Higher June 2023 Paper 2R Q21
21 Work out the coordinates of the points of intersection of
\(y - 2x = 1\;\) and \(\;y^2 + xy = 7\)
Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
Eg \((2x + 1)^2 + x(2x + 1) = 7\) or eg \(y^2 + \left(\dfrac{y - 1}{2}\right)y = 7\) | M1 |
| E.g. \(6x^2 + 5x - 6\;(= 0)\) \(6x^2 + 5x = 6\) or E.g. \(3y^2 - y - 14\;(= 0)\) \(3y^2 - y = 14\) | M1ft |
E.g. \((2x + 3)(3x - 2)\;(= 0)\) or \(x = \dfrac{-5 \pm \sqrt{5^2 - 4 \times 6 \times -6}}{2 \times 6}\) or \(\left(x + \dfrac{5}{12}\right)^2 - \left(\dfrac{5}{12}\right)^2 = 1\) \(\left(x = -\dfrac{3}{2} \text{ and } x = \dfrac{2}{3}\right)\) or E.g. \((y + 2)(3y - 7)\;(= 0)\) or \(y = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 3 \times -14}}{2 \times 3}\) or \(\left(y - \dfrac{1}{6}\right)^2 - \left(\dfrac{1}{6}\right)^2 = \dfrac{14}{3}\) \(\left(y = -2 \text{ and } y = \dfrac{7}{3}\right)\) | M1ft |
\(y = 2\left(\text{``}-\dfrac{3}{2}\text{''}\right) + 1\;(= -2)\) and \(y = 2\left(\text{``}\dfrac{2}{3}\text{''}\right) + 1\left(= \dfrac{7}{3}\right)\) or \(-2 = 2x + 1\) or \(x = -\dfrac{3}{2}\) and \(\dfrac{7}{3} = 2x + 1\) or \(x = \dfrac{2}{3}\) | M1ft |
Working required Answer: \(\left(-\dfrac{3}{2}, -2\right)\) \(\left(\dfrac{2}{3}, \dfrac{7}{3}\right)\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for substitution of \(y = \pm 2x \pm 1\) (or \(x = \dfrac{\pm y \pm 1}{2}\)) into \(y^2 + xy = 7\) to obtain an equation in \(x\) only (or \(y\) only)
M1ft: dep on previous M1 for multiplying out and collecting terms, forming a three term quadratic in any form of \(ax^2 + bx + c\;(= 0)\) where at least 2 coefficients (\(a\) or \(b\) or \(c\)) are correct
M1ft: dep on previous M1 for substituting their 2 found values of \(x\) or \(y\) into one of the two given equations
or fully correct values for the other variable (correct labels for \(x\) / \(y\))
A1: oe dep on M2
allow \(x = -1.5\), \(y = -2\)
\(x = 0.66(6\ldots)\), \(y = 2.33(3\ldots)\) truncated or rounded