Quick questions on Linear and simultaneous equations explained: O-Level E-Maths

A linear equation has the unknown to the first power only. The goal is to isolate the unknown by doing the same operation to both sides: expand any brackets, collect the unknown on one side and the numbers on the other, then divide. For $5x - 3 = 2x + 9$, collecting gives $3x = 12$, so $x = 4$.

Clear fractions first by multiplying every term by the lowest common denominator, or by cross multiplying when each side is a single fraction. This turns a fractional equation into an ordinary linear one before you solve.

When two equations share the same two unknowns, elimination adds or subtracts multiples of the equations so that one unknown cancels. Make the coefficients of one unknown equal in size, then add (if the signs are opposite) or subtract (if the signs are the same).

Substitution rearranges one equation to make one unknown the subject, then puts that expression into the other equation. This works well when one equation already has a unknown with coefficient $1$, such as $y = 2x - 1$.

A solution to a pair of simultaneous equations must satisfy both equations. Substitute your values back into the equation you did not use to find them, as a check.

Subtracting changes the sign of every term in the second equation, a frequent slip.

A quick substitution into the unused equation catches most arithmetic mistakes.

Solve $4(x - 2) = 2x + 6$. [2 marks]

Solve $\dfrac{x}{2} + \dfrac{x}{3} = 5$. [2 marks]

Solve $x + y = 10$ and $x - y = 4$. [3 marks]