Solution of equations (incl. one linear and one quadratic)

Edexcel International GCSE Further Pure Mathematics· Specification section 3C· 25 min read

1. Solving Linear-Quadratic Simultaneous Equations by Substitution★★☆☆☆⏱ 15 min

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📘 Definition

Linear-quadratic simultaneous equations

A system of two equations in two variables where one equation is linear (degree 1) and the second is quadratic (degree 2). Solutions are coordinate pairs $(x, y)$ that satisfy both equations.

The only required method for this syllabus is substitution, which follows a consistent 5-step process: 1. Rearrange the linear equation to isolate one variable as the subject. 2. Substitute this expression into the quadratic equation, expanding to form a single quadratic in one variable. 3. Solve the quadratic to find values of the isolated variable. 4. Substitute each value back into the rearranged linear equation to find the corresponding second variable value. 5. Verify all solutions against both original equations.

📐 Worked Example

Solve the simultaneous equations: $2 x + y = 7$ and $x^{2} + y^{2} = 25$

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Step 1: Rearrange the linear equation to isolate y.
$y = 7 - 2x$
2
Step 2: Substitute $y = 7 - 2 x$ into the quadratic equation.
$x^2 + (7 - 2x)^2 = 25$
3
Step 3: Expand and simplify to form a quadratic in x.
$x^2 + 49 - 28x + 4x^2 = 25 \\ 5x^2 - 28x + 24 = 0$
4
Step 4: Factorise and solve the quadratic.
$(5x - 6)(x - 4) = 0 \\ x = \frac{6}{5} \text{ or } x = 4$
5
Step 5: Substitute each x value back into $y = 7 - 2 x$ to find corresponding y values.
$If $x = 6/5$: $y = 7 - 2(\frac{6}{5}) = \frac{23}{5}$ \\ If $x = 4$: $y = 7 - 8 = -1$$
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Step 6: Verify both pairs satisfy both original equations. Valid solutions are $(\frac{6}{5}, \frac{23}{5})$ and $(4, - 1)$ .

Exam tip:

Always verify solutions: squaring terms during substitution can create invalid values that do not fit the original linear equation, which will cost you marks if included in your final answer.

2. Solving Cubic Equations with Rational Roots★★★☆☆⏱ 15 min

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📘 Definition

Cubic equation

A polynomial equation of degree 3, written in standard form as $a x^{3} + b x^{2} + c x + d = 0$ where $a \neq = 0$ . All 4PM1 cubic equations include at least one rational root.

This skill builds directly on the factor theorem you learned in S3_T02. Follow these steps: 1. List all possible rational roots using the rational root rule. For most exam questions where $a = 1$ , possible roots are positive and negative factors of the constant term $d$ . 2. Test each possible root by substitution until you find a value $k$ where $f (k) = 0$ , meaning $(x - k)$ is a factor. 3. Use polynomial division or equating coefficients to factor the cubic into $(x - k) (A x^{2} + B x + C) = 0$ . 4. Solve the quadratic factor to find the remaining two roots.

📐 Worked Example

Solve the cubic equation $x^{3} - 3 x^{2} - 4 x + 12 = 0$

1
Step 1: List possible rational roots: $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$
2
Step 2: Test $x = 2$ by substitution:
$f(2) = 8 - 12 - 8 + 12 = 0 \\ \text{So } (x - 2) \text{ is a factor}$
3
Step 3: Equate coefficients to factor the cubic into linear × quadratic:
$x^3 - 3x^2 -4x +12 = (x-2)(x^2 + bx + c) \\ = x^3 + (b-2)x^2 + (c - 2b)x - 2c \\ \text{Equate terms: } b-2=-3 \Rightarrow b=-1; -2c=12 \Rightarrow c=-6 \\ \text{Factorised form: } (x-2)(x^2 -x -6) = 0$
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Step 4: Solve the quadratic factor:
$x^2 -x -6 = (x-3)(x+2) = 0 \\ \text{Full roots: } x = -2, x=2, x=3$

3. Equation Solving for Worded Context Questions★★★★☆⏱ 10 min

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Exam questions often require you to set up equations from worded contexts, including geometric scenarios or number problems. Always define variables clearly before setting up your equations, and ensure your final answers match the constraints of the question (e.g. positive values for length).

