = -4$ : $(4)^2 = 16$
These are different expressions and the brackets are what keep them apart.

Tips and Tricks

Show your steps. The specification requires that "algebraic techniques must be shown, as opposed to simply providing the correct answer". A bare answer to a linear equation earns nothing even when it is right.

When you clear fractions, multiply every term — including the whole numerator of each fraction, in brackets. Multiplying only part of a numerator is the most common slip.

If the letter appears twice, factorise. Trying to divide before collecting leaves the letter on both sides and the rearrangement stalls.

Reverse the inequality when you multiply or divide by a negative. A safer habit is to move terms so the variable ends up with a positive coefficient, which avoids the issue entirely.

Brackets when substituting negatives. The report names incorrect substitution and mishandled negatives as two separate Not Achieved behaviours, and both are prevented by the same habit.

Check by substituting your answer back into the original equation. One line, and it catches everything.

Worked ExampleRearranging when the letter appears twice

Make $t$ the subject of the formula $\;S = \dfrac{t + 3}{t - 2}$ .

Step 1 — Clear the fraction first

Nothing can be collected while $t$ is trapped inside a denominator. Multiply both sides by $(t-2)$ :

$S(t - 2) = t + 3$

Step 2 — Expand

$St - 2S = t + 3$

Step 3 — Collect the $t$ terms on one side

$t$ now appears twice, on both sides. Move both to the left and everything else to the right:

$St - t = 3 + 2S$

Step 4 — Factorise out $t$

This is the key step. Without it, $t$ cannot be isolated:

$t(S - 1) = 3 + 2S$

Step 5 — Divide

$t = \frac{3 + 2S}{S - 1}$

Step 6 — State any restriction

The denominator cannot be zero, so:

$S \neq 1$

Interpreting this: $S = 1$ would require $t + 3 = t - 2$ , which is impossible — so the restriction is genuine, not an artefact of the algebra.

Step 7 — Check with a numerical value

Take $t = 5$ in the original formula:

$S = \frac{5+3}{5-2} = \frac{8}{3}$

Now substitute $S = \frac{8}{3}$ into the rearranged formula:

$t = \frac{3 + 2\left(\frac{8}{3}\right)}{\frac{8}{3} - 1} = \frac{3 + \frac{16}{3}}{\frac{5}{3}} = \frac{\frac{25}{3}}{\frac{5}{3}} = 5 \;\checkmark$

Linear equations and rearranging formulae

Covers AS91261

Solving a linear equation

A linear equation has the variable to the power 1 only — no $x^2$ , no $\frac{1}{x}$ .
The principle: whatever you do to one side, do to the other. The equation stays balanced.
The order of operations for solving:
- Expand any brackets.
- Clear any fractions by multiplying every term by the common denominator.
- Collect the variable terms on one side and the numbers on the other.
- Divide by the coefficient of the variable.
Worked through — solve $3(2x - 5) = 4x + 7$ :
- Expand: $6x - 15 = 4x + 7$
- Collect: $6x - 4x = 7 + 15$ , so $2x = 22$
- Divide: $x = 11$
Check by substituting back. $3(22 - 5) = 51$ and $4(11) + 7 = 51$ ✓

Equations with fractions

Multiply every term by the lowest common denominator — including terms that are not fractions.
Worked through — solve $\dfrac{x}{3} + \dfrac{x-1}{4} = 5$ :
- The LCD is 12. Multiply every term: $4x + 3(x-1) = 60$
- Expand: $4x + 3x - 3 = 60$
- Collect: $7x = 63$ , so $x = 9$
The whole numerator gets multiplied, which is why $\frac{x-1}{4} \times 12 = 3(x-1)$ and not $3x - 1$ .

Rearranging a formula

Making a letter the subject uses exactly the same moves as solving an equation — the only difference is that the answer contains letters.
The method:
- Identify the letter you want.
- Clear fractions and roots first.
- Collect every term containing that letter on one side.
- Factorise it out if it appears more than once.
- Divide by whatever multiplies it.
Worked through — make $r$ the subject of $A = \pi r^2 h$ :
- Divide: $\dfrac{A}{\pi h} = r^2$
- Square root: $r = \sqrt{\dfrac{A}{\pi h}}$
- Note that $r$ is a radius, so only the positive root is meaningful — say so.

