Higher June 2019 Paper 3 Q20
20 Solve algebraically the simultaneous equations
\(x^2 - 4y^2 = 9\)
\(3x + 4y = 7\)
(5)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(x = 3\dfrac{2}{5}, y = -\dfrac{4}{5}\) \(x = 5, y = -2\) | M1 | for substitution of a rearrangement eg \(x = \dfrac{7 - 4y}{3}\) or \(y = \dfrac{7 - 3x}{4}\) into \(x^2 - 4y^2 = 9\) or expansion of \(\left(\dfrac{7 - 4y}{3}\right)^2 = \dfrac{49 - 56y + 16y^2}{9}\) or \(\left(\dfrac{7 - 3x}{4}\right)^2 = \dfrac{49 - 42x + 9x^2}{16}\) |
| M1 | for correct expansion and substitution eg \(\dfrac{49 - 56y + 16y^2}{9} - 4y^2 = 9\) or \(x^2 - 4\left(\dfrac{49 - 42x + 9x^2}{16}\right) = 9\) | |
| A1 | for forming quadratic ready for solving eg \(-20y^2 - 56y - 32\ (= 0)\) or \(5y^2 + 14y + 8\ (= 0)\) oe or \(5x^2 - 42x + 85\ (= 0)\) oe | |
| M1 | ft a 3 term quadratic, factorising eg \((5y + 4)(y + 2)\ (= 0)\) or \((5x - 17)(x - 5)\ (= 0)\) or correct use of formula eg (\(y =\)) \(\dfrac{-14 \pm \sqrt{14^2 - 4 \times 5 \times 8}}{2 \times 5}\) or (\(x =\)) \(\dfrac{--42 \pm \sqrt{42^2 - 4 \times 5 \times 85}}{2 \times 5}\) or completing the square, eg \(\left(y + \tfrac{7}{5}\right)^2 - \tfrac{9}{25}\ (= 0)\) or \(\left(x - \tfrac{21}{5}\right)^2 - \tfrac{16}{25}\ (= 0)\) | |
| A1 | correctly pairs \(x\) and \(y\) values: \(x = 3\dfrac{2}{5}, y = -\dfrac{4}{5}\) oe, \(x = 5, y = -2\) |
Additional guidance
Expansion may not be in simplest form but must be correct
Note we do not need to see “= 0”; just the LHS is sufficient.
Can be implied by both \(x\) values correct or both \(y\) values correct.
Answers must be correctly paired.
Accept coordinate pairs