# Simultaneous Equations (Linear and Non-Linear)

> CIE IGCSE Additional Mathematics · 0606 (2025-2027)
> Source: https://www.owlsprep.com/study/cie-0606-u5-simultaneous-equations/

This guide teaches you to solve systems of simultaneous equations in two unknowns where at least one equation is non-linear (and sometimes both), using substitution or elimination, as required for CIE IGCSE Additional Maths 0606. You will learn to interpret solutions as graph intersection points and avoid common exam mistakes.

**Prerequisites:** Rearranging linear equations to isolate a variable; Solving quadratic equations by factorisation, completing the square, or quadratic formula

## Learning objectives

- Solve simultaneous equations in two unknowns where one or both equations are non-linear, using substitution or elimination
- Interpret solutions as points of intersection of two graphs
- Apply the method to solve exam-style structured questions correctly
- Avoid common calculation errors when rearranging and solving resulting quadratics

## Core Substitution Method for Mixed Simultaneous Equations

Substitution is the main method for these systems: rearrange one equation to make a variable the subject, then substitute into the other. Elimination also works — for some systems, especially where both equations are non-linear, subtracting or dividing one equation by the other removes a variable more quickly. Choose whichever route reaches a single-variable equation with the least algebra.

**Mixed Simultaneous Equations** — A system of two equations in two unknowns in which at least one equation is non-linear (degree ≥2, e.g. quadratic, circle, reciprocal); the other may be linear or also non-linear. The solutions are the pairs of $(x,y)$ values that satisfy both equations.

1. Make one variable the subject from the simpler equation (use the linear equation if there is one; otherwise choose the equation and variable that separate most cleanly).
2. Substitute this expression for the isolated variable into the non-linear equation.
3. Simplify the resulting equation to get a quadratic in one variable of the form $ax^2 + bx + c = 0$.
4. Solve the quadratic to find two values (or one repeated value, or no real values) for the first variable.
5. Substitute each value back into the rearranged linear equation to find the corresponding value of the second variable.

> **tip**
>
> Always substitute back into the linear equation, not the non-linear one, to avoid getting extraneous solutions that do not satisfy both equations.

**Worked example:** Identify the linear and non-linear equation in the system: $2x - y = 3$ and $y = x^2 + 2x - 7$

1. Step 1: Check the degree of each equation. The first equation has $x$ and $y$ to the power of 1, so it is linear.
2. Step 2: The second equation has $x^2$ (degree 2), so it is non-linear.
3. Step 3: Isolate $y$ from the linear equation for substitution: $y = 2x - 3$

> **Exam tip:** If the question says 'give your answers to 3 significant figures', use the quadratic formula instead of factorisation for the resulting quadratic.

## Worked Example: Line and Parabola

**Worked example:** Solve the simultaneous equations: $y = 2x + 1$ and $y = x^2 + 3x - 5$

1. Step 1: The linear equation already has $y$ isolated, so no rearrangement needed.
2. Step 2: Substitute $y = 2x + 1$ into the quadratic equation:

   $$2x + 1 = x^2 + 3x - 5$$
3. Step 3: Rearrange to standard quadratic form:

   $$x^2 + x - 6 = 0$$
4. Step 4: Factorise and solve for $x$:

   $$(x + 3)(x - 2) = 0 \implies x = -3 \text{ or } x = 2$$
5. Step 5: Substitute each $x$ value back into $y = 2x + 1$:
6. $$x = -3: y = 2(-3) + 1 = -5$$
7. $$x = 2: y = 2(2) + 1 = 5$$
8. Final solution pairs: $(-3, -5)$ and $(2, 5)$

**Check your understanding**

1. Why do we substitute back into the linear equation instead of the quadratic here?

   - The quadratic will give extra incorrect solutions
   - The linear equation is faster to calculate
   - Both are equally valid

   *Answer:* The quadratic will give extra incorrect solutions

   *Why:* Non-linear equations can produce extraneous solutions that do not satisfy the linear equation, so always use the linear equation to find the second variable.

## Worked Example: Line and Circle

A common non-linear equation in 0606 exams is the equation of a circle, which takes the form $(x - a)^2 + (y - b)^2 = r^2$. The same substitution method applies for these systems.

