Solve linear equations in one unknown, including those with fractions, and solve simultaneous linear equations by substitution and elimination

What this dot point is asking

SEAB wants you to solve linear equations in one unknown, including equations with fractions, and to solve a pair of simultaneous linear equations in two unknowns by substitution or elimination. These are core algebraic skills that appear throughout the syllabus, from coordinate geometry to word problems.

The answer

Solving a linear equation

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 $x = 4$.

Equations with fractions

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.

Simultaneous equations by elimination

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).

Simultaneous equations by substitution

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 $y = 2x - 1$.

Checking the solution

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.

Examples in context

Example 1. Two unknown prices. If $3$ pens and $2$ rulers cost $7.40$ dollars and $1$ pen and $4$ rulers cost $6.80$ dollars, the two relationships form simultaneous equations whose solution gives the price of each item. Many word problems reduce to exactly this.

Example 2. Intersecting lines. Solving $y = 2x + 1$ and $y = -x + 7$ simultaneously gives the point where the two straight lines cross, linking algebra to coordinate geometry. The solution $(2, 5)$ is the intersection point.

Try this

Q1. Solve $4(x - 2) = 2x + 6$. [2 marks]

• Cue. Expand: $4x - 8 = 2x + 6$, so $2x = 14$ and $x = 7$.

Q2. Solve $\dfrac{x}{2} + \dfrac{x}{3} = 5$. [2 marks]

• Cue. Multiply through by $6$: $3x + 2x = 30$, so $5x = 30$ and $x = 6$.

Q3. Solve $x + y = 10$ and $x - y = 4$. [3 marks]

• Cue. Add the equations: $2x = 14$, so $x = 7$, then $y = 3$.