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Simultaneous Equations

Two unknowns need two equations. Solving them together finds the single pair of values that makes both true at once, which is also the point where their graphs cross.

MYP 4-5ExtendedAlgebraCriteria A · C · D~12 min read

Written by Fayek Jawad Haider · MYP student

The core idea

A single equation like \(x + y = 7\) has endless solutions: \((1,6)\), \((2,5)\), \((3,4)\) and so on. Add a second condition and usually only one pair survives. Graphically, each equation is a line, and the solution is the single point where the two lines intersect.

There are two reliable algebraic methods. Both aim to get rid of one unknown so you can solve for the other.

Method 1: elimination

Line the equations up and add or subtract them so that one variable cancels. This works best when a variable has matching coefficients (or you can make them match).

Worked example

Solve \(2x + y = 8\) and \(3x - y = 7\).

1
The \(y\) terms are \(+y\) and \(-y\), opposite signs, so adding the equations eliminates \(y\): \((2x+3x) + (y - y) = 8 + 7\).
2
That gives \(5x = 15\), so \(x = 3\).
3
Substitute \(x=3\) into the first equation: \(2(3) + y = 8 \Rightarrow y = 2\).
\(x = 3,\; y = 2\)

Same signs? Subtract. Opposite signs? Add.

If the matching terms are \(+2y\) and \(+2y\), subtract to cancel. If they are \(+y\) and \(-y\), add. Getting this the wrong way round is the usual elimination error.

Method 2: substitution

Rearrange one equation to make a variable the subject, then substitute that expression into the other. This shines when one equation already says "\(y = \ldots\)".

Worked example

Solve \(y = 2x + 1\) and \(3x + y = 11\).

1
The first equation already gives \(y\). Substitute it into the second: \(3x + (2x + 1) = 11\).
2
Collect terms: \(5x + 1 = 11 \Rightarrow 5x = 10 \Rightarrow x = 2\).
3
Put \(x = 2\) back into \(y = 2x + 1\): \(y = 2(2)+1 = 5\).
\(x = 2,\; y = 5\)

Which method to pick

SituationBest method
A variable is already the subject (\(y = \ldots\))Substitution
Matching or easily-matched coefficientsElimination
Both look equally messyEither, elimination is often tidier

Always finish by checking your pair in the other equation. If it does not balance, you have a slip to hunt down.

A word problem (Criterion D)

The real skill is turning a situation into two equations, then interpreting the answer.

Worked example

At a cinema, 2 adult and 3 child tickets cost £31; 1 adult and 2 child tickets cost £18. Find each price.

1
Let \(a\) be the adult price and \(c\) the child price: \(2a + 3c = 31\) and \(a + 2c = 18\).
2
The second gives \(a = 18 - 2c\). Substitute: \(2(18 - 2c) + 3c = 31\).
3
Expand and solve: \(36 - 4c + 3c = 31 \Rightarrow 36 - c = 31 \Rightarrow c = 5\).
4
Then \(a = 18 - 2(5) = 8\).
Adult tickets cost £8 and child tickets cost £5.

Notice the final line answers the actual question in pounds, not just "\(a=8\)". Naming the values in context is what earns the Criterion D marks.

Check yourself

1. Solve \(x + y = 10\) and \(x - y = 4\). +

Add them: \(2x = 14 \Rightarrow x = 7\). Then \(y = 10 - 7 = \mathbf{3}\), so \(x=7,\ y=3\).

2. Solve \(y = x + 2\) and \(2x + y = 11\). +

Substitute: \(2x + (x+2) = 11 \Rightarrow 3x = 9 \Rightarrow x = 3\), so \(y = 5\). Pair: \((3,5)\).

3. Solve \(3x + 2y = 12\) and \(x + y = 5\). +

From the second, \(y = 5 - x\). Substitute: \(3x + 2(5 - x) = 12 \Rightarrow x + 10 = 12 \Rightarrow x = 2\), so \(y = 3\). Pair: \((2,3)\).

Part of the Extended Maths library. Spotted an error or want a topic added? That feedback makes the notes better.