Higher November 2018 Paper 3 Q19
19 Solve algebraically the simultaneous equations
\(2x^2 - y^2 = 17\)
\(x + 2y = 1\) (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(x = -\dfrac{23}{7}\), \(y = \dfrac{15}{7}\) \(x = 3\), \(y = -1\) | M1 | for substitution of a rearrangement eg for \(2(1 - 2y)^2 - y^2 = 17\) or \(2x^2 - \left(\dfrac{1 - x}{2}\right)^2 = 17\) or expansion of \((1 - 2y)^2 = 1 - 4y + 4y^2\) or \(\left(\dfrac{1 - x}{2}\right)^2 = \dfrac{1 - 2x + x^2}{4}\) |
| M1 | for expansion of bracket and substitution eg \(2(1 - 4y + 4y^2) - y^2\) (= 17) or \(8x^2 - (1 - 2x + x^2)\) (= 68) | |
| A1 | for forming quadratic ready for solving eg \(7y^2 - 8y - 15\) (= 0) or \(7x^2 + 2x - 69\) (= 0) | |
| M1 | ft a 3 term quadratic, factorising eg \((7y - 15)(y + 1)\) (= 0) or \((7x + 23)(x - 3)\) (= 0) or correct use of formula eg \(\dfrac{8 \pm \sqrt{64 + 420}}{14}\) or \(\dfrac{-2 \pm \sqrt{4 + 1932}}{14}\) or completing the square | |
| A1 | \(x = -\dfrac{23}{7}\) oe, \(y = \dfrac{15}{7}\) oe and \(x = 3\), \(y = -1\) |
Additional guidance
M1 (third method mark): Can be implied by both \(x\) values correct or both \(y\) values correct.
Answers must be correctly paired.
(Maybe in the body of the working)
Accept for \(x\) between \(-3.29\) and \(-3.28\) and for \(y\) between 2.14 and 2.15
Answers only award 0 marks