Assign a letter to the unknown quantity, then convert the problem's constraints into algebraic expressions before linking them with an "=" sign. Geometry contexts are especially reliable because they give you a fixed constant to equate to: angles on a straight line sum to $180°$, angles around a point sum to $360°$, and opposite sides of a rectangle are equal.

Example: a straight line is split into three angles $x$, $2x$, and $3x$: $x+2x+3x=180$. Example: three child tickets at $\pounds x$ and five adult tickets at $\pounds9$ total $\pounds63$: $3x+5(9)=63$.

1. Expand any brackets.
2. Collect like terms on each side.
3. Move all variable terms to one side, all constants to the other (performing the same inverse operation on both sides each time — never "jump" a term across the equals sign without writing the operation).
4. Divide by the coefficient of $x$ to isolate it.

When expanding a bracket with a negative number outside, e.g. $-4(x-y)$, distribute the sign to every term inside: the result is $-4x+4y$, not $-4x-4y$. This single sign error is the most common mistake in the whole topic.

Subtract whichever term keeps the resulting coefficient positive — this avoids an unnecessary negative coefficient later. For $5x+3=2x+15$, subtract $2x$ (not $5x$) from both sides to get $3x+3=15$.

Multiply every term by the lowest common denominator to clear the fractions entirely. For $\dfrac{x}{4}+\dfrac{x}{2}=3$, the LCD is 4: multiplying through gives $x+2x=12$, a plain linear equation.

Elimination: scale one or both equations so a variable's coefficients match, then add or subtract to eliminate it — same signs, subtract; opposite signs, add ("same-take, opposite-plus"). Apply the chosen operation to all three parts of each equation (the $x$-term, the $y$-term, and the constant) — forgetting the constant is a frequent slip. Substitute the result back into either original equation to find the second variable.

Substitution: if one equation already isolates a variable (e.g. $y = \dots$), substitute that whole expression — in brackets — into the other equation.

Rearrange into $x^2+bx+c=0$, factorise into two linear brackets, then apply the Zero Product Rule: since the product of the brackets is zero, each bracket individually must equal zero, giving two solutions for $x$.

Solve the simultaneous equations $3x + 2y = 16$ and $5x - 2y = 8$.

1. The $y$-coefficients ($+2$ and $-2$) have opposite signs, so add the equations: $8x = 24$.
2. Solve: $x = 3$.
3. Substitute into the first equation: $3(3)+2y=16 \Rightarrow 9+2y=16 \Rightarrow y=3.5$.
4. Check in the second equation: $5(3)-2(3.5)=15-7=8$ ✓.