Solving simultaneous linear equations in two variables algebraically by elimination and substitution, and graphically as the point of intersection.

What this dot point is asking

The SQA wants you to solve a pair of simultaneous linear equations in two unknowns, by elimination or substitution, and to understand the solution as the point where the two lines cross. You should also be able to set up simultaneous equations from a worded context.

Solving by elimination

Elimination removes one variable by adding or subtracting the equations. If the matching coefficients have opposite signs, add the equations; if they have the same sign, subtract.

Often the coefficients do not match at first. Multiply one or both equations by a suitable number so that one variable has equal-sized coefficients, then eliminate.

Solving by substitution

When one equation already has a variable on its own (or is easy to rearrange), substitution can be quicker: rearrange one equation for a variable, then substitute it into the other.

The graphical meaning

Each linear equation is a straight line. The simultaneous solution is the single point $(x, y)$ that lies on both lines, that is, their point of intersection. If the lines are parallel they never meet, so there is no solution; if they are the same line there are infinitely many.

When both equations need scaling

Sometimes neither variable matches across the two equations, so you multiply both equations to create a matching coefficient. Choose the lowest common multiple of the two coefficients of one variable.

Always check your solution by substituting both values into the equation you did not use to find the last variable; if both sides agree, the answer is correct.

Examples in context

Simultaneous equations solve "two unknowns, two facts" problems. If $3$ coffees and $2$ teas cost $\pounds 13$, and $1$ coffee and $1$ tea cost $\pounds 5$, then with coffee $c$ and tea $t$ you have $3c + 2t = 13$ and $c + t = 5$. Eliminating gives $c = 3$ and $t = 2$, so a coffee is $\pounds 3$ and a tea is $\pounds 2$. Always define your variables clearly before forming the equations.

Try this

Q1. Solve $x + y = 10$ and $x - y = 4$. [3 marks]

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

Q2. Solve $2x + 3y = 12$ and $x = 3$. [2 marks]

• Cue. $6 + 3y = 12$, so $y = 2$.

Q3. Solve $y = x + 2$ and $2x + y = 8$. [3 marks]

• Cue. $2x + x + 2 = 8$, so $x = 2$, $y = 4$.