Simultaneous Equations in IGCSE Maths
Elimination and substitution methods for linear pairs and linear-quadratic systems of equations.
0580 · E2.5 / E2.8What are simultaneous equations?
Simultaneous equations are two (or more) equations that share the same unknowns. The solution is the set of values that makes all equations true at the same time.
Linear + linear: both equations are straight lines (e.g. \(2x + 3y = 12\) and \(x - y = 1\)). Use elimination or substitution.
Linear + quadratic: one equation is a straight line and the other contains \(x^2\), \(y^2\) or \(xy\) (e.g. \(y = x + 3\) and \(x^2 + y^2 = 25\)). Use substitution only — elimination won't work here.
Elimination method
The idea is to add or subtract the equations to remove one variable.
Write them as equation ① and equation ②.
Multiply one or both equations so that the coefficients of one variable are the same (or negatives of each other).
Same sign → subtract. Different signs → add. One variable disappears.
Solve:
\[\begin{aligned}2x + y &= 7 & \text{①} \\[4pt] x - y &= 2 & \text{②}\end{aligned}\]Solution: \(x = 3\), \(y = 1\).
Solve:
\[\begin{aligned}3x + 2y &= 16 & \text{①} \\[4pt] 5x + 3y &= 26 & \text{②}\end{aligned}\]Solution: \(x = 4\), \(y = 2\).
Substitute both values back into the equation you didn't use for the back-substitution. If it works, you're correct. If it doesn't, you've made an error somewhere.
Substitution method (linear)
The idea is to write one variable in terms of the other, then replace it in the second equation.
Pick whichever is easiest — ideally one that already has a coefficient of 1.
Replace the variable with the expression from step 1.
Solve:
\[\begin{aligned}y &= 2x - 1 & \text{①} \\[4pt] 3x + 2y &= 12 & \text{②}\end{aligned}\]Solution: \(x = 2\), \(y = 3\).
Which method should I use?
| You see... | Use | Why |
|---|---|---|
| Both equations in the form \(ax + by = c\) | Elimination | The equations are lined up and ready to add/subtract. |
| One equation already gives \(y = ...\) or \(x = ...\) | Substitution | One variable is already isolated — just plug it in. |
| One equation contains \(x^2\), \(y^2\) or \(xy\) | Substitution | You must use substitution. Elimination won't remove the squared terms. |
| The question says "use an algebraic method" | Either | This just means don't solve it by reading off a graph. Pick whichever suits the equations. |
Linear-quadratic (non-linear) systems
When one equation is quadratic, you must use substitution. The process creates a quadratic equation which you then solve by factorising or the formula.
Non-linear systems usually have two pairs of solutions (because the line crosses the curve at two points).
Present your answers as pairs: \((x_1, y_1)\) and \((x_2, y_2)\).
Solve:
\[\begin{aligned}y &= x + 1 & \text{①} \\[4pt] x^2 + y^2 &= 25 & \text{②}\end{aligned}\]Solutions: \((-4, -3)\) and \((3, 4)\).
Solve:
\[\begin{aligned}y &= 2x + 1 & \text{①} \\[4pt] xy &= 6 & \text{②}\end{aligned}\]Solutions: \((\tfrac{3}{2}, 4)\) and \((-2, -3)\).
Each \(x\) value has a specific \(y\) value that goes with it. Always substitute back into the linear equation (not the quadratic) to find each \(y\). If you mix them up, your answer is wrong even if the individual numbers are correct.
Graphical interpretation
Solving simultaneous equations is the same as finding where two graphs intersect.
Two straight lines cross at one point — one solution. A straight line crossing a curve usually gives two intersection points — two pairs of solutions.
Exam-style questions
Solve the simultaneous equations:
\[\begin{aligned}4x + 3y &= 17 & \text{①} \\[4pt] 2x - 3y &= 1 & \text{②}\end{aligned}\]Show solution
The \(y\) coefficients are \(+3\) and \(-3\) — different signs, so add:
\[\begin{aligned}6x &= 18 \\[4pt] x &= 3\end{aligned}\]Substitute into equation 1:
\[\begin{aligned}4(3) + 3y &= 17 \\[4pt] 3y &= 5 \\[4pt] y &= \tfrac{5}{3}\end{aligned}\]Solution: \(x = 3\), \(y = \tfrac{5}{3}\).
Solve the simultaneous equations:
\[\begin{aligned}y &= 3x - 5 \\[4pt] 2x + 3y &= 7\end{aligned}\]Show solution
Substitute \(y = 3x - 5\) into the second equation:
\[\begin{aligned}2x + 3(3x - 5) &= 7 \\[4pt] 2x + 9x - 15 &= 7 \\[4pt] 11x &= 22 \\[4pt] x &= 2\end{aligned}\]Substitute back: \(y = 3(2) - 5 = 1\).
Solution: \(x = 2\), \(y = 1\).
Solve the simultaneous equations:
\[\begin{aligned}y &= x - 2 \\[4pt] x^2 + y^2 &= 10\end{aligned}\]Show solution
Substitute \(y = x - 2\) into the second equation:
\[\begin{aligned}x^2 + (x - 2)^2 &= 10 \\[4pt] x^2 + x^2 - 4x + 4 &= 10 \\[4pt] 2x^2 - 4x - 6 &= 0 \\[4pt] x^2 - 2x - 3 &= 0 \\[4pt] (x - 3)(x + 1) &= 0\end{aligned}\]\(x = 3\) or \(x = -1\).
Find \(y\) for each:
\[\begin{aligned}\text{When } x = 3: \quad y &= 3 - 2 = 1 \\[4pt] \text{When } x = -1: \quad y &= -1 - 2 = -3\end{aligned}\]Solutions: \((3, 1)\) and \((-1, -3)\).
Two numbers have a sum of 15 and a difference of 3. Find the two numbers.
Show solution
Let the numbers be \(x\) and \(y\) where \(x > y\).
\[\begin{aligned}x + y &= 15 & \text{①} \\[4pt] x - y &= 3 & \text{②}\end{aligned}\]Add:
\[\begin{aligned}\text{① } + \text{ ②}: \qquad 2x &= 18 \\[4pt] x &= 9\end{aligned}\]Substitute into ①: \(y = 15 - 9 = 6\).
The two numbers are 9 and 6.
Common mistakes
If both coefficients have the same sign, you need to subtract the equations. If they have different signs, you add. Mixing these up is the most common elimination error.
When multiplying an equation by a number, every term must be multiplied — including the constant on the right-hand side. \(2(3x + y = 5)\) becomes \(6x + 2y = 10\), not \(6x + 2y = 5\).
In linear-quadratic problems, the substitution step often requires expanding \((x + a)^2\). Remember: \((x + a)^2 = x^2 + 2ax + a^2\), not \(x^2 + a^2\). The middle term is the one students forget.
In non-linear systems with two solutions, each \(x\) value has a specific \(y\) that goes with it. Always present answers as labelled pairs: \((x_1, y_1)\) and \((x_2, y_2)\). Don't just list four separate numbers.
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