Solve a pair of simultaneous linear equations in two unknowns by elimination and by substitution, and form them from word problems

What this dot point is asking

SEAB wants you to solve a pair of linear equations in two unknowns at the same time, using either elimination or substitution, and to form such a pair from a word problem. The solution is the single pair of values that makes both equations true. Graphically, it is the point where the two lines cross.

The answer

What "simultaneous" means

One linear equation in two unknowns has infinitely many solutions. Two equations together usually pin down exactly one pair $(x, y)$ that satisfies both. There are two reliable methods.

The elimination method

The idea is to add or subtract the equations so that one unknown disappears.

1. If a chosen unknown has coefficients that are equal or opposite, add (for opposite signs) or subtract (for equal signs) to remove it.
2. If not, multiply one or both equations by a number so the coefficients match.
3. Solve the resulting single equation, then substitute back for the other unknown.

For $3x + 2y = 16$ and $3x - y = 7$, the $x$ coefficients are equal, so subtract: $3y = 9$, giving $y = 3$. Then $3x - 3 = 7$ gives $x = \dfrac{10}{3}$.

The substitution method

The idea is to make one unknown the subject of one equation, then put it into the other.

1. Rearrange one equation to get one unknown alone, for example $y = \dots$.
2. Substitute that expression into the other equation.
3. Solve, then substitute back.

This method is neat when one equation already has an unknown with coefficient $1$.

Checking

Substitute both values into the equation you did not use for the final substitution. If it balances, the pair is correct.

Forming a pair from words

Define two letters for the two unknowns, then write one equation for each piece of information. Two facts give two equations, which is exactly what you need.

Examples in context

Example 1. Coins in a tin. A tin holds $20$ coins, all either $20$-cent or $50$-cent coins, worth \7.60$in total. Let there be$x$twenty-cent coins and$y$fifty-cent coins. Then$x + y = 20$and$0.20x + 0.50y = 7.60$. Solving gives$x = 8$and$y = 12$. The first equation counts the coins, the second totals their value.

Example 2. Where two lines cross. The lines $y = 2x + 1$ and $y = -x + 7$ meet where both $y$ values are equal: $2x + 1 = -x + 7$, so $3x = 6$ and $x = 2$, giving $y = 5$. The crossing point $(2, 5)$ is the simultaneous solution, which links this topic to straight-line graphs.

Try this

• Cue. Solve $x + y = 10$ and $x - y = 4$. Add to get $2x = 14$, so $x = 7$ and $y = 3$.
• Cue. Solve $y = x + 2$ and $3x + y = 14$ by substitution. Then $3x + x + 2 = 14$, so $x = 3$ and $y = 5$.
• Cue. Solve $2x + 3y = 13$ and $2x + y = 7$. Subtract to get $2y = 6$, so $y = 3$ and $x = 2$.