Higher January 2021 Paper 1R Q17
17 Solve the simultaneous equations
\(x - 6y = 5\)
\(xy - 2y^2 = 6\)
Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
\(y(6y + 5) - 2y^2 = 6\) or \(x\left(\dfrac{x - 5}{6}\right) - 2\left(\dfrac{x - 5}{6}\right)^2 = 6\) | M1 |
| E.g. \(4y^2 + 5y - 6\;(= 0)\) oe \(4y^2 + 5y = 6\) or E.g. \(4x^2 - 10x - 266\;(= 0)\) oe \(4x^2 - 10x = 266\) | A1 |
E.g. \((4y - 3)(y + 2)\;(= 0)\) \((y =)\; \dfrac{-5 \pm \sqrt{5^2 - 4 \times 4 \times -6}}{2 \times 4}\) \(4\left[\left(y + \dfrac{5}{8}\right)^2 - \left(\dfrac{5}{8}\right)^2\right] = 6\) oe or E.g. \((2x - 19)(x + 7)\;(= 0)\) \((x =)\; \dfrac{5 \pm \sqrt{(-5)^2 - 4 \times 2 \times (-133)}}{2 \times 2}\) \(4\left[\left(x - \dfrac{10}{8}\right)^2 - \left(\dfrac{10}{8}\right)^2\right] = 266\) oe | M1 |
\((y =)\; \dfrac{3}{4}\) and \((y =)\; -2\) or \((x =)\; \dfrac{19}{2}\) and \((x =)\; -7\) | A1 |
Working required Answer: \(x = \dfrac{19}{2}\), \(y = \dfrac{3}{4}\) \(x = -7\), \(y = -2\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for substitution of linear equation into quadratic
or
multiplying linear equation by \(y\) e.g. \(xy - 6y^2 = 5y\) and intention to subtract the two equations
A1: (dep on M1) writing the correct quadratic expression in 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 \(\dfrac{-5 \pm \sqrt{25 + 96}}{8}\) or \(\dfrac{5 \pm \sqrt{25 + 1064}}{4}\))
A1: Dep on first M1
for having two correct \(x\) values or two correct \(y\) values
A1: Dep on first M1
Must be paired and labelled correctly