**Worked example:** Solve the simultaneous equations: $x + y = 4$ and $x^2 + y^2 = 10$

1. Step 1: Rearrange the linear equation to isolate $y$: $y = 4 - x$
2. Step 2: Substitute $y = 4 - x$ into the circle equation:

   $$x^2 + (4 - x)^2 = 10$$
3. Step 3: Expand and simplify to standard quadratic form:

   $$x^2 + 16 - 8x + x^2 = 10 \implies 2x^2 - 8x + 6 = 0 \implies x^2 - 4x + 3 = 0$$
4. Step 4: Solve the quadratic:

   $$(x - 1)(x - 3) = 0 \implies x = 1 \text{ or } x = 3$$
5. Step 5: Substitute back into $y = 4 - x$ to find $y$ values:
6. $$x = 1: y = 3; x = 3: y = 1$$
7. Final solutions: $(1, 3)$ and $(3, 1)$

> **warning**
>
> Do not forget to square the entire expression you substitute, e.g. $(4 - x)^2 = 16 - 8x + x^2$, not $16 - x^2$. This is one of the most common errors on this topic.

*Calculator:* allowed

## Interpreting Solutions as Graph Intersections

Every solution pair $(x,y)$ to your system of simultaneous equations corresponds to a point where the graphs of the two equations intersect. This gives you a quick way to check if your solutions make sense.

**Number of Real Solutions** — The number of real solution pairs for a mixed system is equal to the number of intersection points between the linear and non-linear graph.

**Summary**

- 2 distinct real roots: two crossing points, two solution pairs
- 1 repeated real root: line is tangent to the non-linear graph, one solution pair
- No real roots: line does not intersect the non-linear graph, no real solutions

**Worked example:** A line and a parabola have only one solution pair $(2, 5)$. What does this tell you about the graphs?

1. Step 1: A single solution means there is only one intersection point.
2. Step 2: This means the line is a tangent to the parabola at the point $(2,5)$.

> **Exam tip:** If a question asks you to 'find the points of intersection' of a line and a curve, this is just another way of asking you to solve the corresponding simultaneous equations.

## Worked Example: Two Non-Linear Equations

The syllabus also allows systems in which both equations are non-linear. Substitution still works, but eliminating a variable — for example by dividing or subtracting one equation from the other — is often quicker.

**Worked example:** Solve the simultaneous equations $xy = 3$ and $xy^2 = 4$.

1. Step 1: Both equations are non-linear. Since $xy \neq 0$, divide the second equation by the first to eliminate $x$:

   $$\frac{xy^2}{xy} = \frac{4}{3} \implies y = \frac{4}{3}$$
2. Step 2: Substitute $y = \frac{4}{3}$ back into $xy = 3$ to find $x$:

   $$x \cdot \frac{4}{3} = 3 \implies x = \frac{9}{4}$$
3. Step 3: Check in the second equation: $xy^2 = \frac{9}{4} \cdot \left(\frac{4}{3}\right)^2 = \frac{9}{4} \cdot \frac{16}{9} = 4$, as required.
4. Final solution: $\left(\frac{9}{4}, \frac{4}{3}\right)$

A harder and very common exam version has both equations non-linear with an $xy$ cross-term, such as $ax^2 + bxy + cy^2 = d$ paired with a simple product equation like $xy = k$. Use the product equation to replace the cross-term and to write one variable as $y = \frac{k}{x}$, then substitute to reach an equation in $x$ alone. This usually gives a quadratic in $x^2$ with several solution pairs.

**Worked example:** Solve the simultaneous equations $4x^2 + 3xy + y^2 = 22$ and $xy = -6$.

1. Step 1: From $xy = -6$, the cross-term $3xy = 3(-6) = -18$, and $y = -\frac{6}{x}$ so $y^2 = \frac{36}{x^2}$.
2. Step 2: Substitute into the first equation:

   $$4x^2 - 18 + \frac{36}{x^2} = 22 \implies 4x^2 + \frac{36}{x^2} = 40$$
3. Step 3: Divide by 4, then multiply through by $x^2$ to clear the fraction:

   $$x^2 + \frac{9}{x^2} = 10 \implies x^4 - 10x^2 + 9 = 0$$
4. Step 4: This is a quadratic in $x^2$. Factorise it:

   $$(x^2 - 1)(x^2 - 9) = 0 \implies x^2 = 1 \text{ or } x^2 = 9$$
5. Step 5: So $x = \pm 1$ or $x = \pm 3$. Find each $y$ from $y = -\frac{6}{x}$:
6. $$x = 1 \Rightarrow y = -6; \quad x = -1 \Rightarrow y = 6; \quad x = 3 \Rightarrow y = -2; \quad x = -3 \Rightarrow y = 2$$
7. Final solution pairs: $(1, -6)$, $(-1, 6)$, $(3, -2)$ and $(-3, 2)$.

> **Exam tip:** When both equations are non-linear, look for a quick elimination (dividing or subtracting) before resorting to substitution.