📐 Worked Example

The sum of two positive numbers is 12. The sum of their squares is 74. Find the two numbers.

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Step 1: Define variables: let the two positive numbers be $x$ and $y$ .
2
Step 2: Set up equations from the context:
$x + y = 12 \\ x^2 + y^2 = 74$
3
Step 3: Rearrange linear equation to $y = 12 - x$ , substitute into quadratic:
$x^2 + (12 - x)^2 = 74 \\ 2x^2 -24x +70 = 0 \\ x^2 -12x +35 = 0 \\ (x-5)(x-7) = 0 \\ x=5 \text{ or } x=7$
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Step 4: Find corresponding y values: $y = 7$ or $y = 5$ . Both values are positive, so the two numbers are 5 and 7.

Exam tip:

Always discard solutions that do not fit the question context, even if they satisfy the equations you set up (e.g. negative lengths or non-integer counts of objects).

4. Common Pitfalls

Wrong move:

Substituting the rearranged linear equation back into the original linear equation instead of the quadratic

Why:

This only produces an identity, not a solvable equation, wasting valuable exam time

Correct move:

Always substitute the rearranged linear expression into the quadratic equation to get a single-variable quadratic to solve

Wrong move:

Forgetting to check solutions for linear-quadratic systems

Why:

Squaring terms during substitution can introduce extraneous solutions that do not satisfy the original linear equation, leading to lost marks

Correct move:

Substitute every solution pair back into both original equations to verify validity before writing your final answer

Wrong move:

Only testing positive possible roots for cubics

Why:

Negative rational roots are common in exam questions, and skipping them means you will fail to find the first linear factor

Correct move:

Always test both positive and negative factors of the constant term when searching for a rational root of a cubic

Wrong move:

Giving only x-values for simultaneous equation solutions

Why:

Exam questions require complete coordinate pairs for full credit, not just values of one variable

Correct move:

Always present full labelled solution pairs for all simultaneous equation questions

Wrong move:

Making sign errors when rearranging the linear equation for substitution

Why:

Sign errors propagate through all subsequent steps, leading to fully incorrect solutions

Correct move:

Double-check your rearranged linear equation with a test value before substituting into the quadratic

5. Quick Reference Cheatsheet

Skill	Key Steps	Exam Check
Linear-quadratic simultaneous equations	Isolate variable from linear 2. Substitute into quadratic 3. Solve quadratic 4. Find second variable	Verify all pairs fit both original equations
Cubic equation solving (rational root)	List possible rational roots 2. Test to find first factor 3. Factor to linear × quadratic 4. Solve quadratic	Check all roots satisfy the original cubic equation
Worded equation problems	Define variables clearly 2. Set up equations from context 3. Solve using appropriate method 4. Check against context constraints	Discard any solutions that do not fit the question context

6. Frequently Asked

Do I need to check my solutions for simultaneous equations?

Yes. Squaring terms during substitution can introduce extraneous solutions that do not satisfy the original linear equation, so always verify all pairs against both original equations to avoid losing marks.

How many rational roots will I find in 4PM1 cubic equation questions?

All 4PM1 cubic questions include at least one rational root, so you will always be able to factor out one linear term before solving the remaining quadratic for the other two roots.

Going deeper

official_resourceEdexcel 4PM1 SpecificationOfficial syllabus for reference

What's Next

Now that you have mastered core equation solving for Edexcel IGCSE Further Pure Maths, you are ready to move on to inequality solving, the next subtopic in the identities and inequalities unit. These equation-solving skills are also high-frequency across other syllabus areas including coordinate geometry and calculus, so make sure you practice past paper questions to build speed and accuracy, always showing full working to maximise your marks.