Higher June 2018 Paper 2R Q18
18 Solve the simultaneous equations
\(2x^2 + 3y^2 = 14\)
\(x = 2y - 3\)
Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
\(2(2y - 3)^2 + 3y^2 = 14\) or \(2x^2 + 3\left(\dfrac{x + 3}{2}\right)^2 = 14\) | M1 |
| \(11y^2 - 24y + 4 = 0\) or \(11x^2 + 18x - 29 = 0\) | A1 |
\((11y - 2)(y - 2)\) (= 0) or \(\dfrac{24 \pm \sqrt{(-24)^2 - 4 \times 11 \times 4}}{2 \times 11}\) or \((11x + 29)(x - 1)\) (= 0) or \(\dfrac{-18 \pm \sqrt{18^2 - 4 \times 11 \times -29}}{2 \times 11}\) | M1 |
\(y = \dfrac{2}{11}\) or \(y = 2\) (need both) or \(x = \dfrac{-29}{11}\) or \(x = 1\) (need both) | A1 |
\(x = \dfrac{-29}{11}\), \(y = \dfrac{2}{11}\) \(x = 1\), \(y = 2\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: correct first step eg substitution
A1: for a correct simplified quadratic
M1: (dep on M1) first step to solve their 3 term quadratic
A1: Dep on first M1. Must be paired correctly. Must be 2 dp or better