Higher January 2021 Paper 1 Q19
19 Solve the simultaneous equations
\[\begin{aligned} x^2 - 9y - x &= 2y^2 - 12 \\ x + 2y - 1 &= 0 \end{aligned}\]Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
\((1 - 2y)^2 - 9y - (1 - 2y) = 2y^2 - 12\) or \(x^2 - 9\left(\dfrac{1 - x}{2}\right) - x = 2\left(\dfrac{1 - x}{2}\right)^2 - 12\) | M1 |
| e.g. \(2y^2 - 11y + 12\;(= 0)\) oe allow \(2y^2 - 11y = -12\) oe or e.g. \(x^2 + 9x + 14\;(= 0)\) oe allow \(x^2 + 9x = -14\) oe | A1 |
e.g. \((2y - 3)(y - 4)\;(= 0)\) \((y =)\;\dfrac{11 \pm \sqrt{(-11)^2 - 4 \times 2 \times 12}}{2 \times 2}\) e.g. \(2\left[\left(y - \dfrac{11}{4}\right)^2 - \left(\dfrac{11}{4}\right)^2\right] = -12\) oe or e.g. \((x + 7)(x + 2)\;(= 0)\) \((x =)\;\dfrac{-9 \pm \sqrt{9^2 - 4 \times 1 \times 14}}{2}\) e.g. \(\left(x + \dfrac{9}{2}\right)^2 - \left(\dfrac{9}{2}\right)^2 = -14\) | M1 |
\(y = \dfrac{3}{2}\) oe and \(y = 4\) or \(x = -7\) and \(x = -2\) | A1 |
Working required Answer: \(x = -2\), \(y = \dfrac{3}{2}\) oe and \(x = -7\), \(y = 4\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: substitution of linear equation into quadratic
A1: (dep on M1) writing the correct quadratic expression in the form \(ax^2 + bx + c\;(= 0)\)
allow \(ax^2 + bx = c\)
M1: (dep on M1) for a complete method to solve their 3-term quadratic equation (allow one sign error and some simplification – allow as far as \(\dfrac{11 \pm \sqrt{121 - 72}}{4}\) or \(\dfrac{-9 \pm \sqrt{81 - 56}}{2}\))
A1: (dep on M1) both \(x\)-values
or both \(y\)-values
A1: (dep on first M1) must be paired correctly