## Worked Example: Exact Distance Between Two Intersection Points

A very common exam application first solves a system to find two intersection points, then asks for the exact distance between them. Once you have both points $(x_1, y_1)$ and $(x_2, y_2)$, use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ and simplify the surd into the form $a\sqrt{b}$, where $b$ has no square factor.

**Worked example:** The line $y = 2x$ meets the curve $y = x^2 - 3$ at two points $A$ and $B$. Find the exact length of $AB$, giving your answer in the form $a\sqrt{b}$.

1. Step 1: Set the two expressions for $y$ equal and rearrange to standard quadratic form:

   $$2x = x^2 - 3 \implies x^2 - 2x - 3 = 0$$
2. Step 2: Factorise and solve for $x$:

   $$(x - 3)(x + 1) = 0 \implies x = 3 \text{ or } x = -1$$
3. Step 3: Substitute each $x$ back into the line $y = 2x$ to find the points:

   $$x = -1 \Rightarrow y = -2; \quad x = 3 \Rightarrow y = 6$$
4. So the two intersection points are $A(-1, -2)$ and $B(3, 6)$.
5. Step 4: Apply the distance formula with $(x_1, y_1) = (-1, -2)$ and $(x_2, y_2) = (3, 6)$:

   $$AB = \sqrt{(3 - (-1))^2 + (6 - (-2))^2} = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80}$$
6. Step 5: Simplify the surd by taking out the largest square factor ($80 = 16 \times 5$):

   $$\sqrt{80} = \sqrt{16}\,\sqrt{5} = 4\sqrt{5}$$
7. The exact length is $AB = 4\sqrt{5}$.

> **tip**
>
> The distance is symmetric, so it does not matter which point you call $(x_1, y_1)$: squaring removes any sign difference in $(x_2 - x_1)$ and $(y_2 - y_1)$.

> **Exam tip:** Always simplify the surd fully: $\sqrt{80} = 4\sqrt{5}$, not $\sqrt{80}$. Marks for 'exact form' require the largest square factor to be taken outside the root.

## Common pitfalls

- **Wrong:** Isolating a variable with a coefficient not equal to 1 and making arithmetic errors when rearranging
  - Why it fails: Fractions increase the chance of multiplication mistakes later in the calculation
  - Correct: Always isolate the variable with coefficient 1 if possible; if no such variable exists, rearrange carefully and check your substitution step twice
- **Wrong:** Substituting the isolated variable back into the non-linear equation instead of the linear one
  - Why it fails: Non-linear equations can produce extraneous solutions that satisfy the non-linear equation but not the linear one
  - Correct: Always substitute values of the first variable back into the rearranged linear equation to find the second variable
- **Wrong:** Expanding squared terms incorrectly, e.g. writing $(2x + 3)^2 = 4x^2 + 9$ instead of $4x^2 + 12x + 9$
  - Why it fails: Forgetting the cross term when squaring binomials leads to an incorrect quadratic equation, so all subsequent working is wrong
  - Correct: Use the FOIL method to expand binomial squares, or memorize $(a + b)^2 = a^2 + 2ab + b^2$
- **Wrong:** Only giving one solution pair when the quadratic has two distinct roots
  - Why it fails: The question expects all real solutions, so you will lose marks for omitting one pair
  - Correct: Always substitute both roots of the quadratic back into the linear equation to get both solution pairs, unless the question specifies otherwise (e.g. only positive values)
- **Wrong:** Rounding intermediate values when using the quadratic formula, leading to inaccurate final answers
  - Why it fails: Rounding too early propagates errors through your calculation
  - Correct: Keep all intermediate values in exact form (or use the memory function on your calculator) and only round your final answers to the required number of significant figures

## Cheatsheet

| Step | Action |
| --- | --- |
| 1 | Make a variable the subject from the simpler equation (use the linear one if there is one) |
| 2 | Substitute isolated expression into non-linear equation |
| 3 | Simplify to standard quadratic form $ax^2 + bx + c = 0$ |
| 4 | Solve quadratic (factorisation, quadratic formula, completing the square) |
| 5 | Substitute each root back into linear equation to get corresponding $y$/$x$ value |
| 6 | Write all solution pairs as $(x,y)$ coordinates |

## What's next

Now that you have mastered solving one linear one non-linear simultaneous equations, you can apply this skill to a range of other topics in CIE IGCSE Additional Maths 0606. This method is used to find tangent lines to curves, intersection points of circles and lines, and solve problems involving kinematics and geometric shapes. Next, you should practice exam-style questions on this topic, then move on to more advanced algebraic manipulation and coordinate geometry skills that build on this foundation.

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