When the letter appears twice

Factorise it out. This is the step that separates Achieved from Merit on rearranging questions.
Worked through — make $x$ the subject of $ax + b = cx + d$ :
- Collect: $ax - cx = d - b$
- Factorise: $x(a - c) = d - b$
- Divide: $x = \dfrac{d-b}{a-c}$
- Note the restriction $a \neq c$ — otherwise you would divide by zero.

Inequalities

Inequalities are solved the same way, with one exception:

$\textbf{Multiplying or dividing by a negative number reverses the inequality sign.}$

$-2x > 6$ gives $x < -3$ , not $x > -3$ .
Adding and subtracting never reverse it, whatever the sign of the number.

Substituting carefully

The 2025 report lists "substituted incorrectly into algebraic expressions" among the Not Achieved behaviours.
Use brackets around every substituted value, especially negatives:
- To evaluate $-x^2$ at $x = -4$ : $-(-4)^2 = -(16) = -16$
- To evaluate $(-x)^2$ at $x = -4$ : $(4)^2 = 16$
These are different expressions and the brackets are what keep them apart.

Tips and Tricks

When you clear fractions, multiply every term — including the whole numerator of each fraction, in brackets. Multiplying only part of a numerator is the most common slip.

If the letter appears twice, factorise. Trying to divide before collecting leaves the letter on both sides and the rearrangement stalls.

Reverse the inequality when you multiply or divide by a negative. A safer habit is to move terms so the variable ends up with a positive coefficient, which avoids the issue entirely.

Brackets when substituting negatives. The report names incorrect substitution and mishandled negatives as two separate Not Achieved behaviours, and both are prevented by the same habit.

Check by substituting your answer back into the original equation. One line, and it catches everything.

Worked ExampleRearranging when the letter appears twice

Make $t$ the subject of the formula $\;S = \dfrac{t + 3}{t - 2}$ .

Step 1 — Clear the fraction first

Nothing can be collected while $t$ is trapped inside a denominator. Multiply both sides by $(t-2)$ :

$S(t - 2) = t + 3$

Step 2 — Expand

$St - 2S = t + 3$

Step 3 — Collect the $t$ terms on one side

$t$ now appears twice, on both sides. Move both to the left and everything else to the right:

$St - t = 3 + 2S$

Step 4 — Factorise out $t$

This is the key step. Without it, $t$ cannot be isolated:

$t(S - 1) = 3 + 2S$

Step 5 — Divide

$t = \frac{3 + 2S}{S - 1}$

Step 6 — State any restriction

The denominator cannot be zero, so:

$S \neq 1$

Interpreting this: $S = 1$ would require $t + 3 = t - 2$ , which is impossible — so the restriction is genuine, not an artefact of the algebra.

Step 7 — Check with a numerical value

Take $t = 5$ in the original formula:

$S = \frac{5+3}{5-2} = \frac{8}{3}$

Now substitute $S = \frac{8}{3}$ into the rearranged formula:

$t = \frac{3 + 2\left(\frac{8}{3}\right)}{\frac{8}{3} - 1} = \frac{3 + \frac{16}{3}}{\frac{5}{3}} = \frac{\frac{25}{3}}{\frac{5}{3}} = 5 \;\checkmark$

Linear equations and rearranging formulae

Solving a linear equation

Equations with fractions

Rearranging a formula

When the letter appears twice

Inequalities

Substituting carefully

Step 1 — Clear the fraction first

Step 2 — Expand

Step 3 — Collect the ttt terms on one side

Step 4 — Factorise out ttt

Step 5 — Divide

Step 6 — State any restriction

Step 7 — Check with a numerical value

Linear equations and rearranging formulae

Solving a linear equation

Equations with fractions

Rearranging a formula

When the letter appears twice

Inequalities

Substituting carefully

Step 1 — Clear the fraction first

Step 2 — Expand

Step 3 — Collect the ttt terms on one side

Step 4 — Factorise out ttt

Step 5 — Divide

Step 6 — State any restriction

Step 7 — Check with a numerical value

Step 3 — Collect the $t$ terms on one side

Step 4 — Factorise out $t$

Step 3 — Collect the $t$ terms on one side

Step 4 — Factorise out $t$