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Simultaneous Equations

Two unknowns need two facts to pin them down. Solving a pair of equations together finds the single set of values that satisfies both at once, which is also where their graphs cross.

MYP 4Standard MathsAlgebraCriteria A · C · D~11 min read

Written by Fayek Jawad Haider · MYP student

The core idea

A single equation like \(x + y = 6\) has many solutions: \((1,5)\), \((2,4)\), \((3,3)\) and more. Add a second condition and usually only one pair fits both. Each equation is a line, and the solution is the point where the two lines meet.

Simultaneous equations
Two or more equations that share the same unknowns and must all be true together.

Both standard methods have the same goal: remove one unknown so you can solve for the other, then work back to find the first.

Method 1: elimination

Line the equations up and add or subtract them so one variable cancels. This is neatest when a variable has matching coefficients, or you can make them match.

Worked example

Solve \(3x + y = 10\) and \(x - y = 2\).

1
The \(y\) terms are \(+y\) and \(-y\), opposite signs, so adding the equations eliminates \(y\): \((3x + x) + (y - y) = 10 + 2\).
2
That gives \(4x = 12\), so \(x = 3\).
3
Substitute \(x = 3\) into \(x - y = 2\): \(3 - y = 2 \Rightarrow y = 1\).
\(x = 3,\; y = 1\)

Same signs? Subtract. Opposite signs? Add.

If the matching terms are \(+2y\) and \(+2y\), subtract to cancel them. 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 reads "\(y = \ldots\)".

Worked example

Solve \(y = 2x - 1\) and \(3x + y = 14\).

1
The first equation gives \(y\). Substitute it into the second: \(3x + (2x - 1) = 14\).
2
Collect terms: \(5x - 1 = 14 \Rightarrow 5x = 15 \Rightarrow x = 3\).
3
Put \(x = 3\) back into \(y = 2x - 1\): \(y = 2(3) - 1 = 5\).
\(x = 3,\; y = 5\)

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

Whichever you choose, finish by checking the pair in the other equation. For \(x = 3,\ y = 5\) above: \(3(3) + 5 = 14\), which balances, so the solution is correct.

Check yourself

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

Add them: \(2x = 10 \Rightarrow x = 5\). Then \(y = 8 - 5 = 3\). Answer: \(x = 5,\ y = 3\).

2. Solve \(y = x + 1\) and \(2x + y = 7\). +

Substitute: \(2x + (x + 1) = 7 \Rightarrow 3x = 6 \Rightarrow x = 2\), so \(y = 3\). Answer: \(x = 2,\ y = 3\).

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

The \(y\) terms are opposite, so add: \(5x = 15 \Rightarrow x = 3\). Then \(2(3) + y = 7 \Rightarrow y = 1\). Answer: \(x = 3,\ y = 1\).

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