Higher January 2022 Paper 1 Q19
19 Solve the simultaneous equations
\[3x^2 + y^2 - xy = 5\] \[y = 2x - 3\]Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
\(3x^2 + (2x - 3)^2 - x(2x - 3) = 5\) or \(3\left(\dfrac{y + 3}{2}\right)^2 + y^2 - y\left(\dfrac{y + 3}{2}\right) = 5\) | M1 |
| \(5x^2 - 9x + 4\;(= 0)\) oe or \(5x^2 - 9x = -4\) or \(5y^2 + 12y + 7\;(= 0)\) oe or \(5y^2 + 12y = -7\) | M1 |
\((5x - 4)(x - 1)\;(= 0)\) or \((x =)\;\dfrac{9 \pm \sqrt{(-9)^2 - 4 \times 5 \times 4}}{2 \times 5}\) or \(5\left[\left(x - \dfrac{9}{10}\right)^2 - \left(\dfrac{9}{10}\right)^2\right] + 4\;(= 0)\) [leading to \(x\) values of 0.8 and 1] or \((5y + 7)(y + 1)\;(= 0)\) or \((y =)\;\dfrac{-12 \pm \sqrt{12^2 - 4 \times 5 \times 7}}{2 \times 5}\) or \(5\left[\left(y + \dfrac{6}{5}\right)^2 - \left(\dfrac{6}{5}\right)^2\right] + 7\;(= 0)\) [leading to \(y\) values of −1.4 and −1] | M1ft |
(\(y\) =) 2 × “0.8” – 3 and 2 × “1” – 3 or \((x =)\;\dfrac{\text{``}{-1.4}\text{''} + 3}{2}\) and \(\dfrac{\text{``}{-1}\text{''} + 3}{2}\) | M1 |
| Working required Answer: \(x = 0.8\) & \(y = -1.4\) / \(x = 1\) & \(y = -1\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: Correct substitution of \(x\) for \(y\) (or \(y\) for \(x\))
M1: for a correct equation in the form \(ax^2 + bx + c\;(= 0)\) oe or \(ax^2 + bx = -c\)
M1ft: dep on M1 for solving their quadratic equation using any correct method - if factorising, allow brackets which expanded give 2 out of 3 terms correct (if using formula or completing the square allow one sign error and some simplification – allow as far as \(\dfrac{9 \pm \sqrt{81 - 80}}{10}\) oe or \(\dfrac{-12 \pm \sqrt{144 - 140}}{10}\) oe or \(5\left(x - \dfrac{9}{10}\right)^2 - \dfrac{1}{20}\) oe or \(5\left(y + \dfrac{6}{5}\right)^2 - \dfrac{1}{5}\) oe
M1: dep on previous M1
A1: oe, for both solutions dep on M2