Simultaneous Equations: Substitute Back Into the Linear Equation

The question

The straight line $x - 2y = 1$ meets the curve $x^2 + y^2 = 13$ at the points $A$ and $B$ .

(a) Find the coordinates of $A$ and $B$ . [5 marks]

(b) Find the length of $AB$ , giving your answer in the form $\dfrac{p\sqrt{5}}{q}$ , where $p$ and $q$ are integers. [2 marks]

The common wrong answer

The start is usually fine. Make $x$ the subject of the linear equation, $x = 2y + 1$ , and substitute into the curve:

(2y + 1)^2 + y^2 = 13

\begin{aligned} 4y^2 + 4y + 1 + y^2 &= 13 \\ 5y^2 + 4y - 12 &= 0 \\ (y + 2)(5y - 6) &= 0 \end{aligned}

So $y = -2$ or $y = \frac{6}{5}$ . Then comes the slip: to find $x$ , the student substitutes each $y$ back into the curve, $x^2 + y^2 = 13$ :

y = -2: \quad x^2 = 13 - 4 = 9 \quad\Rightarrow\quad x = \pm 3

y = \frac{6}{5}: \quad x^2 = 13 - \frac{36}{25} = \frac{289}{25} \quad\Rightarrow\quad x = \pm\frac{17}{5}

The answer then lists four points: $(3, -2)$ , $(-3, -2)$ , $\left(\frac{17}{5}, \frac{6}{5}\right)$ and $\left(-\frac{17}{5}, \frac{6}{5}\right)$ .

A straight line can cross this curve at most twice, so four points can’t be right. Two of them lie on the curve but not on the line. Test $(3, -2)$ in the line: $3 - 2(-2) = 7$ , not $1$ . Squaring threw away the sign of $x$ , and the curve alone can’t tell you which sign belongs with which $y$ . Part (b) then goes wrong too: a student who picks $(3, -2)$ and $\left(\frac{17}{5}, \frac{6}{5}\right)$ gets $AB = \frac{2\sqrt{65}}{5} \approx 3.22$ , which isn’t even in the form asked for.

A Cambridge examiner report on a similar question noted that, after solving the quadratic correctly, some candidates forgot to substitute back to find the other values. Substituting back is where the marks go, whether the step is skipped or done into the wrong equation.

The correct working

(a) Use the linear equation to make one letter the subject. $x = 2y + 1$ avoids fractions, so it is the easier choice:

x = 2y + 1

Substitute into the curve and solve the quadratic in $y$ :

\begin{aligned} (2y + 1)^2 + y^2 &= 13 \\ 5y^2 + 4y - 12 &= 0 \\ (y + 2)(5y - 6) &= 0 \end{aligned}

y = -2 \quad\text{or}\quad y = \frac{6}{5}

Now find each matching $x$ from the linear equation $x = 2y + 1$ , which gives exactly one $x$ for each $y$ :

y = -2: \quad x = 2(-2) + 1 = -3

y = \frac{6}{5}: \quad x = 2\left(\frac{6}{5}\right) + 1 = \frac{17}{5}

So the points are $A(-3, -2)$ and $B\left(\frac{17}{5}, \frac{6}{5}\right)$ (either way round).

Check $B$ in both equations: $\frac{17}{5} - 2\left(\frac{6}{5}\right) = \frac{5}{5} = 1$ ✓ and $\frac{289}{25} + \frac{36}{25} = \frac{325}{25} = 13$ ✓.

The circle x² + y² = 13 and the straight line x − 2y = 1 on the same axes. The line crosses the circle at exactly two points, A(−3, −2) and B(17/5, 6/5), shown as solid dots. Two hollow grey dots at (3, −2) and (−17/5, 6/5) lie on the circle but not on the line, and are labelled as not on the line. — The line meets the curve at only A and B. The two grey points come from substituting into the curve instead of the line.

(b) Use the distance formula with $A(-3, -2)$ and $B\left(\frac{17}{5}, \frac{6}{5}\right)$ :

\begin{aligned} \Delta x &= \frac{17}{5} - (-3) = \frac{32}{5} \\ \Delta y &= \frac{6}{5} - (-2) = \frac{16}{5} \end{aligned}

AB = \sqrt{\left(\frac{32}{5}\right)^2 + \left(\frac{16}{5}\right)^2} = \sqrt{\frac{1024 + 256}{25}} = \frac{\sqrt{1280}}{5}

Since $1280 = 256 \times 5$ , $\sqrt{1280} = 16\sqrt{5}$ , so

AB = \frac{16\sqrt{5}}{5} \approx 7.16

The answer is $AB = \dfrac{16\sqrt{5}}{5}$ .

Why it works

Each solution of a pair of simultaneous equations is a pair of values: a point that lies on both graphs. Once you know $y$ , the matching $x$ must satisfy both equations, so you can find it from either one in principle. But the curve contains $x^2$ , and squaring hides the sign. For $y = -2$ it says $x^2 = 9$ , which is true at $x = 3$ and at $x = -3$ , because the circle passes through both. Only one of those is also on the line.

The linear equation has no such problem. It gives exactly one $x$ for each $y$ , and that $x$ is automatically on the line. Since the $y$ came from solving the combined equation, the point is on the curve as well.

The number of points also follows from the algebra. Eliminating $x$ gave a quadratic in $y$ , which has at most two roots. Two roots means two points, so any answer with more than two points has gone wrong somewhere.

How to spot it next time

After solving the quadratic, substitute each value into the linear equation to find its partner.
Write the answers as pairs, such as $x = -3, y = -2$ , so nobody (including you) can mix them up.
Count your points: a line meets a quadratic curve at most twice. Then check one point in both original equations.