Introduction to Simultaneous Equations: Linear and Quadratic

Welcome! In your previous math studies, you likely solved simultaneous equations where both were linear (straight lines). In Further Pure Mathematics, we take it a step further. We look at what happens when a linear equation (a straight line) meets a quadratic equation (a curve, like a parabola or a circle).

Finding the solution to these equations is the same as finding the exact coordinates where the line and the curve cross each other on a graph. Don't worry if this seems tricky at first; once you master the "Substitution Method," you will be able to solve these confidently!

1. Understanding the Two Types

Before we start calculating, let’s identify what we are working with:

  • The Linear Equation: The variables (usually \(x\) and \(y\)) only have a power of 1. Example: \(y - 2x = 3\).
  • The Quadratic Equation: At least one variable is squared, or there is an \(xy\) term. Example: \(x^2 + y^2 = 25\) or \(y = x^2 + 5x - 2\).

2. The Method: Substitution

While you might have used "elimination" for two linear equations, substitution is the most reliable method when one equation is quadratic. Here is your step-by-step "battle plan":

  1. Rearrange the linear equation to make either \(x\) or \(y\) the subject (get it on its own).
  2. Substitute this expression into the quadratic equation. This creates a new equation with only one variable.
  3. Solve the resulting quadratic equation. (You might need to factorise or use the Quadratic Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)).
  4. Find the other variable by substituting your answers back into the original linear equation.
  5. Write your answers in pairs, as each \(x\) value belongs to a specific \(y\) value.

Quick Review: Remember that a quadratic equation can have two, one, or zero real solutions! This means a line could cross a curve twice, touch it once (as a tangent), or not touch it at all.

3. Worked Example: Step-by-Step

Question: Solve the simultaneous equations:
(1) \(y - x = 2\)
(2) \(x^2 + y^2 = 10\)

Step 1: Rearrange the linear equation.
From (1), let's get \(y\) on its own: \(y = x + 2\).

Step 2: Substitute into the quadratic.
Replace \(y\) in equation (2) with \((x + 2)\):
\(x^2 + (x + 2)^2 = 10\)

Step 3: Solve for \(x\).
Expand the brackets: \(x^2 + (x^2 + 4x + 4) = 10\)
Simplify: \(2x^2 + 4x + 4 = 10\)
Subtract 10 from both sides: \(2x^2 + 4x - 6 = 0\)
Divide by 2 to make it easier: \(x^2 + 2x - 3 = 0\)
Factorise: \((x + 3)(x - 1) = 0\)
So, \(x = -3\) or \(x = 1\).

Step 4: Find the \(y\) values.
Substitute \(x\) back into our rearranged linear equation \(y = x + 2\):
When \(x = -3\), \(y = -3 + 2 = -1\)
When \(x = 1\), \(y = 1 + 2 = 3\)

Step 5: State the final answers.
The solutions are: \((-3, -1)\) and \((1, 3)\).

4. Common Pitfalls to Avoid

  • Squaring incorrectly: Remember that \((x + 3)^2\) is not \(x^2 + 9\). It is \((x + 3)(x + 3) = x^2 + 6x + 9\). Always write it out!
  • Choosing the wrong equation: When finding the second variable, always plug your first answer back into the linear equation. It is much easier and avoids extra "fake" solutions.
  • Forgetting the second pair: Quadratics usually give two values for \(x\). Make sure you find the corresponding \(y\) value for both of them.
  • Sign errors: Be extremely careful with negative numbers when substituting. Use brackets!
Did you know?

If you substitute the linear into the quadratic and the resulting equation has a discriminant (\(b^2 - 4ac\)) equal to zero, it means the line is a tangent to the curve—it just kisses the curve at exactly one point!

5. Summary Checklist

Key Takeaway: Substitution is your best friend here. If you can rearrange, expand brackets, and solve a basic quadratic, you can solve these!

  • [ ] Did I rearrange the linear equation?
  • [ ] Did I substitute using brackets?
  • [ ] Did I set the quadratic to zero before solving?
  • [ ] Did I find both pairs of coordinates?

Note: For more practice on solving the quadratic part of these problems, refer to the chapter on "The quadratic function" (Section 2). For problems involving cubics, see "Cubic equations and factorising cubics" (Section 3C).